Class 7 Maths - Algebraic Expressions

Add:
$6ax-2by+3cz, 6by-11ax-cz$ and $10cz-2ax-3by$

We have to add

$6ax-2by+3cz, 6by-11ax-cz$ and

$10cz-2ax-3by$

So, simply numeric coefficients of same varaibles will be added.

$\therefore (6ax-2by+3cz)+( 6by-11ax-cz)+(10cz-2ax-3by)\\$

$= (6-11-2)ax+(-2+6-3)by+(3-1+10)cz\\$

$= -7ax+by+12cz\\$

$=by-7ax+12cz$

Add:
$6p+4q-r+3, 2r-5p-6, 11q-7p+2r-1$ and $2q-3r+4$

Given:

$6p+4q-r+3, 2r-5p-6, 11q-7p+2r-1$ and

$2q-3r+4$

To add the given terms we have to add coefficient of like terms:

$\therefore (6p+4q-r+)3+( 2r-5p-6)+( 11q-7p+2r-1)+(2q-3r+4)\\$

$=(6-5-7+0)p+(4+0+11+2)q+(-1+2+2-3)r+(3-6-1+4)\\$

$=-6p+17q+0r+0\\$

$=17q-6p$

Subtract :
$-2a + b + 6d$ from $5a -2b -3c$

The required difference

$= (5a - 2b - 3c) -( -2a + b + 6d)$

$= 5a - 2b - 3c + 2a - b -6d$

Rearranging and collecting like terms

$= (5a + 2a) +(- 2b - b) - 3c - 6d$

$= 7a - 3b - 3c - 6d$

Subtract
$5x^2 -3xy + y^2$ from $7x^2 - 2xy -4y^2$

The required difference

$=(7x^2 - 2xy -4y^2) -(5x^2 -3xy +y^2)$

$=7x^2 -2xy - 4y^2 - 5x^2 +3xy - y^2$

Rearranging and collecting like terms

$=(7x^2-5x^2)+(-4y^2-y^2)+(-2xy+3xy)$

$=2x^2 -5y^2+xy$

Subtract :
$a - 2b -3c$ from $-2a + 5b -4c$

The required difference

$= (- 2a + 5b – 4c) -( a – 2b – 3c)$

$= -2a + 5b – 4c – a + 2b + 3c$

$= -3a + 7b – c$

Subtract
$-11 x^2y^2 + 7xy - 6$ from $9x^2y^2 -6xy +9$

The required difference

$=(9x^2y^2 - 6xy +9) - (-11x^2y^2 +7xy -6)$

$=9x^2y^2 -6xy + 9 + 11x^2y^2 - 7xy + 6$

Rearranging and collecting like terms

$=(9x^2y^2 + 11x^2y^2 )+(- 6xy -7xy )+ (9 + 6 )$

$=20x^2y^2 - 13xy + 15$

Simplify :
$x^4 - 6x^3 + 2x - 7 + 7x^3 - x + 5x^2 + 2 - x^4$

$x^4 - 6x^3 + 2x - 7 + 7x^3 - x + 5x^2 + 2 - x^4$

Rearranging and collecting the like terms, we get:

$=(x^4 - x^4) +(- 6x^3 + 7x^3) + 5x^2 + (2x - x) +(- 7 + 2)$

$= (1 -1) x^4 + (-6 + 7)x^3 +5x^2 + (2 - 1)x +(-7 + 2)$

$= (1)x^3 + 5x^2 + (1) x - 7 + 2$

$= x^3 + 5x^2 + x - 5$

Add :
$2x^3–3x^2+7x–8,\ \ –5x^3+2x^2–4x+1,\ \ 3–6x+5x2–x^3$

The sum of the given expressions

$=(2x^3–3x^2+7x–8)+(–5x^3+2x^2–4x+1)+(3–6x+5x2–x^3)$

Rearranging and collecting like terms

$=(2x^3 - 5x^3 - x^3)+ (-3x^2 + 2x^2 +5x^2) +(7x -4x-6x)+(-8+1+3)$

$=-4x^3 +4x^2 -3x - 4$

Add
$2 + x -x^2+6x^3,\ -6-2x+4x^2-3x^3,\ \ 2 +x^2,\ \ 3-x^3+4x-2x^2$

The sum of the given expressions

$=(2 + x -x^2+6x^3)+(-6-2x+4x^2-3x^3)+(2 +x^2)+(3-x^3+4x-2x^2)$

Rearranging and collecting like terms

$=(2-6+2+3) + (x - 2x + 4x) + (-x^2 + 4x^2 + x^2 - 2x^2) + (6x^3-3x^3-x^3)$

$=1 +3x+2x^2+2x^3$

Subtract :
$6x^3 - 7x^2 + 5x - 3$ from $4 - 5x + 6x^2 -8x^3$

The required difference

$=(4 - 5x + 6x^2 - 8x^3) - (6x^3 - 7x^2 + 5x - 3)$

$= 4 -5x + 6x^2 - 8x^3 - 6x^3 + 7x^2 -5x + 3$

Rearranging and collecting like terms

$= (4 + 3)+( - 5x - 5x) +(6x^2 + 7x^2)+( - 8x^3 - 6x^3)$

$= 7 - 10x + 13x^2 - 14x^3$

Subtract:
$-8xy$ from $7xy$

We know that the difference of two like terms is a like term whose coefficient is the difference of the numerical coefficient of the two like terms.

$\therefore$ The required difference

$=(7xy)-(-8xy)$

$=7xy+8xy$

$=(7+8)xy$

$=15xy$

Subtract $x - 2y - z$ from the sum of $3x - y + z$ and $x + y - 3 z$ .

The value of terms as per the question is calculated as follows

$(3x - y + z) + (x + y - 3z) - (x - 2y - z)$
Adding and subtracting like terms, we get,

$= 3x + x - x - y + y + 2y + z - 3z + z$

$= 3x + 2y - z$
Therefore,

$(3x - y + z) + (x + y - 3z) - (x - 2y - z) = 3x + 2y - z$

Enclose the given terms in brackets as required:
$x^{2}- y^{2} + z^{2} + 3x - 2y = x^{2} - (..............)$

$x^{2} - y^{2} + z^{2} + 3x - 2y = x^{2} - (y^{2} - z^{2} - 3x + 2y)$

Enclose the given terms in brackets as required:
$x^{2} - xy^{2} - 2xy - y^{2} = x^{2} - (..........)$

$x^{2} - xy^{2} - 2xy y^{2} = x^{2} - (xy^{2} + 2xy + y^{2})$

Simplify :
$k / 2 + k / 3 + 2k / 5$

$k / 2 + k / 3 + 2k / 5$
Hear the LCM of

$2,3$ and

$5$ is

$30$

$= (15k + 10k + 12k) / 30$

$= 37k / 30$

Simplify :
$2p / 3 - 3p / 5$

$2p / 3 - 3p / 5$
Taking LCM of

$3,5$ we get

$15$

$= (10p - 9p) / 15$

$= p / 15$

Simplify :
$y / 5 + y - 19y / 15$

$y / 5 + y - 19y / 15$
Here the LCM of

$5,1$ and

$15$ is

$15$

$= (3y + 15y - 19y) / 15$
So we get

$= -y / 15$

Simplify :
$x + x / 2 + x / 3$

$x + x / 2 + x / 3$
Taking LCM of

$1,2$ and

$3$ we get

$6$

$= (6x + 3x + 2x) / 6$

$= 11x / 6$

Simplify :
$3y / 4 - y / 5$

$3y / 4 - y / 5$
Taking LCM of

$4.5$ we get

$20$

$= (15y - 4y) / 20$

$= 11y / 20$

Simplify :
$y / 4 + 3y / 5$

$y / 4 + 3y / 5$
Taking LCM of

$4,5$ we get

$20$

$= (5y + 12y) / 20$

$=17y / 20$

Simplify:

$\dfrac{2x}{5} +\dfrac{3x}{4} - \dfrac{3x}{5}$

$\dfrac{2x}{5} +\dfrac{3x}{4} - \dfrac{3x}{5}$

Here the LCM of

$5$ and

$4$ is

$20$

$=(8x+15x−12x)/20$

$=\dfrac{11x}{20}$

Simplify :
$6k / 7 - (8k / 9 - k / 3)$

$6k / 7 - (8k / 9 - k / 3)$
Here the LCM of

$7, 9$ and

$3$ is

$63$

$= [54k - (56k - 21k)] / 63$

$= 19k / 63$

Simplify :
$4a / 7 - 2a / 3 + a / 7$

$4a / 7 - 2a / 3 + a / 7$
Here the LCM of

$3$ and

$7$ is

$21$

$= (12 a - 14a + 3a) / 21$

$= a / 21$

Simplify :
$x / 2 - x / 8$

$x / 2 - x / 8$
Taking LCM of

$2,8$ we get

$8$

$= (4x - x) / 8$

$= 3x / 8$

Simplify :
$x / 5 + (x + 1)/ 2$

$x / 5 + (x + 1)/ 2$
Here the LCM of

$5$ and

$2$ is

$10$

$= (2x + 5x + 5) / 10$

$= (7x + 5 ) / 10$

Write the number of terms in following polynomial: $ax-by+y \times z$

There are three terms $ax, - by$ and $y \times z$

If $a = 3x - 5y, b = 6x + 3y$ and $c = 2y - 4x$ , then $2a - 3b + 4c.$ $=$

$a= 3x-5y$ then

$2a= 2(3x-5y)$ =

$6x-10y$

$b= 6x+3y$ then

$3b$ =

$3(6x+3y)$ =

$18x+9y$

$c = 2y-4x$ then

$4c$ =

$4(2y-4x)$ =

$8y-16x$
So,

$2a-3b+4c$ =

$6x-10y-18x+9y+8y-16x$
=

$6x-34x-10y+17y$
=

$-28x+7y$

Subtract: $8x - 7y - 8p + 10q$ from $10x + 10y - 7p + 9q$ .

We subtract

$8x-7y-8p+10q$ from

$10x+10y-7p+9q$ as shown below:

$(10x+10y-7p+9q)-(8x-7y-8p+10q)$

$=10x+10y-7p+9q-8x+7y+8p-10q$

$=(10x-8x)+(10y+7y)+(-7p+8p)+(9q-10q)$

$=2x+17y+p-q$

Hence, the difference is

$(2x+17y+p-q)$ .

Subtract: $a- b - 2c$ from $4a + 6b - 2c$ .

Given: Subtract

$a-b-2c$ and

$4a+6b-2c$

Getting the terms together and subtracting them for

$(4a+6b -2c) - (a -b -2c)$

$(4a - a) + (6b +b ) +(-2c + 2c)$

Hence

$(4a+6b -2c) - (a -b -2c)=$

$3a+7b$

If $a = 3x - 5y, b = 6x + 3y$ and $c = 2y - 4x$ , then $a + b - c$ $=$

Given,

$a=3x-5y, b=6x+3y$ and

$c=2y-4x$

We need to find value of

$a+b-c$

Therefore,

$a+b-c=(3x-5y)+(6x+3y)-(2y-4x)$

$=3x-5y+6x+3y-2y+4x$

Combining the like terms, we get:

$(3x+6x+4x)+(-5y+3y-2y)$

$=(13x)+(-4y)$

$=13x-4y$

Hence,

$a+b-c=13x-4y$ .

Subtract the sum of $2x-{x}^{2}+5$ and $-4x-3+7{x}^{2}$ from $5$ is

given equations are

$2x-x^2+5,-4x-3+7x^2$

adding them we get

$(2x-x^2+5)+(-4x-3+7x^2)=6x^2-2x+2$

now subtracting the sum from 5 we get

$5-(6x^2-2x+2)=3+2x-6x^2$

How much does $-5a^{2}+3a-5b^{2}$ exceed $7a^{2}+4a-9b^{2}$ ?

$-5{ a }^{ 2 }+3a-5{ b }^{ 2 }-7{ a }^{ 2 }-4a+9{ b }^{ 2 }$

$-12{ a }^{ 2 }-a+4{ b }^{ 2 }$

Simplify: $4x+3-(2x-5y)$

Let

$\left( {4x + 3y} \right) - \left( {2x - 5y} \right)$

Implies that

$4x + 3y + \left( {\left( { - 2x} \right) - \left( { - 5y} \right)} \right)$

Implies that

$=4x + 3y - 2x + 5y$

Implies that

$=2x + 3y + 5y$

Hence,

$=2x + 8y$

$(n-10)^2 + (10-n)$

The given expression

$(n-10)^2+(10-n)$ can be solved as follows:

$(n-10)^{ 2 }+(10-n)\\ =n^{ 2 }+(10)^{ 2 }-(2\times n\times 10)+(10-n)\quad \quad \quad (\because \quad (a-b)^{ 2 }=a^{ 2 }+b^{ 2 }-2ab)\\ =n^{ 2 }+100-20n+10-n\\ =n^{ 2 }+(-20n-n)+(10+100)\quad \quad \quad (Combining\quad like\quad terms)\\ =n^{ 2 }-21n+110$

Hence,

$(n-10)^{ 2 }+(10-n)=n^{ 2 }-21n+110$ .

Factorize $\left( 1+\dfrac { 1 }{ x } +\dfrac { 1 }{ { x }^{ 2 } } +\dfrac { 1 }{ { x }^{ 3 } } \right)$

$1 + \cfrac{1}{x} + \cfrac{1}{{{x^2}}} + \cfrac{1}{{{x^3}}}$

$= \left( {1 + \cfrac{1}{x}} \right) + \cfrac{1}{{{x^2}}}\left( {1 + \cfrac{1}{x}} \right)$

$= \left( {1 + \cfrac{1}{x}} \right)\left( {1 + \cfrac{1}{{{x^2}}}} \right)$

$-3{x}^{2}y+4{x}^{2}y-5{x}^{2}{y}^{2}$ What is a coficient of $x$

Since the given expression does not contain any term containing

$x$

So, coffecient of

$x$ is

$0$ .

Subtract the sum of $a+2b-3c$ and $2c-3b-4a$ from the sum of $5b-4c+a$ and $2c-3b-4a$ .

$5b-4c+a+2c-3b-4a=-2c+2b-3a$

$a+2b-3c+2c-3b-4a=-3s-b-c$

$(-2c+2b-3s)-(-3a-b-c)$

$-2c+2b-3a+3a+b+c$

$=3b-c$

Simplify
$\left( a+b \right) \left( c-d \right) +\left( a-b \right) \left( c+d \right)$

$\left(a+b\right)\left(c-d\right)+\left(a-b\right)\left(c+d\right)$

$=a\left(c-d\right)+b\left(c-d\right)+a\left(c+d\right)-b\left(c+d\right)$

$=ac-ad+bc-bd+ac+ad-bc-bd$

$=2ac-2bd$

$=2\left(ac-bd\right)$

Simplify : ${ (4m+5n) }^{ 2 }-{ (4m-5n) }^{ 2 }$

${\left(4m+5n\right)}^{2}-{\left(4m-5n\right)}^{2}$ is of the form

${a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)$

where

$a=4m+5n,\,\,b=4m-5n$

$=\left(4m+5n+4m-5n\right)\left(4m+5n-4m+5n\right)$

$=\left(8m\right)\left(10n\right)$

$=80mn$

What is the result when $2x^2+5x-6$ is subtracted from $4x^2-9x+6$

$4x^2-9x+6-2x^2-5x+6\\2x^2-14x+12$

Enclose the given terms in brackets as required:
$4a - 9 + 2b - 6 = 4a - (..........)$

$4a - 9 + 2b - 6 = 4a - (9 - 2b + 6)$

Simplify :
$a/10 + 2a/5$

$a/10 + 2a/5$
Taking LCM of

$10,5$ we get

$10$

$= (a + 4a) / 10$

$= 5a / 10$

$= a / 2$

Enclose the given terms in brackets as required:
$-2a^{2} + 4ab - 6a^{2}b^{2} + 8ab^{2} = -2a (...........)$

$-2a^{2} + 4ab - 6a^{2}b^{2} + 8ab^{2} = -2a (a - 2b + 3ab^{2} - 4b^{2})$

Simplify :
$2x - (x + 2y - z)$

$2x - (x + 2y - z)$
We can write it as

$= 2x - x - 2y + z$
So we get

$= x - 2y + z$

Simplify :
$x / 2 + x / 4$

$x / 2 + x / 4$
Taking LCM of

$2,4$ we get

$4$

$= (2x + x) / 4$

$= 3x / 4$

Simplify :
$8 (2a + 3b - c) - 10 (a + 2b + 3c)$

$8 (2a + 3b - c) - 10 (a + 2b + 3c)$
We can write it as

$= 16a + 24b - 8c - 10a - 20b - 30c$
So we get

$= 6a + 4b - 28c$

Add the following:
$2x^3-9x^2+8,3x^2-6x-5,7x^3-10x+1$ and $3+2x-5x^2-4x^3$

Add the following:

$(2x^3-9x^2+8)+(3x^2-6x-5)+(7x^3-10x+1)+(3+2x-5x^2-4x^3)$
=

$2x^3+7x^3-4x^3-9x^2-5x^2+3x^2-6x-10x+2x+8-5+1+3$

$=5x^3 -11x^2-14x + 7$

Add:
$8ab, -5ab, 3ab, -ab$

$8ab +(-5ab)+ 3ab+( -ab)$

$=8ab -5ab+ 3ab -ab$

$=5ab$

Simplify: $p+2(q- \overline { r+p } )$

$p + 2\left( {q - \overline {r + p} } \right)$

$= p + 2\left( {q - r - p} \right)$

$= p + 2q - 2r - 2p$

$= - p + 2q - 2r$

Add: $p(p-q),q(q-r)$ and $r(r-p)$

$p(p-q),q(q-r),r(r-p)$

$\Rightarrow p(p-q)+q(q-r)+r(r-p)$

$\Rightarrow p^2-pq+q^2-qr+r^2-rp$

Thus the sum

$=p^2+q^2+r^2-pq-qr-pr$

Combine the following : $( x - 3 ) ( x + 3 ) \left( x ^ { 2 } + 9 \right)$

$\left( x-3 \right) \left( x+3 \right) \left( { x }^{ 2 }+9 \right)$

$=\left( { x }^{ 2 }-9 \right) \left( { x }^{ 2 }+9 \right)$

$={ x }^{ 4 }-81$

Simplify: $(3.5e-4.5f)(1.5e+4f+ef)-4.5e+10f$ .

$\begin{array}{l} \left( { 3.5e-4.5f } \right) \left( { 1.5e+4f+ef } \right) -5e+10f \\ 5.25{ e^{ 2 } }+140ef+3.5{ e^{ 2 } }f-6.75ef-18{ f^{ 2 } }-4.5e{ f^{ 2 } }-5e+10f \\ 5.25{ e^{ 2 } }-18{ f^{ 2 } }+3.5{ e^{ 2 } }f-4.5e{ f^{ 2 } }-146.75ef-5e+10f \end{array}$

Identify terms and factors in the expression given.
$-4x+5y$ .

$-4x+5y$

Terms:-

$1)$

$-4x$

$2)$

$5y$

Factors:-

$1)$

$-4x$

$2)$

$1-\dfrac{5y}{4x}$ .

Identify the numerical coefficients of terms (other than constants) in the following expression
$(5-3t)$

The given term is

$5-3t$ .

Here, numerical coefficient of term

$-3t$ is

$-3$ .

Simplify :
$a^{2}+4a^{2}-3a+6-10+5a$

$a^{2}+4a^{2}-3a+6-10+5a$ =

$5a^{2}+2a-4$

Solve:
$z+z=?$

$z+z=2z$ by adding the like terms

Simplify the following expression.
$6xy+13x-2yx-5x$ .

Given:

$6xy+13x-2yx-5x$

Now collecting like terms we get,

$\Rightarrow 6xy-2xy+13x-5x$

Now performing the addition and subtraction of the like terms we get,

$\Rightarrow 4xy+8x$

Subtract: { $4(2p - 3q)$ } - { $4(2p - 3q)$ }

$\left\{ 4\left( 2p-3q \right) \right\} -\left\{ 4\left( 2p-3q \right) \right\} =\left\{ 8p-12q \right\} -\left\{ 8p-12q \right\}$

$=8p-12q-8p+12q=0$

Simplify: $9{\left( {x - 2y} \right)^2} - 13 + 4\left( {x - 2} \right)$

$9(x-2y)^2-13+4(x-2)$

$\Rightarrow 9(x^2+4y^2-4xy)-13+4x-8$

$\Rightarrow 9x^2+36y^2-36xy+4x-21$ .

Find and correct errors of the following mathematical expressions:
$2x+3y =5xy$

The given statement is incorrect as the equality of the LHS and RHS of the equation cannot be determined and is not applicable for all the values of

$x$ and

$y$ ..
The correct statement will be:

$2x+ 3y = 2x+ 3y$

Find and correct errors of the following mathematical expressions:
$x+2x+3x = 5x$

The given statement is incorrect.
As all the terms have the same variable

$x$ , therefore we can add them.

Hence, the correct statement is:

$x+2x+3x= 6x$

Find and correct errors of the following mathematical expressions:
$5y+2y+y-7y =0$

The given statement is incorrect.
As all the terms have same variable, we can perform operation of addition and subtraction.

Therefore, the correct statement is

$5y+2y+y-7y = y$

Simplify the polynomial and write it in standard form:
$2(x+3)-(-3x+7)$

$2\left( x+3 \right) -\left( -3x+7 \right)$

$=2x+6+3x-7$

$=5x-1$

Ans

$=5x-1$

The length and breadth a rectangular field are $( 2 x + 3 y )$ units and $( 3 x - y )$ respectively. Find the perimeter of the field in the form of an algebraic expression .

Length

$=2x+3y$

Breadth

$=(3x-y)$

Perimeter

$=2(l+b)$

$=2(2x+3y+3a-y) \\ =2(5x+2y) \\ =10x+4y$

Solve :
$2a - 3b - [4a - 3b - \{ a - 2c - (a - 2b - c)\} ]]$

$2a-3b-[4a-3b-\{a-2c-(a-2b-c)\}]$

$=2a-3b-[4a-3b-\{a-2c-a+2b+c\}]$

$=2a-3b-[4a-3b-\{2b-c\}]$

$=2a-3b-[4a-5b+c]$

$=2a-3b-4a+5b-c$

$=-2a+2b-c$

Add:-
$(4x^2-12xy+9y^2)+(25x^2+4xy-32y^2)$

$\begin{array}{l} \left( { 4{ x^{ 2 } }-12xy+9{ y^{ 2 } } } \right) +\left( { 25{ x^{ 2 } }+4xy-32{ y^{ 2 } } } \right) \\ =4{ x^{ 2 } }-12xy+9{ y^{ 2 } }+25{ x^{ 2 } }+4xy-32{ y^{ 2 } } \\ =29{ x^{ 2 } }-8xy-23{ y^{ 2 } } \end{array}$

Evaluate: $(l+m) -4m\quad$

Given

$(l+m)-4m$

$=l+m-4m$

$=l-3m$

From the sum of $3x-y+11$ and $-y-11$ subtract $3x-y-11$

1323474_1491896_ans_b169be11b3ad481f99b98c870d43fbce.JPG

Find the difference between $P\left( x \right) ={ x }^{ 4 }-{ 3x }^{ 2 }+4x+5,g\left( x \right) ={ x }^{ 2 }+1-x$

$P(x)-g(x)=x^{4}-3x^{2}+4x+5-x^{2}-1+x$

$=x^{4}-4x^{2}+5x+4$

Identify the terms and factors in the expression given.
$\dfrac{3}{4}x+\dfrac{1}{4}$ .

$\dfrac{3}{4}x+\dfrac{1}{4}$
Terms:

$\dfrac{3}{4}x, \dfrac{1}{4}$
Factors:

$\dfrac{3}{4}, x; \dfrac{1}{4}$ .

Add $14x+10y-12xy-13, 18-7x-10y+8xy, 4xy$ .

$14x+10y-12xy-13, 18-7x-10y+8xy, 4xy=14x+10y-12xy-13+18-7x-10y+8xy+4xy$

$=14x-7x+10y-10y-12xy+8xy+4xy-13+18$

$=7x+0y+0xy+5$

$=7x+5$ .

Add $-7mn+5, 12mn+2, 9mn-8, -2mn-3$ .

$-7mn+5, 12mn+2, 9mn-8, -2mn-3=-7mn+5+12mn+2+9mn-8+(-2mn)-3$

$=-7mn+12mn+9mn-2mn+5+2-8-3$

$=(-7+12+9-2)mn+7-11$

$=12mn-4$ .

Add $4x^2y, -3xy^2, -5xy^2, 5x^2y$ .

$4x^2y, -3xy^2, -5xy^2, 5x^2y=4x^2y+(-3xy^2)+(-5xy^2)+5x^2y$

$=4x^2y+5x^2y-3xy^2-5xy^2$

$=9x^2y-8xy^2$ .

Simplify combining like terms.
$5x^2y-5x^2+3yx^2-3y^2+x^2-y^2+8xy^2-3y^2$ .

$5x^2y-5x^2+3yx^2-3y^2+x^2-y^2+8xy^2-3y^2=5x^2y+3yx^2+8xy^2-5x^2+x^2-3y^2-y^2-3y^2$

$=(5x^2y+3x^2y)+8xy^2-(5x^2-x^3)-(3y^2+y^2+3y^2)$

$=8x^2y+8xy^2-4x^2-7y^2$ .

Add $5m-7n, 3n-4m+2, 2m-3mn-5$ .

$5m-7n, 3n-4m+2, 2m-3mn-5=5m-7n+3n-4m+2+2m-3mn-5$

$=5m-4m+2m-7n+3n-3mn+2-5$

$=(5-4+2)m+(-7+3)n-3mn-3$

$=3m-4n+3mn-3$ .

Add $x^2-y^2-1, y^2-1-x^2, 1-x^2-y^2$ .

$x^2-y^2-1, y^2-1-x^2, 1-x^2-y^2=x^2-y^2-1+y^2-1-x^2+1-x^2-y^2$

$=x^2-x^2-x^2-y^2+y^2-y^2-1-1+1$

$=(1-1-1)x^2+(-1+1-1)y^2-1-1+1$

$=-x^2-y^2-1$ .

Simplify combining like terms.
$(3y^2+5y-4)-(8y-y^2-4)$ .

$(3y^2+5y-4)-(8y-y^2-4)=3y^2+5y-4-8y+y^2+4$

$=(3y^2+y^2)+(5y-8y)-(4-4)$

$=4y^2-3y-0$

$=4y^2-3y$ .

Identify the terms and factors in the expression given.
$0.1p^2+0.2q^2$ .

$0.1p^2+0.2q^2$
Terms:

$0.1p^2, 0.2q^2$
Factors:

$0.1, p, p; 0.2, q, q$ .

Add $a+b-3, b-a+3, a-b+3$ .

$a+b-3, b-a+3, a-b+3=a+b-3+b-a+3+a-b+3$

$=(a-a+a)+(b+b-b)-3+3+3$

$=a+b+3$ .

Subtract $6xy$ from $-12xy$ .

$-12xy-(6xy)=-12xy-6xy=-18xy$ .

Subtract $5a^2-7ab+5b^2$ from $3ab-2a^2-2b^2$ .

$3ab-2a^2-2b^2-(5a^2-7ab+5b^2)=3ab-2a^2-2b^2-5a^2+7ab-5b^2$
Wring like terms aside, we get difference

$=3ab+7ab-2a^2-5a^2-2b^2-5b^2$

$=10ab-7a^2-7b^2$

$=-7a^2-7b^2+10ab$ .

Add:
$x^2 - a^2, - 5x^2 + 2a^2, -4x^2 + 4a^2$

The required sum =

$(x^2 - a^2) + (-5x^2 + 2a^2) + (-4x^2 + 4a^2)$

$= x^2 - a^2 - 5x^2 + 2a^2 -4x^2 + 4a^2$

$= (1 - 5 - 4) x^2 - (1 - 2 - 4) a^2$

$= (1 -9)x^2 - (1 - 6) a^2$

$= - 8x^2 + 5a^2$

Subtract:
$8a + 3ab - 2b + 7$ from $14a - 5ab + 7b - 5$

Subtracting

$8a + 3ab - 2b + 7$ from

$14a - 5ab + 7b - 5$ , we have
=

$(14a - 5ab + 7b - 5) - (8a+ 3ab - 2b + 7)$
=

$14a - 5ab + 7b - 5 - 8a - 3ab + 2b - 7$
=

$6a - 8ab + 9ab - 12$

Subtract: $4p^2q-3pq +5pq^2-8p+7q-10$ from $18 -3p-11q+5pq -2pq^2 + 5p^2q$

Subtract

$4p^2q-3pq +5pq^2-8p+7q-10$ from

$18 -3p-11q+5pq -2pq^2 + 5p^2q$

$=(18 -3p-11q+5pq -2pq^2 + 5p^2q)-(4p^2q-3pq +5pq^2-8p+7q-10)$

$=18 -3p-11q+5pq -2pq^2 + 5p^2q - 4p^2q+3pq -5pq^2+8p-7q+10$

$=28 + 5p - 18q + 8pq - 7pq^2 - p^2q$

Add $ab- bc, bc- ca, ca -ab$ .

$ab-bc,bc-ca,ca-ab$
Adding we get,

$ab-bc+bc-ca+ca-ab$

$=0$

Write the number of terms in following polynomial: $5x^2+3 \times ax$

There are two terms

$5{x}^{2}$ and

$3 \times ax$

Write the number of terms in following polynomial: $ax\div 4 -7$

There are two terms

$ax$ and

$4- 7$

Write the number of terms in following polynomial: $23+ a\times b \div{2}$

There are two terms in the given polynomial first one $23$ and $a \times b \div 2$

$x^3-x-4$ added to $x^4-x^3-x^2+x+3$ to obtain $x^4+x^2-1$ ?
If true then enter $1$ and if false then enter $0$

Let the expression to be added be

$A$

Given,

$({ x }^{ 4 }-{ x }^{ 3 }-{ x }^{ 2 }+x+3) + A = { x }^{ 4 }+{ x }^{ 2 }-1$

$=> A = ({ x }^{ 4 }+{ x }^{ 2 }-1) - ({ x }^{ 4 }-{ x }^{ 3 }-{ x }^{ 2 }+x+3)$

$A = { x }^{ 4 }+{ x }^{ 2 }-1 - { x }^{ 4 }+{ x }^{ 3 }+{ x }^{ 2 }-x-3 = { x }^{ 3 } + 2{ x }^{ 2 } -x- 4$

Simplify the polynomial and write it in standard form:
$-4(8x-10)-3(7x-1)+3(4x)+12$

$-4\left( 8x-10 \right) -3\left( 7x-1 \right) +3\left( 4x \right) +12$

$=-32x+40-21x+3+12x+12$

$=-32x-21x+12x+40+3+12$

$=-41x+55$

Ans

$=-41x+55$

Simplify the polynomial and write it in standard form:
$(-2x-5)(2x-8)-(2x-5)(3x+7)$

$\left( -2x-5 \right) \left( 2x-8 \right) -\left( 2x-5 \right) \left( 3x+7 \right)$

$=-4{ x }^{ 2 }+16x-10x+40-6{ x }^{ 2 }-14x+15x+35$

$=-4{ { x }^{ 2 } }-6{ x }^{ 2 }+16x-10x-14x+15x+40+35$

$=-10{ x }^{ 2 }+17x+75$

Ans

$=-10{ x }^{ 2 }+17x+75$

Simplify the polynomial and write it in standard form:
$(-{x}^{3}+4)(x+2)-(2x-1)(5x)$

$\left( { -x }^{ 3 }+4 \right) \left( x+2 \right) -\left( 2x-1 \right) \left( 5x \right)$

$={ -x }^{ 4 }-{ 2x }^{ 3 }+4x+8-{ 10x }^{ 2 }+5x$

$={ -x }^{ 4 }-{ 2x }^{ 3 }-10{ x }^{ 2 }+4x+5x+8$

$={ -x }^{ 4 }-{ 2x }^{ 3 }-10{ x }^{ 2 }+9x+8$

Ans

$={ -x }^{ 4 }-{ 2x }^{ 3 }-10{ x }^{ 2 }+9x+8$

Simplify the polynomial and write it in standard form:
$2({x}^{2}-3x-2)-(-3{x}^{2}+7x-1)$

$2\left( { x }^{ 2 }-3x-2 \right) -\left( -3{ x }^{ 2 }+7x-1 \right)$

$={ 2x }^{ 2 }-6x-4+{ 3x }^{ 2 }-7x+1$

$={ 2x }^{ 2 }+{ 3x }^{ 2 }-6x-7x-4+1$

$={ 5x }^{ 2 }-13x-3$

Ans

$={ 5x }^{ 2 }-13x-3$

Subtract the following:
$p\left( y \right) =3{ y }^{ 7 }-2{ y }^{ 2 }+3$ and $q\left( y \right) ={ y }^{ 7 }+{ y }^{ 2 }+y$

$p(y)=3y^{7}-2y^{2}+3$

$q(y)=y^{7}+y^{2}+y$

$p(y)-q(y)=3y^{7}-2y^{2}+3-\left ( y^{7}+y^{2}+y \right )$

$\Rightarrow p(y)-q(y)=3y^{7}-2y^{2}+3-y^{7}-y^{2}-y=3y^{7}-y^{7}-2y^{2}-y^{2}+3-y=2y^{7}-3y^{2}-y+3$

Simplify : $3x^2 + 5xy - 4y^2 + x^2 - 8xy - 5y^2$

As Given question

$3{ x }^{ 2 }+5xy-4{ y }^{ 2 }+{ x }^{ 2 }-8xy-5{ y }^{ 2 }$
Simply adding the like terms
We get

$(3{ x }^{ 2 }+{ x }^{ 2 })+(5xy-8xy)-(4{ y }^{ 2 }+5{ y }^{ 2 })$
Hence the solution is

$4x^{ 2 }-3xy-9{ y }^{ 2 }$

Add the given polynomials: $x^3 - x^2y + 5xy^2 +y^3, x^3 - 9xy^2 +y^3$ and $3x^2y + 9xy^2$ .

Simply add The polynomials by clubing the like terms together:

$x^3-x^2y+5xy^2+y^3+x^3-9xy^2+y^3+3x^2y+9xy^2$
Separating the similar terms we get:

$x^3+x^3-x^2y-9xy^2+5xy^2+9xy^2+y^3+y^3$

Adding the like terms together we have:

$(x^3+x^3)-x^2y+(-9xy^2+5xy^2+9xy^2)+(y^3+y^3)$
The Solution to the given question is :

${2x^2-x^2y+5xy^2+2y^2}$

Add the given polynomials: $5a+ 3b, a - 2b$ and $3a+ 5b$

In the given Question
Simply getting all the terms together

$5a+3a+a-2b+2a+5b$
separating Like terms and adding them

$(5a+a+3a)+(3b-2b+5b)$

Hence the correct solution is

$9a+6b$

Simplify: $7x - 9y+3 -3x -5y+8$

Let's combine the

$x$ and

$y$ terms together as :

$7x-3x-9y-5y+8+3$
Now by simply adding the terms comprising of

$x$ and

$y$
we get the solution :

${ 4x-14y+11 }$

Thus the answer is:

${ 4x-14y+11 }$

Add $7x^2- 4x + 5$ and $9x-10$ .

$7x^2- 4x + 5$

$+9x -10$ .
______________

$7x^2+5x-5$
______________

Add $8xy + 4yz -7zx, 6yz + 11zx -6y$ and $-5xz + 6x -2yx$

$+8xy + 4yz -7zx$

$+6yz +11zx$

$- 6y$

$-2xy$

$-5xz$

$+6x$
__________________________________

$6xy$

$+10yz-xz+6x-6y$
__________________________________

Add: $5x^2y, -7x^2y$ , $9x^2y$ .

Given:

$5x^2y, -7x^2y$ ,

$9x^2y$

To find:

$(5x^2y) + (-7x^2y) + (9x^2y)$

$=(5 + (-7) + 9)x^2y$

$=(5 -7 + 9)x^2y$

$= 7x^2y$ .

Subtract $2a - b$ from $3a - b$ .

We need to subtract

$2a-b$ from

$3a-b$

It can be written as

$3a-b-(2a-b)$

$=3a-b-2a+b$

$=a$

Add $5{x}^{2}+3xy+2{y}^{2}$ and $2{y}^{2}-xy+4{x}^{2}$

Write the expression one under another so that like times align in columns. Then add

$5{x}^{2}+3xy+2{y}^{2}$

$+$

$4{x}^{2}\ -xy\ \ +2{y}^{2}$
----------------------------------

$9{x}^{2}+2xy\ +4{y}^{2}$
----------------------------------

Add the following algebraic expressions:
(i) $2x^{2} + 3x + 5, 3x^{2} - 4x - 7$
(ii) $x^{2} - 2x - 3, x^{2} + 3x + 1$
(iii) $2t^{2} + t - 4, 1 - 3t - 5t^{2}$
(iv) $xy - yz, yz - xz, zx - xy$
(v) $a^{2} + b^{2}, b^{2} + c^{2}, c^{2} + a^{2}, 2ab + 2bc + 2ca$

We need to perform addition on the following algebraic expressions:

(i) Given,

$2x^2+3x+5$ and

$3x^2-4x-7$

Thus we have

$(2x^2+3x+5)+(3x^2-4x-7)=5x^2-x-2$

(ii) Given,

$x^2-2x-3$ and

$x^2+3x+1$

Thus we have

$(x^2-2x-3)+(x^2+3x+1)=2x^2+x-2$

(iii) Given,

$2t^2+t-4$ and

$1-3t-5t^2$

Thus we have

$(2t^2+t-4)+(1-3t-5t^2)=-3t^2-2t-3$

(iv) Given,

$xy-yz, yz-xz$ and

$xz-xy$

Thus we have

$(xy-yz)+(yz-xz)+(zx-xy)=0;$

(v) Given,

$a^2+b^2, b^2+c^2, c^2+a^2$ and

$2ab+2bc+2ca$

Thus we have

$(a^2+b^2)+(b^2+c^2)+(c^2+a^2)+(2ab+2bc+2ac)=(a+b)^2+(b+c)^2+(c+a)^2$

substract $4{p^2}q - 3pq + 5p{q^2} - 8p + 7q - 10$ from $18 - 3p - 11q - 2p{q^2} + 5{p^2}q$

Given that:

Say,

$B=4p^2q-3pq+5pq^2-8p+7q-10$

$A=5p^2q-2pq^2-3p-11q+18$

$A-B=5p^2q-2pq^2-3p-11q+18-4p^2q+3pq-5pq^2+8p-7q+10$

$A-B=p^2q-7pq^2+5p-18q+28+3pq$

The two adjacent sides of a rectangle are $2x^2- 5xy+3z^2$ and $4xy-x^2-z^2$ . Find its perimeter

Let

$a$ and

$b$ be two adjacent sides of rectangle.

Then we know perimeter of rectangle will be

$2(a+b)$ .

Here two adjacent sides are

$2x^2-5xy+3z^2$ and

$4xy-x^2-z^2$ .

Thus perimeter

$=$

$2(a+b)=2(2x^2-5xy+3z^2+4xy-x^2-z^2)$

$=2(x^2+2z^2-xy)$

$=2x^2+4z^2-2xy$

Subtract $3x^{2}y - 2xy + 2xy^{2} + 5x - 7y - 10$ from $15 - 2x + 5y - 11xy + 2xy^{2} + 8x^{2}y$

We need to subtract

$3x^{2}y - 2xy + 2xy^{2} + 5x - 7y - 10$ from

$15 - 2x + 5y - 11xy + 2xy^{2} + 8x^{2}y$

Lets combine the like terms.

Therefore,

$(3x^2y+8x^2y)+(-2xy-11xy)+(2xy^2+2xy^2)+(5x-2x)+(-7y+5y)+(-10+15)$

$=11x^2y-13xy+4xy^2+3x-2y+5$

Add the polynomials $2x (x - y - z)$ and $2y(z - y - x)$

We need to add expressions

$2x(x-y-x)$ and

$2y(z-y-x)$

Their addition is

$2x(x-y-z)+2y(z-y-x)$

$=2x^2-2xy-2xz+2yz-2y^2-2xy$

$=2x^2-2y^2-4xy-2xz+2yz$

Subtract the polynomial $2ab + 5bc - 3ca$ from the polynomial $7ab - 2bc + 10ca$

To subtract the polynomial

$2ab+5bc-3ac$ from the polynomial

$(7ab-2bc+10ac)$ , we get

$(7ab-2bc+10ac) - (2ab+5bc-3ac)=5ab-7bc+13ac$

Subtract $-3x + 8y$ from $-7x - 10y.$

We need to subtract

$-3x+8y$ from

$-7x-10y.$

As per given condition, we have

$-7x-10y-(-3x+8y)=-7x-10y+3x-8y$

$= \left( { - 7x + 3x} \right) + \left( { - 10y - 8y} \right)$

$=-4x-18y$

Subtract: $-2x^2y + 3xy^2$ from $8x^2y$ .

To subtract:

$(-2x^2y + 3xy^2)$ from

$8x^2y$

Now,

$8x^2y-(-2x^2y + 3xy^2)$

$=8x^2y+2x^2y - 3xy^2$

$=10x^2y - 3xy^2$

Hence, the result is

$10x^2y - 3xy^2$ .

Subtract the polynomial : $x^{5} - 2x^{2} - 3x$ from $x^{3} + 3x^{2} + 1$

To subtract the polynomial

$x^5-2x^2-3x$ from the polynomial

$x^3++3x^2+1$ , we get

$(x^3++3x^2+1) - (x^5-2x^2-3x)$

$=-x^5+x^3+5x^2+3x+1$

Find out the sum of the polynomials $3x - y, 2y - 2x$ and $x + y$

Given polynomials are

$3x-y, 2y-2x$ and

$x+y.$
Their sum will be,

$\begin{aligned}{}(3x - y) + (2y - 2x) + (x + y)& = \left( {3x - 2x + x} \right) + \left( { - y + 2y + y} \right)\\ &= 2x + 2y\\ &= 2(x + y)\end{aligned}$

What should be added to $(3x^3-2x^2+3x-1)$ to get $(x^3+2x^2-6x+5)?$ .

Let

$y$ be the quantity being added.

$\therefore (3x^3-2x^2+3x-1)+y=x^3+2x^2-6x+5$

$\therefore y=x^3+2x^2-6x+5-(3x^3-2x^2+3x-1)$

$\therefore y=-2x^3+4x^2-9x+6$

Hence, the quantity to be added is

$-2x^3+4x^2-9x+6$ .

Subtract the second term from first term.
$8x , 5x$

$=8x-5x$

$=x(8-5)$

$=3x$

Find the length of the line segment $PR$ in the following figure in terms of $'a'$ .

Length of line segment,

$PR=PQ+QR$

$PR=3a+2a$

$=5a$

Subtract $5xy$ from $8xy$

We need to subtract

$5xy$ from

$8xy$

Thus resultant is

$8xy-5xy=xy(8-5)=3xy$ .

This is the required answer.

Subtract $3c + 7d^{2}$ from $5c - d^{2}$

We need to subtract

$3c+7d^2$ from

$5c-d^2$

Thus we have

$5c-d^2-(3c+7d^2)$

$=5c-d^2-3c-7d^2$

$=2c-8d^2$

$=2(c-4d^2)$

Simplify $3x^{2}+5xy-4y^{2}+x^{2}-8xy-5y^{2}$

$3x^2+5xy-4y^2+x^2-8xy-5y^2$

$=3x^2+x^2-4y^2-5y^2+5xy-8xy$

$=4x^2-9y^2-4xy$

Add : $9p+16q+13p+2q$

$(9p + 16q) + (13p + 2q)$

$= 9p + 13p + 16q + 2q$

$= 22p + 18q$

Simplify: $3x-[5y-\left\{{6y+2(10y-x)} \right\}]$

$3x - \left[ {5y - \left\{ {6y + 2\left( {10y - x} \right)} \right\}} \right]$

$= 3x - \left[ {5y - \left\{ {6y + 20y - 2x} \right\}} \right]$

$= 3x - \left[ {5y - 6y - 20y + 2x} \right]$

$= 3x - \left[ { - 21y + 2x} \right]$

$= 3x + 21y - 2x$

$= x + 21y$

Subtract:
$5x$ from $2x$
$-7x$ from $9y$

$2x-5x=3x$
$9y-(-7x)=9y+7x$

Add: $3{ x }^{ 2 }y-5x{ y }^{ 2 }-6xy$ , $-7x{ y }^{ 2 }+4{ x }^{ 2 }y$ and $-5xy-6{ x }^{ 2 }y+7{ y }^{ 2 }x$

Consider,

$(3{ x }^{ 2 }y-5x{ y }^{ 2 }-6xy)+(4{ x }^{ 2 }y-7x{ y }^{ 2 })+(-6{ x }^{ 2 }y+7x{ y }^{ 2 }-5xy)$

$=({ x }^{ 2 }y-5x{ y }^{ 2 }-11xy)$

Simplify $9x-5(2x-3)$

Given

$9x - 5(2x-3)$

By solving equation

$9x -10x + 15$

Here

$9x+10x$ are like term then by adding then

$\Rightarrow -x +15$

$\Rightarrow 15 -x$

1112487_1037703_ans_fcf10eb8397d47a8817f123222fcca5c.jpg

Solve:-
$ar+br+at+bt$

Solve:-

$ar+br+at+bt$

$\Rightarrow r\left( a+b \right)+t\left( a+b \right)$

$\Rightarrow \left( a+b \right)\left( r+t \right)$

Hence, this is the answer.

$(x^{5}+1),(3-x^{4}),(4+x^{3}+x^{6})$ $-$
$(2a^{3}-3b^{5}),(2a^{3}+3b^{5}),(2a^{4}-3a^{2}b^{2}+b^{4})$

Solution :- Given

$(x^{5}+1), (3-x^{4}), (4+x^{3}+x^{6}),$

$(2a^{3}-3b^{5}), (2a^{3}+3b^{5}), (2a^{4}-3a^{2}b^{2}+b^{4})$

Let subtract and add the terms

$\Rightarrow x^{5}+1+3-x^{4}+4+x^{3}+x^{6}+2a^{3}-3b^{5}+2a^{3}+3b^{5}$

$+2a^{4}-3a^{2}b^{2}+b^{4}$

$\Rightarrow x^{6}+x^{5}-x^{4}+x^{3}+8+4a^{3}+2a^{4}-3a^{2}b^{2}+b^{4}$

1106503_1037492_ans_3443d3d002f54a09a291c34e33f6a54d.jpg

Simplify: $x-(y-z)+x+(y-z)+y-(z+x)$

$x - \left( {y - z} \right) + x + \left( {y - z} \right) + y - \left( {z + x} \right)$

$= x - y + z + x + y - z + y - z - x$

$= x + y - z$

Evaluate ${a^2} - {b^2} - {(a + b)^2}$

${a^2} - {b^2} - {(a + b)^2}$

${a^2} - {b^2} - \left( {{a^2} + {b^2} + 2ab} \right)$

${a^2} - {b^2} - {a^2} - {b^2} - 2ab$

$- 2{b^2} - 2ab$

$= - 2b(b + a)$

Solve : $5a - 9b - 7a + 10b$

Given expression is

$5a-9b-7a+10b$

On solving it further, we get:

$(5a-7a)+(10b-9b)$

$=-2a+b$

Hence,

$5a-9b-7a+10b=-2a+b$ .

Simplify:
$(4x^{2} + 7x^{3}y^{2}) -(-6x^{2} - 7x^{3}y^{2} -4x) - (10x + 2x^{2})$

$(4x^2 + 7x^3y^2) - (-6x^2 -7x^3y^2 - 4x) - (10x + 2x^2)$

$= 4x^2 + 7x^3y^2 + 6x^2 + 7x^3y^2 + 4x - 10x - 2x^2$

$= (4+6-2)x^2 + (7 + 7)x^3y^2 + (4 - 10)x$

$= 8x^2 + 14x^3y^2 - 6x$

Subtract
i) $4a - 2b - c$ from $-a + 2b + 3c$
ii) $-3x^2 + 7y^2 - z^2 \,$ from $\, 2x^2 - 5y^2 - 7z^2$

$(1)$

$(-a+2b+3c)-(4a-2b-c)$

$=-a+2b+3c-4a+2b+c$

$=-a-4a+2b+2b+3c+c$

$=-5a+4b+4c$

$(2)$

$\left ( 2x^{2} -5y^{2}-7z^{2}\right )-\left ( -3x^{2}+7y^{2}-z^{2} \right )$

$=2x^{2} -5y^{2}-7z^{2}+3x^{2}-7y^{2}+z^{2}$

$=2x^{2} +3x^{2}-5y^{2}-7y^{2}-7z^{2}+z^{2}$

$=5x^{2}-12y^{2}-6z^{2}$

Simplify:
$-4x\left(5x^{2}-12\right)+ 7(x+5)$

Given that:

$-4x(5x^2-12)+7(x+5)$

On simplifying, we get

$=-20x^3+48x+7x+35$

$=-20x^3+45x+35.$

Subtract : $9b\left(-2a^{2}b^{2}+2b^{2}+3a\right)$ from $2a\left(7a^{2}b^{2}+ab^{2}+5b\right)$

We have,

$2a\left( 7{{a}^{2}}{{b}^{2}}+a{{b}^{2}}+5b \right)-9b\left( -2{{a}^{2}}{{b}^{2}}+2{{b}^{2}}+3a \right)$

$=14{{a}^{3}}{{b}^{2}}+2{{a}^{2}}{{b}^{2}}+10ab+18{{a}^{2}}{{b}^{3}}-18{{b}^{3}}-27ab$

$=14{{a}^{3}}{{b}^{2}}+18{{a}^{2}}{{b}^{3}}+2{{a}^{2}}{{b}^{2}}-17ab-18{{b}^{3}}$

Hence, this is the answer.

Add the expressions in each of the following.
i) $2l + 3m - 6n + 4p, \, 3l - 5m + 16n - 4p , \, 12l - 6m - 4n - 2p \, and \, l - 2m + 3n - 4p$
ii) $7x^3 + 3x + 9 , \, -2x^3 -3x^2 - 15, \, 3x^3 - 6x^2 + 4x - 6 \, and \, 12x^2 - 6$
iii) $5a^2 - 7ab + 9b^2 , \, 4a^2 - 2b^2 - 9ab - 6, \, 4 - 3b^2 + 2ab + 6a^2 \, and \, 12 ab - 3a^2 - 9b^2$

(i)

$2l+3m-6n+4p$

$3l-5m+16n-4p$

$12l-6m-4n-2p$

$l-2m+3n-4p$

---------------------------------

$18l-10m+9n-6p$

(ii)

$7x^3+0+3x+9$

$-2x^3-3x^2+0-15$

$3x^3-6x^2+4x-6$

$0+12x^2+0-6$

---------------------------------

$8x^3+3x^2+7x-18$

(iii)

$5a^2-7ab+9b^2$

$4a^2-9ab-2b^2$

$6a^2+2ab-3b^2+4$

$-3a^2+12ab-9b^2$

---------------------------------

$12a^2-2ab-5b^2+4$

Subtract $(7 pqr-8q+6)$ from $(10-4pqr+3q)$

Let $a=7pqr-8q+6$ and $b=10-4pqr+3q$

According to the question,

$b-a=\left( 10-4pqr+3q \right)-\left( 7pqr-8q+6 \right)$

$b-a=10-4pqr+3q-7pqr+8q-6$

$b-a=4-11pqr+11q$

Hence, this is the answer.

Express the polynomial $(1, 2, 3)$ in index form.

Polynomial

$(1,2,3)$ in index form is

$1x^{2}+2x^{1}+3x^{0}$

$=x^{2}+2x+3$

What must be subtracted from $2{x^4} + 4{x^2} - 3x + 7$ to get $3{x^3} - {x^2} + 2x + 1$ ?

Let $p\left( x \right)$ be subtracted

Then $2{x^4} + 4{x^2} - 3x + 7 - p\left( x \right) = 3{x^3} - {x^2} + 2x + 1$

$\Rightarrow p\left( x \right) = 2{x^4} - 3{x^3} + 5{x^2} - 5x + 6$

Relation between $m_{1}$ and $m_{2}$ : $-4m_{1}-4m_{2}=3m_{1}-6m_{2}$

Given relation:

$-4m_1-4m_2=3m_1-6m_2$

$-4m_1-3m_1=4m_2-6m_2$

$-7m_1=-2m_2$

$m_2=\dfrac{7}{2}m_2.$

Hence, this is the simplified relation.

Simplify $(5p^{2}-3)+(2p^{2}-3p^{3})$

We have,

$\left( 5{{p}^{2}}-3 \right)+\left( 2{{p}^{2}}-3{{p}^{3}} \right)$

$=5{{p}^{2}}-3+2{{p}^{2}}-3{{p}^{3}}$

$=7{{p}^{2}}-3-3{{p}^{3}}$

$=-3{{p}^{3}}+7{{p}^{2}}-3$

Hence, this is the answer.

Simplify $4{x^2} + 7 - \left( {{x^2}y + 3 - \overline {2{x^2}y + 2} } \right)$

Consider the given expression.

$4{{x}^{2}}+7-\left( {{x}^{2}}y+3-\overline{2{{x}^{2}}y+2} \right)$

$=4{{x}^{2}}+7-\left( {{x}^{2}}y+3-\left( -\left( 2{{x}^{2}}y+2 \right) \right) \right)$

$=4{{x}^{2}}+7-\left( {{x}^{2}}y+3+2{{x}^{2}}y+2 \right)$

$=4{{x}^{2}}+7-\left( 3{{x}^{2}}y+5 \right)$

$=4{{x}^{2}}+7-3{{x}^{2}}y-5$

$=4{{x}^{2}}-3{{x}^{2}}y+2$

Hence, this is the answer.

Add $(6ab^{3}-5ab)(2a^{2}+b-5),\ ab(a^{2}+1)$ and $ab^{3}(3a^{3}+2b+1).$

Given that,

$\left( 6a{{b}^{3}}-5ab \right)\left( 2{{a}^{2}}+b-5 \right)+ab\left( {{a}^{2}}+1 \right)+a{{b}^{3}}\left( 3{{a}^{3}}+2b+1 \right)$

$=12{{a}^{3}}{{b}^{3}}+6a{{b}^{4}}-30a{{b}^{3}}-10{{a}^{2}}b-5a{{b}^{2}}-25ab+{{a}^{3}}b+ab+3{{a}^{4}}{{b}^{3}}+2a{{b}^{4}}+a{{b}^{3}}$

$=12{{a}^{3}}{{b}^{3}}+8a{{b}^{4}}-29a{{b}^{3}}-10{{a}^{2}}b-5a{{b}^{2}}-2$ $ab+{{a}^{3}}b+3{{a}^{4}}{{b}^{3}}$

$=3{{a}^{3}}{{b}^{3}}(4+a)-5ab(2a+b)+a{{b}^{3}}(8b-29)+ab({{a}^{2}}-2$ $)$

Hence, this is the answer.

Add all of them $x-8y, 3xy-y$ and $y+1$

We have,

$x-8y,3xy-y\,and\,y+1$

On adding,

$\left( x-8y \right)+\left( 3xy-y \right)+\left( y+1 \right)$

$=x-8y+3xy-y+y+1$

$=x+3xy-8y+1$

Subtract $a^{3}-4a^{2}+5a-6$ from the sum of $3a^{3}+a^{2}+1$ and $a^{2}-2$

Let,

$p={{a}^{3}}-4{{a}^{2}}+5a-6$

$q=3{{a}^{3}}+{{a}^{2}}+1$

$r={{a}^{2}}-2$

According to the question,

$\Rightarrow 3{{a}^{3}}+{{a}^{2}}+1+{{a}^{2}}-2-\left( {{a}^{3}}-4{{a}^{2}}+5a-6 \right)$

$\Rightarrow 3{{a}^{3}}+2{{a}^{2}}-1-{{a}^{3}}+4{{a}^{2}}-5a+6$

$\Rightarrow 2{{a}^{3}}+6{{a}^{2}}-5a+5$

Hence, this is the answer.

Subtract $(x-y)$ from $(4y-5x)$

$4y-5x-\left( x-y \right)$

$\Rightarrow 4y-5x-x+y$

$\Rightarrow 5y-6x$

If $(3a^{2}+1)-(4+2a^{2})=0$ , find the value of $a$ .

We have,

$\left( 3{{a}^{2}}+1 \right)-\left( 4+2{{a}^{2}} \right) =0$

$3{{a}^{2}}+1-4-2{{a}^{2}} =0$

${{a}^{2}}-3 =0$

$a=\pm 3$

Hence, this is the answer.

Simplify $x + y - \left[ {x + y - \left\{ {x + y - \left( {\overline {x + y} } \right)} \right\}} \right]$

Consider the given expression.

$x+y-\left[ x+y-\left\{ x+y-\overline{\left( x+y \right)} \right\} \right]$

$=x+y-\left[ x+y-\left\{ x+y-\left( -\left( x+y \right) \right) \right\} \right]$

$=x+y-\left[ x+y-\left\{ x+y+x+y \right\} \right]$

$=x+y-\left[ x+y-\left\{ 2x+2y \right\} \right]$

$=x+y-\left[ x+y-2x-2y \right]$

$=x+y-\left[ -x-y \right]$

$=x+y+x+y$

$=2\left( x+y \right)$

Hence, this is the answer.

Subtract the following :
$4z^2xy+x^2zy-5xyz+2xy^2z$ from $x^2zy+xyz+xy^2z-z^2xy$

$(x^2zy + xyz + xy^2z - z^2xy)-(4z^2xy+x^2zy-5xyz+2xy^2z)$

$=x^2zy + xyz + xy^2z - z^2xy -4z^2xy-x^2zy+5xyz-2xy^2z$

$=x^2zy - x^2zy + xyz + 5xyz +xy^2z-2xy^2z-z^2xy-4z^2xy$

$={6xyz-xy^2x-5z^2xy}$

Add : $3x - 5y - 5z$ and $-5x - 2y - 3$

add

$3x+5y-5z$ and

$-5x-2y-3$

$\Rightarrow (3x-5y-5z)$

$+ (-5x-2y-3)$

$\overline{(3-5)x-(5+2)y-5z-3}$

$\Rightarrow -\boxed{2x-7y-5z-3}$

Subtract the following :
$3m^2-2+1m$ from $m^3-9$

$(m^3-9) - (3m^2-2+1m)$

$=m^3-9 - 3m^2 +2 -m$

$={m^3-3m^2-m-7}$

Subtract $2p - 2q + 2 pq$ from $7p - 8q - pq$ .

$(7p-8q-pq)-(2p-2q+2pq)$

$\Rightarrow (7p-2p)+(-8q+2q)+(-pq-2pq)$

$\therefore \boxed{5p-6q-3pq}$

Add the followings: $3ab,-4ab,7ab,-5ab$ .

To find:

$3ab-4ab+7ab-5ab$

$=(3-4+7-5)ab$

$=(1)(ab)$

$=ab$

Subtract the following :
$3r+2t$ from $2r-3t$

$(2r-3t) - (3r+2t)$

$=2r - 2t - 3r - 2t$

$=2r - 3r - 2t -2t$

$=(2-3)r + (-2-2)t$

$={-r - 4t}$

Add : $3a + 4b - 2c$ and $-6a - 2b + 3c$

add

$3a+4b-2c$ and

$-6a-2b+3c$

$(3a+4b-2c)+(-ba-2b+3c)$

$\Rightarrow (3a-6a)+(4b-2b)+(3c-2c)$

$\Rightarrow \boxed{-3a+2b+c}$

Find
$3a{ b }^{ 2 }$ + $7a{ b }^{ 2 }$

$8ab^{2}+7ab^{2}= 10ab^{2}$
1196328_1111133_ans_7275b0a78c444ff581b3bcc3c7c65a46.jpg

Add: ${2x^2}-{4y^2}-{z^2},\,{8x^2}+{5y^2}+{4z^2},\,{5x^2}-{5y^2}+{2z^2}$

Say

$A=2x^2-4y^2-z^2$

$B=8x^2+5y^2+4z^2$

$C=5x^2-5y^2+2z^2$

$A+B+C=15x^2-4y^2+5z^2.$

Solve: $\dfrac{11a^2 + 9ac}{11b^2 + 9bd} = \dfrac{a^2 + 3ac}{b^2 + 3bd}$

$\cfrac{11a^{2}+9ac}{11b^{2}+9bd}=\cfrac{a^{2}+3ac}{b^{2}+3bd}$

$\cfrac{a(11a+9c)}{b(11b+9d)}=\cfrac{a(a+3c)}{b(b+3d)}$

$(11a+9c)(b+3d)=(a+3c)(11b+9d)$

$11 ab+33ad+9cb+27cd=11ab+9ad+33bc+27cd$

$33ad-9ad+9cb-33bc=0$

$24ad-24bc=0$

$24(ad-bc)=0$

Identify terms and factors in the expression given.
$pq+q$ .

Terms in the given expression are

$pq$ and

$q$

Take

$q$ common from the given two terms

$pq+q=q(p+1)$

Thus factors are

$p+1$ and

$q$ .

Identify terms and factors in the expression given.
$xy+2x^2y^2$ .

${x}{y}+2{x^{2}}{y^{2}}={x}{y}(1+2{x}{y})$

terms are

${x}{y},2{x^{2}}{y^{2}}$

factors are

$xy,1+2{x}{y}$

Identify the terms in the expression given below:
$2x - 3$ .

A term in a given expression can be a signed number, a variable, or a constant multiplied by a variable or variables.

Term in the given expression

$=2x,-3$

Identify terms and factors in the expression given.
$-4x+5$ .

$-4x+5$

Terms:-

$1)$

$-4x$

$2)$

$5$

Factors:

$1)$

$-4$

$2)$

$\left(x-\dfrac{5}{4}\right)$ .

If $x = 1, y = 2$ and $z = 5$ , find the value of $x^{2} + y^{2} + z^{2}$ .

1*1+2*2+5*5

1+4+25

Simplify:
$15pq + 15 + 9q + 25p$ .

$15pq+9q+25p+15\\ 3q(5p+3)+5(5p+3)\\ (3q+5)(5p+3)$

Subtract $3x (x - 4y + 5z)$ from $4x (2x - 3y + 10z)$

$3x(x-4y+5z)$ from

$4x(2x-3y+10z)$

$3x(x-4y+5z)$

$\Rightarrow (8x^2-12xy+40xz)-(3x^2-12xy+15xz)$

$\Rightarrow (8x^2-3x^2)+[-12xy+12xy]+40xz-15xz$

$\Rightarrow 5x^2+25xz$

$\Rightarrow \boxed{5x[x+57]}$

$\left(x-6\right)$ + $\left(3x-4\right)$ + $\left(x-1\right)$

$(x-6)+(3x-4)+(x-1)$

$=x-6+3x-4+x-1$

$=5x-11$

Leo bought a shit for $Rs(5x+20)$ and a belt for $Rs(2x-10)$ . How much did he spend?

Cost of Shirt $$=5x+20

Cost of Belt

$=2x-10$

Total money spent

$=5x+20+2x-10$

$=7x+10$

Identify the terms in the expression $ab+2b^{2}-3a^{2}$ .

The given expression is :

$ab+2b^2-3a^2$

Therefore, terms are :

$ab,2b^2,-3a^2$

Simplify $x(a-3)+y(3-a)$

Solving this we get,

$x(a-3)+y(3-a)$

$\implies ax-3x+3y-ay$

$\implies a(x-y)-3(x-y)$

$\implies (x-y)(a-3)$

$\dfrac{3+5x}{2x}-\dfrac{36}{x}+\dfrac{5}{2x}$

$=\dfrac{3+5x-72+5}{2x}$

$=\dfrac{5x-64}{2x}$

$=\dfrac{5}{2}-\dfrac{32}{x}$

Add the following like terms
$13x,\ -27x$ and $-39x$

According to question,

$13x,\ -27x$ and

$-39x$

Adding terms,

$\implies 13x-27x-39x$

$\implies -53x$

Add the following like terms
$7xyz$ and $-4xyz$

In case of addition of like terms numerical coefficient gets added and the variable part remains as it is.

$7xyz - 4xyz = \left( {7 - 4} \right)xyz$

$=3 xyz$

$\dfrac{2}{3x}+\dfrac{4}{3x(2x)}$

$=\dfrac{4x+4}{6x^2}$

$=\dfrac{4(x+1)}{6x^2}$

$=\dfrac{2(x+1)}{3x^2}$

Add the following like terms
$19x,\ -55x$ and $\dfrac{2}{3}x$

According to question,

$19x,-55x,\dfrac{2}{3}x$

Adding terms,

$19x-55x+\dfrac{2}{3}x$

$\implies -36x+\dfrac{2}{3}x$

$\implies \dfrac{-106}{3}x$

Identify terms and factors in the expressions given.
$1.2ab-2.4b+3.6a$ .

Terms:-

$1)$

$1.2$ ab

$2)$

$-2.4$ b

$3)$

$3.6$ a

Factors:-

$1)$

$1.2$

$2)$

$ab-2b+3a$ .

Subtract :
$-x^{2}+y^{2}-x^{2}y+5xy^{2}$ from $x^{2}+x^{2}y-5xy^{2}-y^{2}$

$\dfrac { { \quad \quad x }^{ 2 }+{ x }^{ 2 }y-5{ xy }^{ 2 }-{ y }^{ 2 }\\ -\quad -{ x }^{ 2 }-{ x }^{ 2 }y+{ 5xy }^{ 2 }+{ y }^{ 2 }\\ +\quad \quad +\quad \quad -\quad \quad - }{ { 2x }^{ 2 }+{ 2x }^{ 2 }y-10{ xy }^{ 2 }-2{ y }^{ 2 } }$

$2x^2+2x^2y-10xy^2-2y^2$

Substract $x^{3}+2y^{2}$ from $2y^{3}+x^{2}$ .

$\left( 2 {y}^{3} + {x}^{2} \right) - \left( {x}^{3} + 2{y}^{2} \right)$

$= 2 {y}^{3} + {x}^{2} - {x}^{3} - 2 {y}^{2}$

$= \left( 2 {y}^{3} - 2 {y}^{2} \right) + \left( {x}^{2} - {x}^{3} \right)$

$= 2 {y}^{2} \left( y - 1 \right) - {x}^{2} \left( x - 1 \right)$

Hence the correct answer is

$2 {y}^{2} \left( y - 1 \right) - {x}^{2} \left( x - 1 \right)$ .

From the sum of $4 + 3x$ and $5 - 4x + 2{x^2}$ , subtract the sum of $3{x^2} - 5x$ and $- {x^2} + 2x + 5$ .

given,

$[(4+3x)+(5-4x+2x^{2})]-[(3x^{2}-5x)+(-x^{2}+2x+5)]$

$=(2x^{2}-x+9)-(2x^{2}-3x+5)$

$=2x+4$

Add the following like terms
$-15pq,\ 12qp,\ -23pq$ and $7pq$

$- 15pq + 12pq - 23pq + 7pq = \left( { - 15 + 12 - 23 + 7} \right)pq$

$= - 19pq$

(i)Subtract $4 a - 7 a b + 3 b + 12$ from $12 a - 9 a b + 5 b - 3$ .
(ii)Subtract $3 x y + 5 y z - 7 z x$ from $5 x y - 2 y z - 2 z x + 10 x y z$ .
(iii)Subtract $4 p ^ { 2 } q - 3 p q + 5 p q ^ { 2 } - 8 p + 7 q - 10$ from $18 - 3 p - 11 q + 5 p q - 2 p q ^ { 2 } + 5 p ^ { 2 } q$ .

$1) (12a-9ab+5b-3)-(4a-7ab+3b+12)=8a-2ab+2b-15$

$2) (5xy-2yz-2zx+10xyz)-(3xy+5yz-7zx)=2xy-7yz+5zx+10xyz$

$3) (18-3p-11q+5pq-2pq^2+5p^2q)-(4p^2q-3pq+5pq^2-8p+7q-10) = 28+5p-18q+8pq-7pq^2+p^2q$

Add $p^{2}-5pq-q^{2}$ and $-3p^{2}+2pq-2p^{2}$

$\left( {p}^{2} - 5pq - {q}^{2} \right) + \left( -3 {p}^{2} + 2 pq - 2 {p}^{2} \right)$

$= {p}^{2} - 5pq - {q}^{2} - 3 {p}^{2} + 2pq - 2 {p}^{2}$

$= \left( {p}^{2} - 3 {p}^{2} - 2 {p}^{2} \right) + \left( 2pq - 5 pq \right) - {q}^{2}$

$= - 4{p}^{2} - 3 pq - {q}^{2}$

$= - \left( 4 {p}^{2} + 3pq + {q}^{2} \right)$

Solve ${ ab }^{ 2 }+\left( a-1 \right) b-1$

According to question,

$ab^2+(a-1)b-1$

$\implies ab^2+ab-b-1$

$\implies ab(b+1)-1(b+1)$

$\implies (ab-1)(b+1)$

Add $-10v+6v$

According to question,

$-10v+6v$

$\implies v(-10+6)$

$\implies -4v$

Substract $xyz+2xy$ from $3zxy-5yx$ .

$\left( 3zxy - 5yx \right) - \left( xyz + 2xy \right)$

$= 3zxy - 5yx - xyz - 2xy$

$= \left( 3zxy - xyz \right) - \left( 5yx + 2xy \right)$

$= 2xyz - 7xy$

Hence the correct answer is

$\left( 2 xyz - 7 xy \right)$ .

Substract $x^{2}$ from $x^{2}+y^{2}-3y$

$\left( {x}^{2} + {y}^{2} - 3y \right) - {x}^{2}$

$= {x}^{2} + {y}^{2} - 3y - {x}^{2}$

$= \left( {x}^{2} - {x}^{2} \right) + {y}^{2} - 3y$

$= 0 + {y}^{2} - 3y$

$= {y}^{2} - 3y$

Hence the correct answer is

$\left( {y}^{2} - 3y \right)$ .

Add: $2x(z-x-y)$ and $2y(z-y-x)$

$2x(z-x-y)+2y(z-y-x)$

$2xz-2x^2-2xy+2yz-2y^2-2xy$

$-2x^2-2y^2-4xy+2yz+2xz$

1082295_1173089_ans_434cd08224034a5eb466de86994b3356.png

Add $a - b + ab,b - c + bc,c - a + ac$

To add:

$a-b+ab,\ b-c+bc,\ c-a+ac$

$a-b+b-c+c-a\ +ab+bc+ac$

$=0+ab+bc+ac$

$=ab+bc+ac$

Subtract $3a(a+b+c)-2b(a-b+c)$ from $4c(-a+b+c)$

$4c(-a+b+c)-[3a(a+b+c)-2b(a-b+c)]$

$4c(-a+b+c)-[3a^{2}+3ab+3ac-2ab+2b^{2}-2bc]$

$4c(-a+b+c)-[3a^{2}+2b^{2}+ab+3ac-2bc]$

$-4ac+4bc+4c^{2}-3a^{2}-2b^{2}-ab-3ac+2bc$

$4c^{2}-3a^{2}-2b^{2}-ab+6bc-7ac$

Solve the following:
$2\left(5x^{2}+7x+8\right)-\left(4x^{2}-7x+3\right) =$

Given,

$2(5x^2 + 7x + 8) - (4x^2 - 7x + 3)$

$= 10x^2 + 14x + 16 - 4x^2 + 7x - 3$

$= 6x^2 + 21x +13$

Add the following:
$ab-bc,bc-ca,ca-ab$

$ab-bc+bc-ca+ca-ab=0$

Add the following :-
$ab-ba,bc-ca,ca-ab$

$ab - ba + bc - ca + ca - ab = -ba + bc$

$=$

$b (c - a)$

Add:
$2{p^4} - 3{p^3} + {p^2} - 5p + 7, - 3{p^4} - 7{p^3} - 3{p^2} - p - 12$

Given that:

$2p^4-3p^3+p^2-5p+7\, \& -3p^4-7p^3-3p^2-p-12$

now add both of them,

$=(2p^4-3p^3+p^2-5p+7)+(-3p^4-7p^3-3p^2-p-12)$

$=(2p^4+(-3p^4))+((-3p^3)+(-7p^3))+((p^2)+(-3p^2))+(-5p-p)+(7-12)$

$=-p^4-10p^3-2p^2-6p-5$

Subtract $4a-7ab+3b+12$ from $12a-9ab+5b-3$ .

$\rightarrow (12a-9ab+5b-3)-(4a-7ab+3b+12)$

$=(12a-4a)+(-9ab+7ab)+(5b-3b)+(-3-12)$

$=8a-2ab+2b-15$

Subtract $3xy+5yz-7zx$ from $5xy-2yz-2zx+10xyz$ .

$(5xy-2yz-2zx+10xyz)-(3xy+5yz-7zx)$
On pairing the like terms, we get

$(5xy-3xy)+(-2yz-5yz)+(-2zx+7zx)+10xyz$

$2xy-7yz+5zx+10xyz$

Solve :
$1 + x y - x - y$

$1 + x y - x - y=1-x+xy-y=1-x-y+xy=1(1-x)-y(1-x)=(1-y)(1-x)$

Add $x+3,-3x^{2}-x-4,1+3$

$(x+3)+(-3x^{2}-x-4)+(1+3)=-3x^{2}+3$

Simplify: $(-10{a}^{3}b+12{a}^{2}{b}^{2}-6a{b}^{3})-(8{a}^{3}b+6{a}^{2}{b}^{2}-9a{b}^{3})$

Given

$(-10a^3b+12a^2b^2-6ab^3)-(8a^3b+6a^2b^2-9ab^3)$

$=-10a^3b+12a^2b^2-6ab^3-8a^3b-6a^2b^2+9ab^3$

$=-10a^3b-8a^3b+12a^2b^2-6a^2b^2-6ab^3+9ab^3$

$=-18a^3b+6a^2b^2+3ab^3$

Subtract:
$a+2b-c$ from $3a-b+2c$

$3a-b+2c-(a+2b-c)$

$=3a-b+2c-a-2b+c$

$=3a-a-b-2b+2c+c$

$=2a-3b+3c$

Subtract $(2x - 4y + 3z)$ from the sum of $(x + 3y - 4z), (2x + y - z)$ and $(-2x + 3y - z)$ .

$(x+3y-4z)+ (2x+y-z)+ (-2x+3y-z)- (2x-4y+3z)$

$= (x+2x+3y+y-4z-z)+(-2x+3y-z)-(2z-4y+3z)$

$= (3x+4y-5z)+(-2x+3y-z)-(2x-4y+3z)$

$= (x+7y-6z)-(2x-4y+3z)$

$= x-2x+7y+4y-6z-3z$

$= -x+11y-9z$

$\therefore~$ Ans is

$(-x+11y-9z)$

Solve:
$- [3{a^2} - 2{a^2} + 9{a^2} - \{ 6{a^2} - ( - 2{a^2} + 3{a^2})\} ]$

$-[{ 3a }^{ 2 }-{ 2a }^{ 2 }+{ 9a }^{ 2 }-\{ { 6a }^{ 2 }-(-{ 2a }^{ 2 }+{ 3a }^{ 2 })\} ]\\ =-[{ 3a }^{ 2 }-{ 2a }^{ 2 }+{ 9a }^{ 2 }-\{ { 6a }^{ 2 }-({ a }^{ 2 })\} ]\\ =-[{ 3a }^{ 2 }-{ 2a }^{ 2 }+{ 9a }^{ 2 }-{ 5a }^{ 2 }]\\ =-[{ a }^{ 2 }+{ 4a }^{ 2 }]\\ =-[{ 5a }^{ 2 }]\\ =-{ 5a }^{ 2 }$

Find the sum of $-3ab+7cd-5qr;\,\,2ry+8qr-cd;\,\,2cd-3qr+ab-2ry$

given,

$-3ab+7cd-5qr+2ry+8qr-cd+2cd-3qr+ab-2ry$

Grouping the like terms,

$=\left(-3ab+ab\right)+\left(7cd-cd\right)+\left(-5qr+8qr-3qr\right)+\left(2ry-2ry\right)$

$=-2ab+6cd+0+0$

$=-2ab+6cd$

Simplify $(3x-11y)-(17x+13y)$

$(3x-11y)-(17x+13y)=3x-11y-17x-13y$

$=3x-17x-11y-13y$

$=-14x-24y$

$=-2(7x+12y)$

Group the like terms together from the following : $12x,12,-25x,-25,-25y,1,x,12y,y$

forms with one variable.

$12 x , - 25 x , - 25 y , x , 12 y , y$

constant terms

$12,1$

Add
$7{x^3} + 2{x^2} - 5x - 7\,\& \, - 5{x^2} + {x^3} + 4x - 5$

$(7x^3+2x^2-5x-7)+(-5x^2+x^3+4x-5)=(7x^3+x^3)+(2x^2-5x^2)+(-5x+4x)+(-7-5)$

$=8x^3-3x^2-x-12$

Subtract $2x-3y+4z$ from $4x-6y-z$ .

$4x-6y-z-(2x-3y+4z)=4x-6y-z-2x+3y-4z$

$=(4x-2x)+(-6y+3y)+(-z-4z)$

$=(2x)+(-3y)+(-5z)$

$=2x-3y-5z$

Factorize:
$16a^4-9b^4$

We have,

$16a^4-9b^4$

$\Rightarrow (4a^2)^2-(3b^2)^2$

$\Rightarrow (4a^2+3b^2)(4a^2-3b^2)$

Hence, this is the answer.

$(7x^{2}-2y^{2}-6xy+y)-(2y^{2}+2x^{2}-1+2xy)$

$(7x^2-2y^2-6xy+y)-(2y^2+2x^2-1+2xy)$

$=7x^2-2x^2-2y^2-2y^2-6xy-2xy+y+1$

$=5x^2-4y^2-8xy+y+1$ .

1128140_1244837_ans_8cf9b73e1a3149c598f90cdcac80e899.jfif

$(x^{3}+ x^{2}-3x +5)+(x-1)$

$(x^{3}+x^{2}-3x+5)+(x-1)$

$= x^{3}+x^{2}-3x+x+5-1$

$= x^{3}+x^{2}-2x+4$

1130969_1248361_ans_b8f3f64df5b74f99bb7ca61619dbe206.jpg

$A=\dfrac { { 3x }^{ 4 } }{ 4 } -\dfrac { { x }^{ 4 } }{ 2 } +5x-7;B=\dfrac { { 2x }^{ 3 } }{ 3 } +\dfrac { { 3x }^{ 3 } }{ 2 } -\dfrac { x }{ 6 } ;C=\dfrac { { 2x }^{ 4 } }{ 3 } -\dfrac { { 3x }^{ 4 } }{ 4 } +5$ find
$A+B+C$ ?

Given that,

$A = \dfrac{3x^4}{4} - \dfrac{x^4}{2} + 5x - 7 = \dfrac{x^4}{4} + 5x - 7$

$B = \dfrac{2x^3}{3} + \dfrac{3x^3}{2} - \dfrac{x}{6} = \dfrac{13x^3}{6} - \dfrac{x}{6}$

$C = \dfrac{2x^4}{3} - \dfrac{3x^4}{4} + 5 = - \dfrac{x^4}{12} + 5$

$A + B + C = \dfrac{x^4}{4} + 5x - 7 + \dfrac{13x^3}{6} - \dfrac{x}{6} - \dfrac{x^4}{12} + 5$

$= \dfrac{x^4}{4} - \dfrac{x^4}{12} + \dfrac{13x^3}{6} + 5x - \dfrac{x}{6} - 7 + 5$

$= \dfrac{x^4}{6} + \dfrac{13x^3}{6} + \dfrac{29x}{6} - 2$

$(4x-5y)-(2x-3y)$

$=4x-5y-2x+3y$

$=(4x-2x)-(5y-3y)$

$=2x-2y$

Add and subtract
$(i)m-n,m+n$
$(ii)mn+5-2,mn+3$

$Part (i)$

$Add$

$=m-n+m+n=2m$

$Subtract$

$=m-n-m-n=-2n$

$Part (ii)$

$Add$

$=mn+5-2+mn+3=2mn+6$

$Subtract$

$=mn+5-2-mn-3=0$

Hence, this is the answer.

If $\frac { a } { b } = \frac { c } { d }$ prove that
$( 5 a + 7 b ) ( 2 c - 3 d ) = ( 5 c + 7 d ) ( 2 a - 3 b )$

Given

$\dfrac { a }{ b } =\dfrac { c }{ d } \\ ad=bc$

LHS:

$\left( 5a+7b \right) \left( 2c-3d \right) \\ =10ac+14bc-15ad-21bd\\ =10ac+14ad-15ad-21bd\\ =10ac-ad-21bd$

RHS:

$\left( 5c+7d \right) \left( 2a-3b \right) \\ =10ac+14ad-15bc-21bd\\ =10ac+14ad-15ad-21bd\\ =10ac-ad-21bd$

$LHS=RHS$ (Hence proved)

$A=\dfrac { { 3x }^{ 4 } }{ 4 } -\dfrac { { x }^{ 4 } }{ 2 } +5x-7;b=\dfrac { { 2x }^{ 3 } }{ 3 } +\dfrac { { 3x }^{ 3 } }{ 2 } -\dfrac { x }{ 6 } ;C=\dfrac { { 2x }^{ 4 } }{ 3 } -\dfrac { { 3x }^{ 4 } }{ 4 } +5$ find
$-A+B-C$

Given that,

$A = \dfrac{3x^4}{4} - \dfrac{x^4}{2} + 5x - 7 = \dfrac{x^4}{4} + 5x - 7$

$B = \dfrac{2x^3}{3} + \dfrac{3x^3}{2} - \dfrac{x}{6} = \dfrac{13x^3}{6} - \dfrac{x}{6}$

$C = \dfrac{2x^4}{3} - \dfrac{3x^4}{4} + 5 = - \dfrac{x^4}{12} + 5$

$- A + B - C = - \left(\dfrac{x^4}{4} + 5x - 7 \right) + \left(\dfrac{13x^3}{6} - \dfrac{x}{6} \right ) - \left( - \dfrac{x^4}{12} + 5 \right)$

$= - \dfrac{x^4}{4} + \dfrac{x^4}{12} + \dfrac{13x^3}{6} - 5x - \dfrac{x}{6} + 7 - 5$

$= - \dfrac{x^4}{6} + \dfrac{13x^3}{6} - \dfrac{31x}{6} + 2$

Addition of $(x^{2}+y^{2})$ and $(x^{2}-y^{2})$ is______________

Sum

$=({ x }^{ 2 }+{ y }^{ 2 })+({ x }^{ 2 }-{ y }^{ 2 })$

$={ x }^{ 2 }+{ y }^{ 2 }+{ x }^{ 2 }-{ y }^{ 2 }\\ =2{ x }^{ 2 }$

Subtract $-a-b-9c$ from $-a+b-9c$

$-a+b-9c-\left(-a-b-9c\right)$

$=-a+b-9c+a+b+9c$

Grouping the like terms,

$=\left(-a+a\right)+\left(b+b\right)+\left(-9c+9c\right)$

$=2b$

The sum of two expressions is $3 a^2 + 2 ab - b^2$ . If one of them is $2 a^2 + 3 b^2$ , find the other.

The other expression is calculated as follows

$(3 a^2 + 2 ab - b^2) - (2 a^2 + 3 b^2)$
On simplification, we get

$= 3 a^2 - 2 a^2 - b^2 - 3 b^2 + 2 ab$

$= a^2 4 b^2 + 2 ab$

multiple $A=\dfrac { { 3x }^{ 4 } }{ 4 } -\dfrac { { x }^{ 4 } }{ 2 } +5x-7;B=\dfrac { { 2x }^{ 3 } }{ 3 } +\dfrac { { 3x }^{ 3 } }{ 2 } -\dfrac { x }{ 6 } ;C=\dfrac { { 2x }^{ 4 } }{ 3 } -\dfrac { { 3x }^{ 4 } }{ 4 } +5$ find
$3A-2B-C$

Given that,

$A = \dfrac{3x^4}{4} - \dfrac{x^4}{2} + 5x - 7 = \dfrac{x^4}{4} + 5x - 7$

$B = \dfrac{2x^3}{3} + \dfrac{3x^3}{2} - \dfrac{x}{6} = \dfrac{13x^3}{6} - \dfrac{x}{6}$

$C = \dfrac{2x^4}{3} - \dfrac{3x^4}{4} + 5 = - \dfrac{x^4}{12} + 5$

$3A - 2B - C = 3 \left(\dfrac{x^4}{4} + 5x - 7 \right) - 2 \left(\dfrac{13x^3}{6} - \dfrac{x}{6} \right ) - \left( - \dfrac{x^4}{12} + 5 \right)$

$= \dfrac{3x^4}{4} + 15x - 21 - \dfrac{13x^3}{3} + \dfrac{x}{3} + \dfrac{x^4}{12} - 5$

$= \dfrac{3x^4}{4} + \dfrac{x^4}{12} - \dfrac{13x^3}{3} + 15x + \dfrac{x}{3} - 21 - 5$

$= \dfrac{5x^4}{6} - \dfrac{13x^3}{3} + \dfrac{46x}{3} - 26$

Add the given expressions:
$a + b - 3, 6 - a + 3, a - b + 3$

1216598_1248588_ans_fea818de7c0b4efc8cc4e39bc35c2537.jpg

Find sum $a(b-c)+b(c-d)+c(a-b)$

a(b - c) + b(c - d) + c(a - b) = ab - ac + bc - bd + ac - bc

= ab - bd

= b (a - d)

From the sum of $4+3x$ and $5-4x+2x^{2}$ , subtract the sum of $3x^{3}-5x$ and $-x^{2}+2x+5$

$(5-4x+2x^2)+(4+3x)=9-x+2x^2$ ....A

$(3x^3-5x)+(-x^2+2x+5)=3x^3-x^2-3x+5$ ....B

A-B gives

$9-x+2x^2-3x^3+x^2+3x-5$

$=-3x^3+3x^2+2x+4$

Subtract: $-5y^2$ from $y^2$ .

Subtract:

$-5y^{2}$ from

$y^{2}$

$\therefore y^{2}-(-5y^{2}) = y^{2}+5y^{2} = 6y^{2}$

Subtract $-6x^{2}+4a^{2}-1$ from $25x^2+16xy-3b^2-2$

$(25x^{2}+16xy-3b^{2}-2)-(-6x^{2}+4a^{2}-1)$

$31x^{2} + 16xy- 3b^{2} - 4a^{2}-1$

What should be added to ${x^2} + \dfrac{1}{{25}}{x^2}{y^2}$ make it a perfect square

$x^2+\dfrac{1}{25}x^2y^2$

$=(x)^2+\left(\dfrac{xy}{5}\right)^2+2(x)\left(\dfrac{xy}{5}\right)$

$=\left(x+\dfrac{xy}{5}\right)^2$

So after adding

$\dfrac{2x^2y}{5}$

We get perfect square.

If $P = 7 x ^ { 2 } + 5 x y - 9 y ^ { 2 } , Q = 4 y ^ { 2 } - 3 x ^ { 2 }$ and $R = 4 x ^ { 2 } + x y + 5 y ^ { 2 }$ show that $P + Q + R = 8x^2+6xy$

We have,

$P=7x^2+5xy-9y^2$

$Q=4y^2-3x^2$

$R=4x^2+xy+5y^2$

Since,

$=P+Q+R$

$=7x^2+5xy-9y^2+4y^2-3x^2+4x^2+xy+5y^2$

$=8x^2+6xy$

Hence, proved.

Simplify: $1\dfrac{3}{5}+5\dfrac{4}{7}$

1258000_1309174_ans_9f3863ae303b4714aea3d12673f4d84c.jpeg

Solve:-
$(5x^2+2y)+(7x^2-y)$

We have,

$(5x^2+2y)+(7x^2-y)$

$=5x^2+2y+7x^2-y$

$=12x^2+y$

Hence, this is the answer.

Add: $3x$ and $7x$

add 3x & 7xFor adding like terms, add the the coefficients of the terms for both the terms, i.e...

$\Rightarrow 3x+7x$

$= 10 x$

$2x^2+4xy-3x$ subtracted from $3x - xy + 8{y^2}$

According to the question,

$3x-xy+8y^2-(2x^2+4xy-3x)$

$\Rightarrow 3x-xy+8y^2-2x^2-4xy+3x$

$\Rightarrow 6x-5xy+8y^2-2x^2$

Hence, this is the answer.

Simplify combining like terms:
$p-(p-q)-q-(q-p)$

p-(p-q)-q-(q-p)

p-p+q-q-q+p

p-q

1211787_1363013_ans_b5630243d8ff492ca383af30a9e26447.JPG

Subtract $8a^{2}b^{2}$ from $3a^{2}b^{2}$

$3a^{2}b^{2}-8a^{2}b^{2}$

$=-5a^{2}b^{2}$

1211695_1362901_ans_33c8467137174141801b86f52a93f83d.JPG

Simplify combining like terms:
$(3y^{2}+5y-4)-(8y-y^{2}-4)$

$3y^{2}+5y-4-(8y-y^{2}-4)$

$\Rightarrow 3y^{2}+5y-4-8y+y^{2}+4$

$\Rightarrow 3y^{2}+y^{2}+5y-8y-4+4$

$\Rightarrow 4y^{2}-3y$

$\Rightarrow y[4y-3]$

1211792_1363025_ans_8103878acd41423b868730957ecd6b06.JPG

Simplify :
$(a-3b+2c)-(a+3b-2c)$

$(a-3b+2c)-(a+3b-2c)$ =

$a-3b+2c-a-3b+2c$ =

$-6b+4c$

Find the value of -
$({ x }^{ 2 }-3x+5)+(2{ x }^{ 2 }-x-2)-(3{ x }^{ 2 }+7x-3)$

$(x^{2}-3x+5) +(2x^{2}-x-2)-(3x^{2}+7x-3)$ =

$-11x+6$ .

Add:
$5 x ^ { 2 } y , - 7 x ^ { 2 } y , 9 x ^ { 2 } y$

We have,

$5x^2y, -7x^2y, 9x^2y$

$\Rightarrow 5x^2y-7x^2y+9x^2y$

$\Rightarrow 7x^2y$

Hence, this is the answer.

Add the product: $x ( x + y - r ) , y ( x - y + r ) , z ( x - y - z )$

1324308_1321900_ans_e9d67edcac33469e92fba13fca2c2d11.jpg

Subtract $4abc$ from $-6abc$

From the question,

$-6abc-4abc=-10abc$

Simplify: $12m-\left(-7m\right)$

$12m-\left(-7m\right)$

$=12m+7m$

$=19m$

simplify :-
$3(x+y)+3(x-y)$

$3(x+y)+3(x-y) = 3x+3y+3x-3y = 6x$

Find the sum of ${x}^{2}+2xy+3{y}^{2},\,\,3{z}^{2}+2yz+{y}^{2},\,\,{x}^{2}+3{z}^{2}+2xz,\,\,{z}^{2}-3xy-3yz$

given,

${x}^{2}+2xy+3{y}^{2}+3{z}^{2}+2yz+{y}^{2}+{x}^{2}+3{z}^{2}+2xz+{z}^{2}-3xy-3yz$

Grouping the like terms,

$=\left({x}^{2}+{x}^{2}\right)+\left(2xy-3xy\right)+\left(3{y}^{2}+{y}^{2}\right)+\left(3{z}^{2}+3{z}^{2}+{z}^{2}\right)+\left(2yz-3yz\right)+\left(+2xz\right)$

$=2{x}^{2}-xy+4{y}^{2}+7{z}^{2}-yz+2xz$

$=2{x}^{2}+4{y}^{2}+7{z}^{2}-xy-yz+2xz$

Solve: $11\left(5-4x\right)=7\left(5-6x\right)$

given,

$11\left(5-4x\right)=7\left(5-6x\right)$

$\Rightarrow \,55-44x=35-42x$

$\Rightarrow \,55-35=-42x+44x$

$\Rightarrow \,20=2x$

$\Rightarrow \,\dfrac{20}{2}=\dfrac{2x}{2}$

$\therefore\,x=10$

In the expression given below identify all the terms.
$5x-7$

Terms in expression are quantities separated by

$+1 -$

$\Rightarrow$ Terms in given expression

$= 5x, 1$

1226941_1398346_ans_f95fd2adb54d48b5b3d1c68c6e4f2aa1.jpg

Simplify
$\left( { 7x }^{ 2 }-2xy-{ 6y }^{ 2 } \right) +\left( { 9y }^{ 2 }+12xy-{ 12x }^{ 2 } \right) -\left( 8xy+{ 2y }^{ 2 }-{ 3x }^{ 2 } \right)$

$7x^2-2xy-6y^2+9y^2+12xy-12x^2-8xy-2y^2+3x^2$

$\Rightarrow -2x^2+y^2+2xy$ .

1209690_1463900_ans_b48863f63c5f4b19bce26615961a522a.jpg

Subtract the second expression from the first expression
$2a + b, a + b$

Given First expression is

$2 a+b$

Second expression is

$a+b$

Subtracting second expression from first one will give

$2 a+b-a-b=a$

Simplify :
$x^{3}(a-2b)+x^{2}(a-2b)$

$x^{3}(a-2b)+x^{2}(a-2b)$ =

$ax^{3}-2bx^{3} +ax^{2}-2bx^{2}$

Simplify :
$a(a+b)+8a+8b$

$a(a+b) + 8a+ 8b$ =

$a^{2} + ab + 8a +8b$

Solve $(4{ x }^{ 4 }-5{ x }^{ 3 }-7x+1)+(4x-1)$ .

$(4x^{4}-5x^{3}-7x+1) + (4x-1)$ =

$4x^{4}-5x^{3}-3x$

Simplify: $12x^3-3(2x^3+4x-1)-5x+7$

First expand the term

$-3(2x^3+4x-1)$

$12x^3 - 3(2x^3 + 4x -1) - 5x + 7 = 12x^3 - 6 x^3 - 12 x + 3 - 5x + 7$
Group like terms

$=6x^3-17x+10$

$\text{Subtract } a(b - 5) \text{ from } b(5 - a).$

Given expressions are:

$a(b-5)\quad ........(i)$

and

$b(5-a)\quad ....(ii)$

Subtracting

$(i)$ from

$(ii)$ , we get:

$b(5-a) - a(b-5)$

$\Rightarrow 5b - ab - ab + 5a = 5a+5b - 2ab$

$\Rightarrow 5(a+b)-2ab$

Hence,

$b(5-a)-a(b-5)=5(a+b)-2ab$

The sum of three expression is $8 + 73x+{ 7x }^{ 2 }$ and two of then are ${ 2x }^{ 2 }+9x+2$ and ${ 3x }^{ 2 }-4x+1$

$2x^2+9x+2+3x^2-4x+1+p(x)=8+73x+7x^2$

$5x^2+5x+3+p(x)=7x^2+73x+8$

$p(x)=7x^2+73x+8-5x^2-5x-3$

$p(x)=2x^2+68x+5$ .

1209748_1461162_ans_2a25ab3cb9f448c4a4f336bde0e8ef65.jpg

Subtract the second expression form the first expression
$2a+b,a+b$

Given First expression is

$2 a+b$

Second expression is

$a+b$

Subtracting second expression from first one will give

$2 a+b-a-b=a$

What is the sum of $-2 m, +2 m ,-8 m$ .

1320195_1542574_ans_ae780be0e6a6449186edfec3251de55d.jpeg

By how much does 3x-4xy+2z exceed 8x+5z-7xy?

$3x-4xy+2z-8x-5z+7xy-5x+3xy-3z$

$\therefore$ It exceeds by

$-5x+3xy-3z$ .

1227046_1501955_ans_48cd6a7d512841c68db80a0ceba2a79e.jpeg

Simplify combining like terms.
$3a-2b-ab-(a-b+ab)+3ab+b-a$ .

$3a-2b-ab-(a-b+ab)+3ab+b-a=3a-2b-ab-a+b-ab+3ab+b-a$

$=3a-a-a-2b+b-ab-ab+3ab$

$=(3a-a-a)-(2b-b-b)-(ab+ab-3ab)$

$=a-0-(-ab)$

$=a+ab$ .

Simplify combining like terms:
$-z^2+13z^2-5x+7z^2-15z$ .

$-z^2+13z^2-5z+7z^2-15z=7z^2+(-z^2+13z^2)-(5z+15z)$

$=7z^2+12z^2-20z$ .

Simplify combining like terms.
$21b-32+7b-20b$ .

In the given expression there are only two type of terms one with variable

$b$ and other is constant,

$\begin{aligned}{}21b - 32 + 7b - 20b &= 21b + 7b - 20b - 32\\ &= \left( {21 + 7 - 20} \right) - 32\\& = \left( {28 - 20} \right)b - 32\\ &= 8b - 32\end{aligned}$

Simplify:
$(24m^{4} - 15 m^{3} +10 m +8) + (5m^{3}+4)$

1499434_1489775_ans_00629bf050374004ac43951dba16da96.jpg

Simplify combining like terms:
21 b - 32 + 7 b - 20 b

1318614_1490646_ans_392e9a58ee754439b7c6936f1523bf5b.JPG

Simplify combining like terms.
$p-(p-q)-q-(q-p)$ .

$p-(p-q)-q-(q-p)=p-p+q-q-q+p$

$=p-p+p+q-q-q$

$=p-q$ .

Solve:
(5p-8q)-(3p-7q)=?

1326048_1566040_ans_e42529070c7f44cc87480a453ff110d1.jpeg

Subtract:
$\begin{array} { l } { \text { (i) } - 5 y ^ { 2 } \text { from } y ^ { 2 } } \\ { \text { (ii) } 6 x y \text { from } - 12 x y } \\ { \text { (ii) } ( a - b ) \text { from } ( a + b ) } \\ { \text { (iv) } a ( b - 5 ) \text { from } b ( 5 - a ) } \end{array}$

1318645_1490391_ans_502d8df31f6742f3923b9abc28f5d9f5.JPG

Add $3mn, -5mn, 8mn-4mn$ .

$3mn, -5mn, 8mn, -4mn=3mn+(-5mn)+8mn+(-4mn)$

$=(3-5+8-4)mn=2mn$ .

Subtract:
$-8pq$ from $6pq$

We have to subtract

$-8pq$ from

$6pq$ .

Both terms have same variable

$pq$ .

So, simply their numeric coefficients will be subtracted.

$\therefore 6pq-(-8pq)\\$

$= (6+8)pq\\$

$=14pq$

Add $t-8tz, 3tz-z, z-t$ .

$t-8tz, 3tz-z, z-t=t-8tz+3tz-z+z-t$

$=t-t-8tz+3tz-z+z$

$=(1-1)t+(-8+3)tz+(-1+1)z$

$=0-5tz+0=-5tz$ .

Add $3p^2q^2-4pq+5, -10p^2q^2, 15+9pq+7p^2q^2$ .

$3p^2q^2-4pq+5, -10p^2q^2, 15+9pq+7p^2q^2=3p^2q^2-4pq+5+(-10p^2q^2)+15+9pq+7p^2q^2$

$=3p^2q^2-10p^2q^2+7p^2q^2+4pq+9pq+5+15$

$=(3-10+7)p^2q^2+(-4+9)pq+20$

$=0p^2q^2+5pq+20=5pq+20$ .

What should be added to $3x^2-4y^2+5xy+20$ to obtain $-x^2-y^2+6xy+20$ ?

Let q should be subtracted.
Then according to question,

$3x^2-4y^2+5xy+20-q=-x^2-y^2+6xy+20$

$\Rightarrow q=3x^2-4y^2+5xy+20-(-x^2-y^2+6xy+20)$

$\Rightarrow q=3x^2-4y^2+5xy+20+x^2+y^2-6xy-20$

$\Rightarrow q=3x^2+x^2-4y^2+y^2+5xy-6xy+20-20$

$\Rightarrow q=4x^2-3y^2-xy+0$
Hence,

$4x^2-3y^2-xy$ should be subtracted.

Add $ab-4a, 4b-ab, 4a-4b$ .

$ab-4a, 4b-ab, 4a-ab=ab-4a+4b-ab+4a-ab$

$=-4a+4a+4b-4b+ab-ab$

$=0+0+0=0$ .

Subtract:
$3a^2b$ from $-5a^2b$

We have to subtract

$3a^2b$ from

$-5a^2b.$

Both terms have the same set of variables

$a^2b$ .

So, simply their numeric coefficients will be subtracted.

$-5a^2b-3a^2b= (-5-3)a^2b$

$= -8a^2b\\$

Subtract $-5y^2$ from $y^2$ .

$y^2-(-5y^2)=y^2+5y^2=6y^2$ .

From the sum of $3x-y+11$ and $-y-11$ , subtract the sum of $3x^2-5x$ and $-x^2+2x+5$ .

According to question,

$=[(3x-y+11)+(-y-11)]-[(3x^2-5x)+(-x^2+2x+5)]$

$=[3x-y+11-y-11]-[3x^2-5x-x^2+2x+5]$

$=[3x-2y]-[2x^2-3x+5]$

$=3x-2y-2x^2+3x-5$

$=-2x^2+6x-2y-5$

Subtract $(a-b)$ from $(a+b)$ .

$(a+b)-(a-b)=a+b-a+b=a-a+b+b=2b$ .

Add
$2x^2,\ -3x^2,\ 7x^2$

The required sum =

$2x^2 + (-3x^2) + 7x^2$

$=2x^2 - 3x^2 + 7x^2$

$=(2 - 3 + 7) x^2$

$=6x^2$

Subtract:
$-2abc$ from $-8bac$

We have to subtract

$-2abc$ from

$-8abc$

Both terms have same variable

$abc$ .

So, simply their numeric coefficients will be subtracted.

$\therefore -8abc-(-2abc)\\$

$= (-8-(-2))abc\\$

$= (-8+2)abc\\$

$=-6abc$

Subtract :
$5x$ from $2x$

The required difference

$=2x -5x$

$= (2 - 5) x$

$=-3x$

Add
$7xyz,\ -5xyz,\ 9xyz,\ -8xyz$

The required sum =

$7xyz + (-5xyz) + 9xyz + (-8xyz)$

$= 7xyz – 5xyz + 9xyz – 8xyz$

$= (7 – 5 + 9 – 8) xyz$

$= (16 – 13) xyz$

$= 3xyz$

Add
$7y$ , $-9y$

The required sum

$= 7y + (-9y)$

$= 7y\ –\ 9y$

$= (7 -9) y$

$= -2y$

Subtract:
$-16p$ from $-11p$

We have to subtract

$-16p$ from

$-11p$

$\because$ their variables are same, so simply numeric coefficient will be subtracted.

$-11p-(-16p)$

$=-11p+16p$

$=(-11+16)p$

$=5p$

Hence, the required answer is

$5p$ .

Subtract:
$xy$ from $6xy$

The required difference

$=6xy-(-xy)$

$=6xy + xy = (6 + 1) xy$

$=7xy$

Subtract
$7x$ from $9y$

The required difference

$=9y-(-7x)$

$= (9y + 7x)$

Add
$3x$ , $7x$

The required sum

$= 3x + 7x$

$= (3 + 7) x$

$= 10x$

Add
$2xy,\ 5xy,\ -xy$

The required sum =

$2xy + 5xy + (-xy)$

$= 2xy + 5xy – xy$

$= (2x + 5x –x) y$

$= 6xy$

Add the following expressions:
$(3/5)x, (2/3)x, (-4/5)x$

We add the like terms:

$=(3/5)x+(2/3)x+(-4/5)x$
Now, L.C.M of

$5,3,5=15$
So,

$=(9x+10x-12x)/15$

$=(7/15)x$

Subtract:
$10x^2$ from $-7x^2$

The required difference

$=-7x^2-10x^2$

$= (-7-10) x^2$

$= - 17x^2$

Simplify
$a - (b -2a)$

Here,

$(-)$ sign precedes the second parenthesis, so we remove it and change the sign of each term within

$\therefore$

$a – (b – 2a)$

$= a – b + 2a$

$= 3a – b$

Subtract :
$5a + 7b - 2c$ from $3a -7b + 4c$

The required difference

$= (3a – 7b + 4c) - ( 5a + 7b - 2c)$

$= 3a – 5a – 7b – 7b + 4c + 2c$

$= -2a – 14b + 6c$

How much is 3p-4q+r less than 4p+3q-5r?

Required expression

$(4p+3q-5r)-(3p-4q+r)$

$=4p+3q-5r-3p+4q-r$

$=p+7q-6r$

Using the horizontal method:
Subtract $-x^2+y^2+2xy$ from $2x^2-3y^2$

$(2x^2-3y^2)-(-x^2+y^2+2xy)$

$=2x^2-3y^2-x^2-y^2-2xy$

$=3x^2-4y^2-2xy$

Subtract:
$- 8 x - 12 y + 17 z$ from $x - y - z$

$- 8 x - 12 y + 17 z$ from

$x - y - z$
The value of the subtraction is calculated as follows

$(x - y - z) - (- 8 x - 12 y + 17 z)$

$= x + 8 x + 12 y - y - z - 17 z$
We get,

$= 9 x + 11 y - 18 z$

Simplify the following by combining the like terms and then write whether the expression is a monomial, a binomial or a trinomial.
$2a+2b+2c-2a-2b-2c-2b+2c+2a$

$\textbf{Step 1:Combine the like terms and simplify}$

$\begin{aligned}&2 a+2 b+2 c-2 a-2 b-2 c-2 b+2 c+2 a \\&\quad=(2 a-2 a+2 a)+(2 b-2 b-2 b)+(2 c-2 c+2 c) \\&2 a+2 b+2 c-2 a-2 b-2 c-2 b+2 c+2 a=2 a-2 b+2 c\end{aligned}$

$\textbf{Step 2:Identify the number of terms in the expression}$

$2a$ ,

$-2b$ and

$2c$ are the three unlike terms in the simplified expression.
So, the given expression is a trinomial.

$\textbf{Hence, } 2a+2b+2c-2a-2b-2c-2b+2c+2a \textbf{ will result in a trinomial.}$

Subtract :
$a^2 - b^2$ from $b^2 - a^2$

The required difference

$= (b^2- a^2)-(a^2 - b^2)$

$=(b^2 - a^2 - a^2 + b^2)$

$=2b^2 - 2a^2$

Add: $5x^2,-9x^2$

$5x^2+(-9x^2)$

$=5x^2-9x^2$

$=-4x^2$

Add: $3ab^2,-5ab^2$

$3ab +(-5ab^2)$

$=3ab^2-5ab^2$

$=-2ab^2$

Add: $7x,-3x$

$7x+(-3x)$

$=7x-3x$

$=4x$

Verify the following: $(a + b + c) (a^{2} + b^{2} +c^{2} - ab- bc - ca) = a^{2} + b^{2}+ c^{2} - 3abc$

LHS-

$(a + b + c) (a^{2} + b^{2} +c^{2} - ab- bc - ca)$

$\implies a(a^{2}+b^{2}+c^{2}-ab-bc-ca)+b(a^{2}+b^{2}+c^{2}-ab-bc-ca)+c(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

=

$a^{3}+ab^{2}+ac^{2}-a^{2}b-abc-a^{2}c+ba^{2}+b^{3}+bc^{2}-b^{2}a-b^{2}c-bca+ca^{2}+cb^{2}+c^{3}-cab-c^{2}b-c^{2}a$

=

$a^{2}+b^{2}+c^{2}-3abc$

Subtract: $5a^{2}b^{2}c^{2}$ from $-7a^{2}b^{2}c^{2}$

To find:

$-7a^{2}b^{2}c^{2}-5a^{2}b^{2}c^{2}$

$= -(7+5)a^{2}b^{2}c^{2}$

$= -(12)a^{2}b^{2}c^{2}$

Add: $\dfrac{1}{2}pq,\dfrac{-1}{3}pq$

$\dfrac{1}{2}pq+(-\dfrac{1}{3}pq)$

$=\dfrac{1}{2}pq-\dfrac{1}{3}pq$

$=\dfrac{3pq-2pq}{6}$

$=\dfrac{pq}{6}$

$=\dfrac{1}{6}pq$

Add: $6x,-11 x$

$6x+(-11x)$

$=6x-11x$

$=-5x$

Subtract: $7xy+5x^2-7y^2+3$ from $7x^2-8xy+3y^2-5$

$7xy+5x^2-7y^2+3$ from

$7x^2-8xy+3y^2-5$

$=(7x^2+3y^2-8xy-5)-(7xy+5x^2-7y^2+3)$

$=7x^2+3y^2-8xy-5-7xy-5x^2-7y^2-5-3$

$=2x^2+10y^2-15xy-8$

Add: $5x^3y,\dfrac{-2}{3}x^3y$

$5x^3y+(-\dfrac{2}{3}x^3y)$

$=5x^3y-\dfrac{2}{3}x^3y$

$=\dfrac{15x^3y-2x^3y}{3}$

$=\dfrac{13x^3y}{3}$

$=\dfrac{13}{3}x^3y$

Add: $2x^3,3x^3,-4x^3,-5x^3$

$2x^3,3x^3,-4x^3,-5x^3$

$2x^3+3x^3-4x^3-5x^3$

$(2+3-4-5)x^3$

$=(5-9)x^3$

$=-4x^3$

Find the sum of the following algebraic expressions: $5xy,-7xy,3x^2$

$5xy,-7xy,3x^2$

$=5xy-7xy+3x^2$

$=3x^2-2xy$

Add: $7xy,2xy,-8xy$

$7xy,2xy,-8xy$

$=7xy+2xy-8xy$

$=(7+2-8)xy$

$=(9-8)xy$

$=xy$

Add: $3x,-5x,7x$

$3x,-5x,7x$

$=3x-5x+7x$

$=(3-5+7)x$

$(10-5)x$

$=5x$

Subtract: $-m^2+5mn$ from $4m^2-3mn+8$

$-m^2+5mn$ from

$4m^2-3mn+8$

$=4m^2-3mn+8-(-m^2+5mn)$

$=4m^2-3mn+8+m^2-5mn$

$=5m^2-8mn+8$

Add: $-2abc,3abc,abc$

Given:

$-2abc,3abc,abc$
To find

$=-2abc+3abc+abc$

$=(-2+3+1)abc$

$=(4-2)abc$

$=2abc$

Subtract:
$7xy(x^2 - 2xy + 3y^2) - 8x(x^2y - 4xy + 7xy^2)$ from $3y(4x^2y - 5xy + 8xy^2)$

Subtracting,

$7xy(x^2 - 2xy + 3y^2) - 8x(x^2y - 4xy + 7xy^2)$ from

$3y(4x^2y - 5xy + 8xy^2)$

$\Rightarrow \ \ 7x^3y - 14x^2y^2 + 21xy^3 - 8x^3y + 32x^2y - 56x^2y^2$ from

$12x^2y^2 - 15xy^2 + 24xy^3$
=

$(12x^2y^2 - 15xy^2 + 24xy^3) - (7x^3y - 14x^2y^2 + 21xy^3 - 8x^3y + 32x^2y - 56x^2y^2$
=

$12x^2y^2 - 15xy^2 + 24xy^3 - 7x^3y + 14x^2y^2 - 12xy^3 + 8x^3y - 32x^2y + 56x^2y^x2$
=

$82x^2y^2 + 3xy^3 + x^3y - 15xy^2 - 32x^2y$

Evaluate: $(23 15) + 4$ .

$(23 – 15) + 4$ ,
The value of the given expression

$(23 – 15) + 4$ is calculated as follows,

$(23 – 15) + 4 = 8 + 4$
We get,

$= 12$ .
Hence, the value of the given expression

$(23 – 15) + 4 = 12$ .

Take $- ab + bc - ca\,\, from\,\, bc - ca + ab$

The value of the subtraction is calculated as,

$(bc - ca + ab) - (-ab + bc - ca)$

$= bc - bc - ca + ca + ab + ab$
We get,

$= 2 ab$
Hence,

$(bc - ca + ab) - (- ab + bc - ca)$

$= 2 ab$

Take $(-3 / 2) p + q - r\,\, from\,\, (1 / 2)p - (1 / 3)q - (3 / 2) r$

The value of the subtraction is calculated as,

$[(1 / 2)p - (1 / 3)q - (3 / 2) r] - [(-3 / 2) p + q - r]$

$= (1 / 2)p + (3 / 2)p - (1 / 3)q - q - (3 / 2)r + r$
We get,

$= [(3p + 9p - 2q - 6q - 9r + 6r) / 6]$
On further calculation, we get

$= 2p - (4 / 3)q - (1 / 2)r$
Hence,

$[(1 / 2)p - (1 / 3)q - (3 / 2) r] - [(-3 / 2) p + q - r] = 2p - (4 / 3)q - (1 / 2)r$

$5 + 4 =$ .. and $5x + 4x =$

$5 + 4 = 9$ and

$5x + 4x = 9x$

Add the following expressions:
$ab - bv, bv - ca, ca - ab$

$ab - bc, bc - ca, ca - ab$
On adding the expressions, we have

$\Rightarrow \ \ ab - bc + bc - ca + ca - ab = 0$

Subtract
(i) $4a-7ab+3b+12$ from $12a-9ab+5b-3$
(ii) $3xy+5yz-7zx$ from $5xy-2yz-2zx+10xyz$
(iii) $4p^2q-3pq+5pq^2-8p+7q-10$ from $18-3p-11q+5pq-2pq^2+5p^2q$

$12a-9ab+5b-3-(4a-7ab+3b+12)\\ =12a-9ab+5b-3-4a+7ab-3b-12\\ =8a+2b-2ab-15$

ii)

$5xy-2yz-2zx+10xyz-(3xy+5yz-7xz)\\ =5xy-2yz-2zx+10xyz-3xy-5yz+7xz\\ =2xy-7yz+5xz$

iii)

$18-3p-11q+5pq-2p{ q }^{ 2 }+5{ p }^{ 2 }q-(4{ p }^{ 2 }q-3pq+5{ q }^{ 2 }p-8p+7q-10)\\ =18-3p-11q+5pq-2p{ q }^{ 2 }+5{ p }^{ 2 }q-4{ p }^{ 2 }q+3pq-5{ q }^{ 2 }p+8p-7q+10\\ =28+5p-18q+8pq-7p{ q }^{ 2 }+{ p }^{ 2 }q$

Add following.
(i) $ab-bc,\,bc-ca,\,ca-ab$
(ii) $a-b+ab,\,b-c+bc,\,c-a+ac$
(iii) $2p^2q^2-3pq+4,\,5+7pq-3p^2q^2$
(iv) $l^2+m^2,\,m^2+n^2,\,n^2+l^2,\,2lm+2mn+2nl$

(i)

Given are 3 terms i.e.,

$=ab-bc+bc-ca+ca-ab$

$=ab-ba+bc-bc+ca-ca$

$=0$

(ii)

Three terms i.e,

$=a-b+ab+b-c+bc+c-a+ac$

$=a-a+b-b+c-c+ab+bc+ca$

$=ab+bc+ac$

(iii)

Adding the given two terms,

$=2{ p }^{ 2 }{ q }^{ 2 }-3pq+4+5+7pq-3{ p }^{ 2 }{ q }^{ 2 }$

$=2p^2 q^2 - 3p^2 q^2 +7pq-3pq+5+4$

$=-{ p }^{ 2 }{ q }^{ 2 }+4pq+9$

(iv)

Addition of the four terms results in

$={ l }^{ 2 }+{ m }^{ 2 }+n^{ 2 }+{ m }^{ 2 }+{ l }^{ 2 }+{ n }^{ 2 }+2lm+2mn+2nl\\=2{ l }^{ 2 }+2{ m }^{ 2 }+2{ n }^{ 2 }+2(lm+mn+nl)\\ = 2({ l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 }+lm+mn+nl)\\$

Identify the terms, their coefficients for each of the following expressions.

(i) $5xyz^2-3zy$

(ii) $1+x+x^2$

(iii) $4x^2y^2-4x^2y^2z^2+z^2$

(iv) $3-pq+qr-rp$

(v) $\dfrac{x}{2}+\dfrac{y}{2}-xy$

(vi) $0.3a-0.6ab+0.5b$

Expression	Terms	Coefficients
$5xyz^2-3zy$	$5xyz^2$ , $-3zy$	$5$ $-3$
$1+x+x^2$	$1$ , $x$ , $x^2$	No coefficient, $1$ , $1$
$4x^2y^2-4x^2y^2z^2+z^2$	$4x^2y^2$ , $-4x^2y^2z^2$ , $z^2$	$4$ , $-4$ , $1$
$3-pq_qr-rp$	$3$ , $-pq$ , $-rp$	No coefficient, $-1$ , $-1$
$\dfrac x2+\dfrac y2-xy$	$\dfrac x2$ $\dfrac y2$ $-xy$	$\dfrac 12$ $\dfrac 12$ $-1$
$0.3a-0.6ab+0.5b$	$0.3a$ $-0.6ab$ $0.5b$	$0.3$ , $-0.6$ , $0.5$

(a) What should be added to $2{ x }^{ 2 }+6x-5$ to get $3{ x }^{ 2 }-2x+6$ ?
(b) What must be subtracted from ${ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }-3abc$ to get $2{ a }^{ 2 }-{ b }^{ 2 }-3{ c }^{ 2 }+abc$

(I). In order to get $(3x^2−2x+6) \; from \; (2x^2+6x−5)$ we need to:

$(3x^2−2x+6)- (2x^2+6x−5)$

$= 3x^2−2x+6- 2x^2-6x+5$

$= x^2-4x+11$

(ii) In order to get $(2a^2−b^2−3c^2+abc) \; from \; (a^2+b^2+c^2−3abc)$ we need to:

$(a^2+b^2+c^2−3abc)- (2a^2−b^2−3c^2+abc)$

$= a^2+b^2+c^2−3abc - 2a^2+b^2+3c^2-abc$

$= a^2+2b^2+4c^2-4abc$

Find the sum of the following:
(a) $P\left( x \right) ={ x }^{ 2 }-2x+1$ and $Q\left( x \right) ={ x }^{ 3 }+3{ x }^{ 2 }+2x-1$
(b) $P\left( t \right) =4{ t }^{ 3 }-3{ t }^{ 2 }+2$ , $Q\left( t \right) ={ t }^{ 4 }-2{ t }^{ 3 }+6$ and $R\left( t \right) ={ t }^{ 3 }+4{ t }^{ 2 }+4$

We need to find sum of the given expressions.

(a) Given,

$P(x)=x^2-2x+1$ and

$Q(x)=x^3+3x^2+2x-1$

Their sum

$=P(x)+Q(x)=x^2-2x+1+x^3+3x^2+2x-1$

$=x^3+4x^2$

(b) Given,

$P(t)=4t^3-3t^2+2, Q(t)=t^4-2t^3+6, R(t)=t^3+4t^2+4$

Their sum

$=P(t)+Q(t)+R(t)={4t^3-3x^2+2}+{t^4-2t^2+6+t^3+4t^2+4=t^4+3t^3+t^2+12}$

(a) Add $p(p-q),\,q(q-r)$ and $r(r-p)$

(b) Add $2x(z-x-y)$ and $27(z-y-x)$

(c) Subtract $3l(l-4m+5n)$ from $4l(10n-3m+2l)$

(d) Subtract $3a(a+b+c)-2b(a-b+c)$ from $4c(-a+b+c)$

$p(p-q)+q(q-r)+r(r-p)=p^2-pq+q^2-qr+r^2-pr$

$={ p }^{ 2 }+{ q }^{ 2 }+{ r }^{ 2 }-pq-qr-rp$

$2x(z-x-y)+27(z-y-x) =2xz-2{ x }^{ 2 }-2xy+27z-27y-27z$

$4l(10n-3m+2l)-3l(l-4m+5n) =40ln -12lm+8{ l }^{ 2 }-3{ l }^{ 2 }+12lm-15ln$

$=25ln+5{ l }^{ 2 }$

$4c(-a+b+c)-[3a(a+b+c)-2b(a-b+c)] =-4ac+4bc+4{ c }^{ 2 }-[3{ a }^{ 2 }+3ab+3ac-2ab+2{ b }^{ 2 }-2bc]$

$=-4ac+4bc+4{ c }^{ 2 }-3{ a }^{ 2 }-ab-3ac-2{ b }^{ 2 }+2bc$

Simplify the polynomial and write it in standard form:
$({x}^{4}-3)(3x-4)+2{x}^{3}(2{x}^{2}-2)$

$\left( { x }^{ 4 }-3 \right) \left( 3x-4 \right) +2{ x }^{ 3 }\left( 2{ x }^{ 2 }-2 \right)$

$={ 3x }^{ 5 }-4{ x }^{ 4 }-9x+12+4{ x }^{ 5 }-4{ x }^{ 3 }$

$={ 3x }^{ 5 }+4{ x }^{ 5 }-4{ x }^{ 4 }-4{ x }^{ 3 }-9x+12$

$={ 7x }^{ 5 }-4{ x }^{ 4 }-4{ x }^{ 3 }-9x+12$

Ans

$={ 7x }^{ 5 }-4{ x }^{ 4 }-4{ x }^{ 3 }-9x+12$

Subtract $2x^3- x^2 + 4x- 6$ from $x^3 + 5x^2 -4x + 6$ .

$(x^3 + 5x^2 -4x + 6)- (2x^3 -x^2 + 4x -6)$

$= x^3 + 5x^2 -4x + 6- 2x^3 + x^2 -4x + 6$

$= (1- 2)x^3 + (5 + 1)x^2 + (-4 -4)x + (6 + 6)$

$= -1x^3 + 6x^2 -8x + 12$

$=- x^3 + 6x^2 -8x + 12$

Hence, $$(x^3 + 5x^2 -4x + 6)- (2x^3 -x^2 + 4x -6)=- x^3 + 6x^2 -8x + 12$$.

The perimeter of a triangle is $15x^2 - 23x + 9$ and two of its sides are $5x^2+8x-1$ and $6x^2-9x+4$ . Find the third side

Formula For Perimeter of triangle =

$(side)_1+(side)_2+(side)_3$
Using formula We have :-

$(15x^2-23x+9)$ =

$5x^2+8x-1 + 6x^2-9x+4 +(side)_3$

$\Rightarrow$

$(15x^2-23x+9)$ =

$11x^2-x+3+(side)_3$

$\Rightarrow$

$(side)_3 = 15x^2-23x+9-11x^2+x-3$

$\Rightarrow$

$(side)_3$ =

$4x^2-22x+6$
The answer is

${4x^2-22x+6}$

Subtract $2x^{2} + 2y^{2} - 6$ from $3x^{2} - 7y^{2} + 9$

Given expressions are

$2x^{2} + 2y^{2} - 6$ and

$3x^{2} - 7y^{2} + 9$ .

To find out,

The result when we subtract

$2x^{2} + 2y^{2} - 6$ from

$3x^{2} - 7y^{2} + 9$ .

On subtracting, we get:

$(3x^{2} - 7y^{2} + 9)-(2x^{2} + 2y^{2} - 6)$

$=3x^{2} - 7y^{2} + 9-2x^{2} - 2y^{2} + 6$

On combining the like terms, we get:

$3x^{2}-2x^{2} - 7y^{2}- 2y^{2} +9+ 6$

$=x^{2} - 9y^{2}+15$

Hence, the result when we subtract

$2x^{2} + 2y^{2} - 6$ from

$3x^{2} - 7y^{2} + 9$ is

$x^{2} - 9y^{2}+15$ .

If $A=4x^2+y^2-6xy$ ;
$B=3y^2+12x^2+8xy$ ;
$C=6x^2+8y^2+6xy$

Find the (i) $A+B+C$ (ii) $(A-B)-C$

(i)

$22x^2+12y^2+8xy$

(ii)

$-14x^2-10y^2-20xy$ or

$-(14x^2+10y^2+20 xy)$

Add the product: $x(x+y-r), y(x-y+r), z(x-y-z)$

${\textbf{Step - 1: Simplify the given terms by multiplication}}{\text{.}}$

${\text{So}}{\text{, }}x \times \left( {x + y - r} \right) = \left( {x \times x} \right) + \left( {x \times y} \right) - \left( {x \times r} \right) = {x^2} + xy - xr.$

$y \times \left( {x + y - r} \right) = y \times x + y \times y - y \times r = xy + {y^2} - ry.$

$z \times \left( {x - y - z} \right) = z \times x - z \times y - z \times z = xz - yz - {z^2}.$

${\textbf{Step - 2: Add those terms}}{\text{.}}$

$\therefore x\left( {x + y - r} \right) + y\left( {x - y + r} \right) + z\left( {x - y - z} \right).$

$= {x^2} + xy - rx + xy + {y^2} - ry + xz - yz - {z^2}.$

$= {x^2} + {y^2} - {z^2} + 2xy - r\left( {x + y} \right) + z\left( {x - y} \right).$

${\textbf{Hence, After adding the product will be = }}\mathbf{{x^2} + {y^2} - {z^2} + 2xy - r\left( {x + y} \right) + z\left( {x - y} \right)}.$

Subtract $2xy+9{x}^{2}$ from $12xy+4{x}^{2}-3{y}^{2}$ .

Given Two expression;

(1)

$2xy+9x^2$

(2)

$12xy+4x^2-3y^2$

To find = subtraction of 1st expression from 2nd expression

$\Rightarrow$

$12xy+4x^2-3y^2$

$-$ (

$2xy+9x^2$ )

$\Rightarrow$

$12xy+4x^2-3y^2$

$-$

$2xy-9x^2$

$\Rightarrow$

$12xy-2xy+4x^2-9x^2-3y^2$

$\Rightarrow$

$10xy-5x^2-3y^2$

The additive inverse of $x+\displaystyle\frac{1}{x}$ will be ________.

Given expression is

$x+\dfrac{1}{x}$ .

We know that, additive inverse means an addition of a term or expression to the given expression, such that their sum equals to zero.

Let

$P$ be such an expression.

Hence,

$x + \cfrac 1x +P = 0$

$\Rightarrow P = -x-\displaystyle\frac{1}{x}$ .

Hence, the additive inverse of

$x+\dfrac{1}{x}$ is

$-x-\dfrac{1}{x}$ .

Subtract $3a(a - 2b + 3c)$ from $4a(5a + 2b - 3c)$

Subtraction of

$3a\left( a-2b+3c \right)$ from

$4a\left( 5a+2b-3c \right)$

$\Rightarrow 4a\left( 5a+2b-3c \right)=20a^2+8ab-12ac$

$\Rightarrow 3a\left( a-2b+3c \right)=3a^2-6ab+9ac$

Thus,

$20a^2+8ab-12ac-\left( 3a^2+9ac-6ab \right)$

$=20a^2+8ab-12ac-3a^2-9ac+6ab$

$=17a^2+14ab-21ac$

Hence, the answer is

$17a^2+14ab-21ac.$

Prove that :
$\large{\frac{1}{1+x^{a-b}}}+\large{\frac{1}{1+x^{b-a}}}=1$

Now,

$\dfrac{1}{1+x^{a-b}}+\dfrac{1}{1+x^{b-a}}$

$=\dfrac{x^b}{x^b+x^{a}}+\dfrac{x^a}{x^a+x^{b}}$

$=\dfrac{x^a+x^b}{x^a+x^b}$

$=1$ .

solve: $ab\left( {{x^2} + 1} \right) + x\left( {{a^2} + {b^2}} \right)$

$ab\left( { x }^{ 2 }+1 \right) +x\left( a^2+b^2 \right)$

$\Rightarrow abx^2+ab+xa^2+xb^2$

$abx^2+a^2x+bx^2+ab$

$ax\left( xb+a \right) +b\left( bx+a \right)$

$\left( ax+b \right) \left( xb+a \right).$

Hence, the answer is

$\left( ax+b \right) \left( xb+a \right).$

Simplify: $\{(3x^2-6x+5)+(2x-2x^2+5)\}-(x^2-4x+10)$

Let us solve the given expression

$[(3x^2−6x+5)+(2x−2x^2+5)]−(x^2−4x+10)$ as shown below:

$[(3x^{ 2 }−6x+5)+(2x−2x^{ 2 }+5)]−(x^{ 2 }−4x+10)\\ =(3x^{ 2 }−6x+5+2x−2x^{ 2 }+5)−(x^{ 2 }−4x+10)\\ =[(3x^{ 2 }−2x^{ 2 })+(−6x+2x)+(5+5)]−(x^{ 2 }−4x+10)\quad \quad \quad \quad (Combining\quad like\quad terms)\\ =[(3−2)x^{ 2 }+(−6+2)x+10]−(x^{ 2 }−4x+10)\\ =(x^{ 2 }-4x+10)−(x^{ 2 }−4x+10)\\ =0$

Hence,

$[(3x^2−6x+5)+(2x−2x^2+5)]−(x^2−4x+10)=0$ .

Subtract the following:
(1) $16x^3$ from $15x^3 + 3x^2 - 25$
(2) $5m^2 + 3mn + n^2$ from $m^2 + 6mn - 7n^2$

(1)

$15x^3+3x^2-25-16x^3=$

$-x^3 +3x^2 -25$
(2)

$m^2+6mn-7n^2-5m^2-3mn-n^2=$

$-4m^2 + 3mn - 8n^2$

Add the following:
(1) $a - b + ab, b - c + bc, c - a + ac$
(2) $2lm + 2mn, -7lm + 2nl$

(1)

$a-b+ab+b-c+bc+c-a+ac=$

$ab+bc+ac$
(2)

$2lm+2mn-7lm+2nl=$

$-5lm+2mn+2nl$

simplify $(\cfrac { 1 }{ 3 } { y }^{ 2 }-\cfrac { 4 }{ 7 } y+5)-(\cfrac { 2 }{ 7 } { y }-\cfrac { 2 }{ 3 } { y }^{ 2 }+2)-(\cfrac { 1 }{ 7 } { y }-3+2{ y }^{ 2 })$

We have,

$(\dfrac{1}{3}y^2-\dfrac{4}{7}y+5)-(\dfrac{2}{7}y-\dfrac{2}{3}y^2+2)-(\dfrac{1}{7}y-$

$3+2y^2)$

$=(\dfrac{1}{3}+\dfrac{2}{3}-2)y^2-(\dfrac{4}{7}+\dfrac{2}{7}+\dfrac{1}{7})y+5-2+3$

$=-y^2-{1}+6$

This is the required solution.

ExampleAdd $(2{x}^{2}+6y+3xy),(3{x}^{2}-5xy-x)$ and $(6xy+5)$ .

$(2x^2+6y+3xy)+(3x^2-5xy-x)+(6xy+5)$

$(2x^2+3x^2)+(3xy-5xy+6xy)+6y-x+5$

$5x^2+4xy-x+6y+5$

Add $(a + 1) (a^2 - a + 1) \, and \, (a - 1) (a^2 + a + 1)$

$\textbf{Step 1: Open the brackets and add the alike terms.}$

$\text{(a+1) (a$^2$ - a+1) = }$

$\text{ a$^3$ - a$^2$ + a + a$^2$ - a+1}$

$\Rightarrow \text{(a+1) (a$^2$ - a+1) = }$

$\text{ a$^3$ + 1}$

$\text{( a - 1) (a$^2$+a+1) = }$

$\text{a$^3$ + a$^2$ +a - a$^2$ - a - 1}$

$\Rightarrow \text{( a - 1) (a$^2$+a+1) = }$

$\text{a$^3$ - 1}$

$\text{Adding both terms}$

$\text{(a + b) (a$^2$ - a+1) + (a - 1) (a$^2$ + a+1)}$

$\text{= a$^3$ + 1 + a$^3$ - 1}$

$\text{= a$^3$ + a$^3$}$

$\text{= 2a$^3$}$

$\textbf{Hence the sum is 2a$^3$ }$

$\sqrt [ 3 ]{ \left( x+1 \right) \left( y+2 \right) \left( z+3 \right) } =7$ the find $xyz$

$\sqrt [ 3 ]{ \left( x+1 \right) \left( y+2 \right) \left( z+3 \right) } =7$

Cubing both sides, we get

$\left( x+1 \right) \left( y+2 \right) \left( z+3 \right)={7}^{3}=343$

$\Rightarrow \left[x\left(y+2\right)+1\left(y+2\right)\right]\left( z+3 \right)=343$

$\Rightarrow \left[xy+2x+y+2\right]\left( z+3 \right)=343$

$\Rightarrow xy\left( z+3 \right)+2x\left( z+3 \right)+y\left( z+3 \right)+2\left( z+3 \right)=343$

$\Rightarrow xyz+3xy+2xz+6x+yz+3y+2z+6=343$

$\Rightarrow xyz=343-\left(3xy+2xz+6x+yz+3y+2z+6\right)$

$\Rightarrow xyz=343-3xy-2xz-6x-yz-3y-2z-6$

$\therefore xyz=-6x-3y-2z-3xy-2xz-yz+337$

What should be added to $6x +y$ to get $-6x + 4y$ ?

Let 'P' should be added to

$6x+y$ to get

$-6x+4y$

I.e.,

$6x+y+P= -6x+4y$

$P= -6x +4y -6x -y$

$P=-12x+3y$

The required expression is

$-12x+3y$

Subtract $\dfrac{7}{4}y^3+\dfrac{3}{5}y^2+\dfrac{1}{2}y+\dfrac{7}{2}$ from $\dfrac{9}{2}-\dfrac{y}{3}-\dfrac{y^2}{5}$ .

According to the question,

$\Rightarrow \dfrac{9}{2}-\dfrac{y}{3}-\dfrac{y^2}{5}$

$- \left(\dfrac{7}{4}y^3+\dfrac{3}{5}y^2+\dfrac{1}{2}y+\dfrac{7}{2}\right)$

$\Rightarrow \dfrac{9}{2}-\dfrac{y}{3}-\dfrac{y^2}{5}$

$- \dfrac{7}{4}y^3-\dfrac{3}{5}y^2-\dfrac{1}{2}y-\dfrac{7}{2}$

$\Rightarrow \dfrac{9-7}{2}-\left(\dfrac{2+3}{6}\right)y-\left(\dfrac{1+3}{5}\right)y^2$

$- \dfrac{7}{4}y^3$

$\Rightarrow \dfrac{2}{2}-\left(\dfrac{5}{6}\right)y-\left(\dfrac{4}{5}\right)y^2$

$- \dfrac{7}{4}y^3$

$\Rightarrow 1-\dfrac{5y}{6}-\dfrac{4y^2}{5}$

$- \dfrac{7}{4}y^3$

Hence, this is the answer.

Simplify: $(a + b) (c - d) + (a - b) (c + d) + 2(ac + bd)$

Now,

$(a + b) (c - d) + (a - b) (c + d) + 2(ac + bd)$

$=(ac+bc-ad-bd)+(ac+ad-bc-bd)+2(ac+bd)$

$=4ac$ .

Subtract $3xy + 2z^2$ from $5xy + 3z^2 - xz$

Now,

$(5xy + 3z^2 - xz)-$

$(3xy + 2z^2)$

$=2xy+z^2-xz$

Add:
$2, \dfrac{2}{3}x -\dfrac{5}{3}x^2+\dfrac{5}{2}x^3, \dfrac{-4}{3}+\dfrac{2x^2}{3}-\dfrac{x}{2}, \dfrac{5x^3}{3}+3x^2+3x+\dfrac{6}{5}$ .

So we are given,

$2, \dfrac{5}{2}x^{2}-\dfrac{5}{3}x^{2}+\dfrac{2}{3}x, \dfrac{2}{3}x^{2}-\dfrac{x}{1}-\dfrac{4}{3},\dfrac{5}{3}x^{3}+3x^{2}+3x+\dfrac{6}{5}$

$=2+\left(\dfrac{5}{2}x^{3}-\dfrac{5}{3}x^{2}+\dfrac{2}{3}x\right)+\left(\dfrac{2}{3}x^{2}-\dfrac{x}{2}-\dfrac{4}{3}\right)+\left(\dfrac{5}{3}x^{3}+3x^{2}+3x+\dfrac{6}{5}\right)$

$=\left(\dfrac{5}{2}+\dfrac{5}{3}\right)x^{3}+\left(-\dfrac{5}{3}+\dfrac{2}{3}+3\right)x^{2}+\left(\dfrac{2}{3}-\dfrac{1}{2}+3\right)x+\left(2-\dfrac{4}{3}+\dfrac{6}{5}\right)$

$=\left(\dfrac{5(3)+5(2)}{3.2}\right)x^{2}+\left(\dfrac{-5(1)+2(1)+3(3)}{3}\right)x^{2}+\left(\dfrac{2(2)-1(3)+3(6)}{3.2}\right)x+\left(\dfrac{2(15)-4(5)+6(3)}{3.5}\right)$

$=\dfrac{25}{6}x^{3}+2x^{2}+\dfrac{19}{6}x+\dfrac{28}{15}$

Subtract $(4x^2+2x+3)$ from $(x^3+5x^2+3x+5)$ .

If we subtract

$(4x^2+2x+3)$ from

$(x^3+5x^2+3x+5)$ we get

$(x^3+5x^2+3x+5)-(4x^2+2x+3)=x^3+x^2+x+1$ .

Add $6x^2 + 3y - 2z$ and $3x^2 - y$ .

Now,

$(6x^2 + 3y - 2z)+$

$(3x^2 - y)$

$=9x^2+2y-2z$

Add the following expressions:
$6ax-2by+3cz,\ 6by-11ax-cz$ and $10cz-2ax-3by$ .

Given:

$6ax-2by+.3cz\dots (1)$

$6by-11ax-cz\dots (2)$

$10cz-2ax-3by\dots (3)$

Adding

$(1), (2)$ and

$(3)$ :

$(6a-11a-2a)x+(-2b+6b-3b)y+(3c-c+10c)z$

$\ =-7ax+by+12cz$

By how much must $2{p^2}{q^2} - 3{p^2}q + 4pq$ be increased to get $8{p^2}{q^2} + 4{p^2}q - 3{p^2}q - 2pq + 4?$

$A=8p^2q^2+4p^2q-3p^2q-2pq+4$

$B=2p^2q^2-3p^2q+4pq$

$A-B=8p^2q^2+4p^2q-3p^2q-2pq+4-2p^2q^2+3p^2q-4pq$

$A-B=6p^2q^2+7p^2q-3p^2q-6pq+4$

$(6p^2q^2+7p^2q-3p^2q-6pq+4)$ should be added.

Find sum of : $8ab, -5ab, 3ab, -ab$ .

Find sum of :

$8ab-5ab, 3ab, -ab$ .

$\Rightarrow 8ab+(-5ab)+(3ab)+(-ab)$

$=(8-5+3-1)ab$

$=5ab$

If a, b , c, d are positive real numbers such that $a^2 + b^2 = c^2 + d^2$ and $a^2 + d^2 - ad = b^2 + c^2 + bc$ , find the value of $\dfrac{ab + cd}{ad + bc}$

$a^2 + b^2 = c^2 + d^2 \, \&\, a^2 + d^2 - ad = b^2 + c^2 + bc$

$\boxed {\dfrac{ab + cd}{ad + bc} = ?}$

$bc^2 + d^2 - b^2 + d^2 - ad = b^2 + c^2 + bc$

$2d^2 - ad = 2b^2 + bc$

$\boxed {2d^2 - 2b^2 = ad + bc}$

$\boxed {ad + bc = 2cd^2 - b^2}$

similarly

$ab + cd = 2 (a^2 - c^2)$

$\boxed {\dfrac{ab + cd}{ad + bc} = \dfrac{a^2 - c^2}{d^2 - b^2}}$

A rope of length $(2x+5)$ metre. It is cut into two parts, one measuring $(6x-4)$ metre, what will be the length of the other part?

The length of the other part

$=$ (total length of the rope)-(length of the cut part)

$\begin{array}{l} = \left( {2x + 5} \right) - \left( {6x - 4} \right)\\ = 2x + 5 - 6x - 4\\ = - 4x + 9\end{array}$
The length of the other part will be

$-4x+9$ metre.

Add: $3a-4b+4c, 2a+3b-8c, a-6b+c$ .

$(3a-4b+4c)+(2a+3b-8c)+(a-6b+c)\\ =3a-4b+4c+2a+3b-8c+a-6b+c\\ =(3a+2a+a)+(-4b+3b-6b)+(4c-8c+c)\quad \quad \quad (Combining\quad like\quad terms)\\ =(3+2+1)a+(-4+3-6)b+(4-8+1)c\\ =6a-7b-3c$

Hence,

$(3a-4b+4c)+(2a+3b-8c)+(a-6b+c)=6a-7b-3c$ .

To expression to be subtracted from :
$x^3 + x-1$ to get $2x^3 - x^2 + 3x - 7$ is :

$(x^{3}+x-1)-(p)=2x^{3}-x^{2}+3x-7$

Since both the given polynomials are cibc.

we assume

$p$ tio be a circle polynomials are of form

$A\ x^{3}+B\ x^{2}+C\ x+D$

Put this in

$1$

$(x^{3}+x-1)-(A\ x^{3}+B\ x^{2}+C\ x+D)=2x^{3}-x^{2}+3x-7$

compare the efficient of

$x^{3},x^{2},x$ and constant.

$\left. \begin{matrix} 1-A=2 \\ \Rightarrow A=-1 \end{matrix} \right| \left. \begin{matrix} O-B=-1 \\ \Rightarrow B=1 \end{matrix} \right| \left. \begin{matrix} 1-C \\ \Rightarrow =-2 \end{matrix} \right| \left. \begin{matrix} -1-D=-7 \\ \Rightarrow D=6 \end{matrix} \right|$

$\therefore p\Rightarrow \boxed{-x^{3}-x^{2}-2x+6}$

Subtract ${2ab + 5bc + 3ca}$ from $(7ab+ 2bc + 10ca)$

Here, we are subtracting an algebraic expression from another

$7ab-2bc+10ca$

$-$

$2ab+5bc-3ca$

$----------$

$4ab-7bc-7ca$

$----------$

The difference of

$7ab-2bc+10ca$ and

$2ab+5bc-3ca$ is

$4ab-7bc-7ca$

Subtract:
$- {c^2} + 5c$
$\underline { - 4{c^2} + 8c}$

Solution -

$- {c^2} + 5c$

$\underline { - 4{c^2} + 8c}$

$= -c^{2}-(-4c^2)+5c -8c$

$= -c^2 + 4c^2 + 5c-8c$

$= 3c^2 -3c$

$= 3c(c-1)$

Identify the terms and factors in the expression given below:
$-2x + 3$ .

terms in the expression

$=-2x,3$

factors in the expression

$=-2,3$

Subtract : $3l (l - 4m + 5n)$ from $4l (10n - 3m + 2l)$ .

$4l(10n-3m+2l)-3l(l-4m+5n)$

$\Rightarrow 40ln-12lm+8l^2-(3l^2-12lm+15ln)$

$\Rightarrow 40ln-12lm+8l^2-3l^2+12lm-15ln$

$\Rightarrow 5l^2+25ln$

Subtract the sum of $\left( 8m-7n+6 \right)$ and $\left( -3m-4n-{ p }^{ 2 } \right)$ from the sum of $\left( 2m+4n-3{ p }^{ 2 } \right)$ and $\left( -m-n-{ p }^{ 2 } \right)$ ?

$(8m-7n+6{ p }^{ 2 })+(-3m-4n-{ p }^{ 2 })=(8-3)m-(7+4)n+(6-1){ p }^{ 2 }$

$=5m-11n+5{ p }^{ 2 }\Longrightarrow (1)$

$(2m+4n-3{ p }^{ 2 })+(-m-n-{ p }^{ 2 })=(2-1)m+(4+1)n-(3+1){ p }^{ 2 }$

$=(2-1)m+(4-1)n-(3+1){ p }^{ 2 }\Longrightarrow (2)$

$Subtracting(1)from(2)$

$(m+3n-4{ p }^{ 2 })-(5m-11n+5{ p }^{ 2 })=(1-5)m+(3+11)n-(4+5){ p }^{ 2 }$

$=-4m+14n-9{ p }^{ 2 }$

What polynomial added to $x^{2}+4x-5$ gives $2x^{2}-3x+1$ ?

Let the polynomial that should be added to the polynomial

$x^2+4x-5$ be

$A$ , so that the resulting polynomial is

$2x^2-3x+1$ . Therefore, we have:

$(x^{ 2 }+4x-5)+A=2x^{ 2 }-3x+1\\ \Rightarrow A=(2x^{ 2 }-3x+1)-(x^{ 2 }+4x-5)\\ \Rightarrow A=2x^{ 2 }-3x+1-x^{ 2 }-4x+5\\ \Rightarrow A=(2x^{ 2 }-x^{ 2 })+(-3x-4x)+(1+5)\\ \Rightarrow A=x^{ 2 }-7x+6$

Hence,

$x^{ 2 }-7x+6$ should be added to the polynomial

$x^2+4x-5$ , so that the resulting polynomial is

$2x^2-3x+1$ .

Solve:
$\left\{ {\left( {3{x^2} - 6x + 5} \right) + \left( {2x - 2{x^2} + 5} \right)} \right\} - \left( {{x^2} - 4x + 10} \right)$

$(3x^2-6x+5)+(2x-2x^2+5)-(x^2-4x+10)$

$3x^2-6x+5+2x-2x^2+5-x^2+4x-10$

$3x^2-6x+6x-3x^2+0$

$=0$ .

Simplify :
$12a - 4a - 9 ab + 7 ab + 5b - 3b - 3 - 12$

$12a-4a-9ab+7ab+5b-3b-3-12$

$\Rightarrow (12a-4a)-(9ab-7ab)+(5b-3b)-(3+12)$

$\Rightarrow (8a)-(2ab)+2b-15$

$\Rightarrow 2a(a-b)+2b-8-7$

$\Rightarrow (2a)(a-b)-2(a-b)-7$

$\Rightarrow (2a-2)(4-b)-7$

$\Rightarrow 2(a-1)(4-b)-7$

Find the sum:
(i) $7x-4y,2x+6y$
(ii) $9{x^2} + 5x,6{x^2} - 2x$
(iii) $2a +b,6{a^2} -7a-b$

(i)

$(7x-4y)+(2x+6y)=7x+2x-4y+6y=9x+2y$

(ii)

$(9x^2+5x)+(6x^2-2x)=9x^2+6x^2+5x-2x=15x^2+3x$

(iii)

$(2a+b)+(6a^2-7a-b)=6a^2+2a-7a+b-b=6a^2-5a$

Find the sum of the following.
$-b-4a$ , $7b-6a$ .

$\left( { -b-4a } \right) +\left( { 7b-6a } \right) \\ \Rightarrow \left( { -b+7b } \right) +\left( { -6a-4a } \right) \, \, \, \, \, \, \, \, \, \, \, \, \, \, \\ \Rightarrow 6b+\left( { -10a } \right) \\ \Rightarrow 6b-10a\, \, Ans. \\$

Solve:
$a\left( x-2y+1 \right) ^{ 2 }+9\left( 2x+y+2 \right) ^{ 2 }=25$

We have

${\left(a+b+c\right)}^{2}={a}^{2}+{b}^{2}+{c}^{2}+2ab+2bc+2ca$

Using this, we have

$a\left({x}^{2}+4{y}^{2}+1-4xy-4y+2x\right)+9\left(4{x}^{2}+{y}^{2}+4+4xy+4y+8x\right)-25=0$

$\Rightarrow \left(a{x}^{2}+36{x}^{2}\right)+\left(4a{y}^{2}+9{y}^{2}\right)+\left(a+36\right)+\left(-4axy+36xy\right)+\left(-4ay+36y\right)+\left(2ax+72x\right)-25=0$

$\Rightarrow \left(a+36\right){x}^{2}+\left(4a+9\right){y}^{2}+\left(a+36-25\right)+\left(-4a+36\right)xy+\left(-4a+36\right)y+\left(2a+72\right)x=0$

$\Rightarrow \left(a+36\right){x}^{2}+\left(4a+9\right){y}^{2}+\left(a+11\right)-4\left(a-9\right)xy-4\left(a-9\right)y+2\left(a+36\right)x=0$

$\Rightarrow \left(a+36\right){x}^{2}+\left(4a+9\right){y}^{2}-4\left(a-9\right)\left(xy+x\right)+2\left(a+36\right)x+\left(a+11\right)=0$

$\Rightarrow \left(a+36\right){x}^{2}+\left(4a+9\right){y}^{2}-4x\left(a-9\right)\left(y+1\right)+2\left(a+36\right)x+\left(a+11\right)=0$

what should be taken away from $3x^2+4x+1$ to get $2x^2-4x+35$ ?

$3x^2+4x+1-(2x^2-4x+35)=x^2+8x-34$

Hence,

$x^2+8x-34$ should be taken away from

$3x^2+4x+1$ to get

$(2x^2-4x+35)$

Add the following terms:
$5xy , \;-2xy \;, -11xy , \;8xy$

On adding,

$\begin{aligned}{}5xy + ( - 2xy) + ( - 11xy) + 8xy &= 5xy - 2xy - 11xy + 8xy\\ &= \left( {5 - 2 - 11 + 8} \right)xy\\ &= 0 \times xy\\ &= 0\end{aligned}$

Simplify: $(m+n)(3m+n)+(m+2n)(m-n)$ .

$\begin{array}{l} \Rightarrow (m+n)(3m+n)+(m+2n)(m-n) \\ \Rightarrow 3{ m^{ 2 } }+3mn+mn+{ n^{ 2 } }+{ m^{ 2 } }+2mn-mn-2{ n^{ 2 } } \\ \Rightarrow 4{ m^{ 2 } }+5mn-{ n^{ 2 } } \\ \Rightarrow 4{ m^{ 2 } }-{ n^{ 2 } }+5mn \end{array}$

Simplify the following
$(2x-3y)(3x+y)+(x+2y)(x-y)$ .

Given expression is

$\left( { 2x-3y } \right) \left( { 3x+y } \right) +\left( { x+2y } \right) \left( { x-y } \right)$

Simplifying the expression by multiplying the terms as per the given operation,

$[2x( 3x+y)-3y(3x+y)] +[x(x-y)+2y(x-y)]$

$\Rightarrow [6x^2+2xy-9xy-3y^2]+[x^2-xy+2xy-2y^2]$

$\Rightarrow [6x^2-3y^2-7xy]+[x^2-2y^2+xy]$

$\Rightarrow 6x^2-3y^2-7xy+x^2-2y^2+xy$

$\Rightarrow 7x^2-5y^2-6xy$

Hence,

$\left( { 2x-3y } \right) \left( { 3x+y } \right) +\left( { x+2y } \right) \left( { x-y } \right)=7x^2-5y^2-6xy$ .

Solve the following expression:
$(x+y)(a+b)+(x-y)(a+b)$

1355276_1133186_ans_30ae132333cb4804bc9caaa9a9a67ea7.jpg

$\left( a+2b \right) \left( 3a+b \right) -\left( a+b \right) \left( a+2b \right) +{ \left( a+2b \right) }^{ 2 }$

$\left(a+2b\right) \left(3a+b\right)- \left(a+b\right) \left(a+2b\right)+{\left(a+2b\right)}^{2}$

$\Rightarrow \left(a+2b\right) \left[\left(3a+b\right)- \left(a+2b\right)+ \left(a+2b\right)\right]$

a king

$\left(a+2b\right)$ as common term in the

above equation

now solving the terms in the bracket , we get

$=\left(a+2b\right)\left[3a+b-a-2b+a+2b\right]$

$\Rightarrow\left(a+2b\right) \left(3a+b\right)$

$\Rightarrow \ a \left(3a+b\right)+2b\left(3a+b\right)$

$\Rightarrow3{a}^{2}+ab+6ab+2{b}^{2}$

$\boxed { 3{ a }^{ 2 }+2{ b }^{ 2 }+7ab }$ answer

Subtract.
$5a^{2} - 7ab + 5b^{2}$ from $3ab - 2a^{2} - 2b^{2}$ .

$3ab-2{ a }^{ 2 }-2{ b }^{ 2 }-(5{ a }^{ 2 }-7ab+5{ b }^{ 2 })\\ 3ab-2{ a }^{ 2 }-2{ b }^{ 2 }-5{ a }^{ 2 }+7ab-5{ b }^{ 2 }\\ 10ab-7{ a }^{ 2 }-7{ b }^{ 2 }$

Simplify:
$15xy - 6x + 5y - 2$ .

$15xy-6x+5y-2\\ 3x(5y-2)+1(5y-2)\\ (3x+1)(5y-2)$

Subtract $(-x^{2})+10x-5$ from $5x-10$

Substracting

$(-x^2)+10x-5$ from

$5x-10$

So,

$(5x-10)-(-x^2+10x-5)$

$=5x-10+x^2-10x+5$

$=x^2-5x-5$

$z^{2}-(x^{2}-2xy+y^{2})$

$z^{2}(x^{2}-2xy+y^{2})$

$\Rightarrow z^{2}-(x-y)^{2}$

$((x-y)^{2}=x^{2}+y^{2}-2xy) \: (a^{2}-b^{2}=(a-b)(a+b))$

$\Rightarrow (z-(x-y))(z+(x-y))$

$\Rightarrow (z-x+y)(z+x-y)$

$\therefore \Rightarrow (-x+y+z)(x-y+z)$

1147094_1144070_ans_b5411c60ef6e45b58a6bca82bb921483.jpg

Prove:
$(a+b)^3=a^3+3a^2b+3ab^2+b^3$ .

$(a+b)^{3}= (a+b)^{2}(a+b)$

$=(a^{2}+2ab+b^{2})(a+b)$

$=a^{3}+2a^{2}b+ab^{2}+a^{2}b+2ab^{2}+b^{3}$

$=a^{3}+3a^{2}b+3ab^{2}+b^{3}$

Hence Proved.

${5}^{2}-{a}^{2}$ , ${7x}^{2}-{8a}^{2}$ , ${-7x}^{2}-{9a}^{2}$

$\left(25-{a}^{2}\right)+\left(7{x}^{2}-8{a}^{2}\right)+\left(-7{x}^{2}-9{a}^{2}\right)$

$=25-{a}^{2}+7{x}^{2}-8{a}^{2}-7{x}^{2}-9{a}^{2}$

$=25-{a}^{2}-17{a}^{2}$ on simplification

$=25-18{a}^{2}$

Subtract $x^{5}-2x^{2}-3x$ form $x^{3}+3x^{2}+1$

$(x^3+3x^2+1)-(x^5-2x^2-3x)$

open the brackets and change the sign

$x^3+3x^2+1-x^5+2x^2+3x$

Rearrange into standard form and combine like terms

$-x^5+x^3+5x^2+3x+1$

Evaluate:
$2a+6b-3(a+3b)^{2}$

$2a+6b-3(a+3b)^2$

$=2(a+3b)-3(a+3b)^2$

$=(a+3b)(2-3(a+3b))$

$=(a+3b)(2-3a-9b)$

Evaluate
$12(2x-3y)^{2}-16(3y-2x)$

$12(2x-3y)^2-16(3y-2x)=12(3y-2x)^2-16(3y-2x)$

$=4(3y-2x)(3(3y-2x)-4)$

$=4(3y-2x)(9y-6x-4)$

Find the sum of the following.
$6d-4c$ , $-7c+6d$ .

$\left( { 6d-4c } \right) ,\left( { -7c+6d } \right) \\ \text{Adding} \\ \Rightarrow \left( { 6d-4c+6d-7c } \right) \\= \left( { 6d+6d } \right) +\left( { -4c-7c } \right) \\ = 12d+\left( { -11c } \right) \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \\ = 12d-11c\, \,$

$-3(a+b)+4(2a-3b)-(2a-b)$

$=-3a-3b+8a-12-2a+b$

$=3a-14b$

If $\sqrt{3}[\sqrt{7}-\sqrt{3}]=\sqrt{a}+b$ , then find the values of $a$ and $b$ .

$\sqrt{3} \left[ \sqrt{7} - \sqrt{3} \right] = \sqrt{a} + b \; \left( \text{Given} \right)$

$\sqrt{3} \times \sqrt{7} - \sqrt{3} \times \sqrt{3} = \sqrt{a} + b$

$\sqrt{3 \times 7} - \sqrt{3 \times 3} = \sqrt{a} + b$

$\sqrt{21} - 3 = \sqrt{a} + b$

$\sqrt{21} + \left( -3 \right) = \sqrt{a} + b$

Comparing both sides, we have

$a = 21$

$b = -3$

Hence the values of

$a$ and

$b$ are

$21$ and

$-3$ respectively.

${9x}^{2}-{y}^{2}+4y-4$

$9{x}^{2}-{y}^{2}+4y-4$

$=9{x}^{2}-\left({y}^{2}-4y+4\right)$ where

${y}^{2}-4y+4={y}^{2}-2\times y\times 2+{2}^{2}={\left(y-2\right)}^{2}$

$={\left(3x\right)}^{2}-{\left(y-2\right)}^{2}$ is of the form

${a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)$

$=\left(3x+y-2\right)\left(3x-y+2\right)$

${x}^{2}-{y}^{2}-2y-1$

$={x}^{2}-\left({y}^{2}+2y+1\right)$ the second expression is of the form

${a}^{2}+2ab+{b}^{2}$

$={x}^{2}-{\left(y+1\right)}^{2}$ the second expression is of the form

${a}^{2}-{b}^{2}=\left(a-b\right)\left(a+b\right)$

$=\left(x-y-1\right)\left(x+y+1\right)$

Solve the following:
$a-\left( 2a- 4a- 3a \right)$

$a-(2a-4a-3a) =a-(-5a)$

$= a+5a$

$= 6a$

Simplify: $\left(\dfrac{x^2}{3} - \dfrac{4x}{7} + 11\right) - \left(\dfrac{x}{7} - 3 + 2x^2\right) - \left(\dfrac{2x}{7} - \dfrac{2x^2}{3} + 2\right)$

We have,

$\left(\dfrac{x^2}{3} - \dfrac{4x}{7} + 11\right) - \left(\dfrac{x}{7} - 3 + 2x^2\right) - \left(\dfrac{2x}{7} - \dfrac{2x^2}{3} + 2\right)$

$=\dfrac{x^2}{3} - \dfrac{4x}{7} + 11- \dfrac{x}{7} + 3 -2x^2- \dfrac{2x}{7} + \dfrac{2x^2}{3} -2$

$=\dfrac{x^2+2x^2}{3} -\left (\dfrac{4x+x+2x}{7} \right)+ 11+ 3 -2x^2 -2$

$=\dfrac{3x^2}{3} -\left (\dfrac{7x}{7} \right) -2x^2+12$

$=x^2-x -2x^2+12$

$=-x^2-x+12$

Hence, this is the answer.

If $A = (16x + 8) + 4$ and $B = (15x - 10) + -5$ then $A - B$ is ____.

$A=(16x+8)+4$

$B=(15x-10)+(-5)\\A=16x+12 ; B=15x-15\\A-B=16x+12-15x+15\\ \quad=x+27$

Factorise $x - 1 - ( x - 1 ) ^ { 2 } + a x - a$

$\left( x-1 \right) -{ \left( x-1 \right) }^{ 2 }+a\left( x-1 \right)$

$=\left( x-1 \right) \left( 1-\left( x-1 \right) +a \right)$

$=\left( x-1 \right) \left( 1-x+1+a \right)$

$=\left( x-1 \right) \left( 2+a-x \right)$

The perimeter of a triangle is $8 + 13a + 7a^{2}$ and two of its sides are $2a^{2} + 3a + 2$ and $3a^{2} - 4a - 1$ . Find the third side of the triangle.

Two sides of triangle are

$2a^2+3a+2$ and

$3a^2-4a-1$

Let the third side be

$s$

Perimeter

$=8+13a+7a^2=$

$2a^2+3a+2$

$+3a^2-4a-1+s$

$\implies s=2a^2+15a+7$

So, the third side of triangle is

$=2a^2+15a+7$

From the sum of $3y^{2}-5x^{2}-3yx$ and $3x^{2}-3y^{2}+4xy-1$ subtract the sum of $11x^{2}-2y^{2}-3xy.$ Write the expression one below the corner for adding and subtracting.

We have,

$3{{y}^{2}}-5x-3xy\,sum\,3{{x}^{2}}-3{{y}^{2}}+4xy-1$

$\Rightarrow 3{{y}^{2}}-5x-3xy+3{{x}^{2}}-3{{y}^{2}}+4xy-1$

$\Rightarrow 3{{x}^{2}}+xy-5x-1$

According to given question.

$3{{x}^{2}}+xy-5x-1-\left( 11{{x}^{2}}-2{{y}^{2}}-3xy \right)$

$=3{{x}^{2}}+xy-5x-1-11{{x}^{2}}+2y+3xy$

$=-8{{x}^{2}}+4xy-5x+2y-1$

Hence, this is the answer.

Simplify the expression and find its value, when $x=-1$ and $y=2$ .
${ 2x }^{ 2 }-3y+{ 3x }^{ 2 }+{ 5x }^{ 2 }+5x-7-7x+2y$

According to question,

$2x^2-3y+3x^2+5x^2+5x-7-7x+2y$

$= 10x^2-2x-y-7$

$x= -1$ and

$y =2$ ,

$10(-1)^2-2(-1)-2-7$

$=$

$10+2-2-7$

$= 3$

Find the coefficient of $x^{2}$ in $-15x^{2}y^{2}$ .

Here, coefficient of

$x^2$ is 0,

as, given term is not only considering the variable

$x^2$

Subtract the first term from the second term:
$2 x y , 7 x y$

To subtract

$2xy$ from

$7xy$ .

Now,

$7xy-2xy$

$=(7-2)xy$

$=5xy$

Hence, it is the required answer.

Find the value of the addition of $(x - 3y + 4z), (y - 2x - 8z), (5x - 2y - 3z)$

We have,

$x - 3y + 4z, y - 2x - 8z, 5x - 2y - 3z$

According to the question,

$\Rightarrow x - 3y + 4z+y - 2x - 8z+5x - 2y - 3z$

$\Rightarrow 4x-4y-7z$

Hence, this is the answer.

Show that :
$n ^ { 2 } + ( n + 1 ) ^ { 2 } + \left( n ^ { 2 } + n \right) ^ { 2 } = \left( n ^ { 2 } + n + 1 \right) ^ { 2 }$

LHS

$={ n }^{ 2 }+{ \left( n+1 \right) }^{ 2 }+{ \left( { n }^{ 2 }+n \right) }^{ 2 }$

$={ n }^{ 2 }+{ n }^{ 2 }+2n+1+{ n }^{ 4 }+{ n }^{ 2 }+{ 2n }^{ 3 }$

$={ n }^{ 4 }+{ 2n }^{ 3 }+{ 3n }^{ 2 }+2n+1$

RHS

$={ \left( { n }^{ 2 }+n+1 \right) }^{ 2 }$

$={ n }^{ 4 }+{ n }^{ 2 }+1+{ 2n }^{ 3 }+{ 2n }^{ 2 }+2n$

$={ n }^{ 4 }+{ 2n }^{ 3 }+{ 3n }^{ 2 }+2n+1=$ LHS

Subtract $4x-2y+z$ from the sum of $3x+4y+z$ and $-6x+7y-8z$

Sum :

$(3x+4y+z)+(-6x+7y-8z)=-3x+11y-7z$

Now,

$-3x+11y-7z-\ (4x-2y+z)=-7x+13y-8z$

Add the following expressions :
$m - 7mn + 6, 2m + 9mn - 4$ and $-11m - 8mn - 5$

To add:

$m - 7mn + 6, 2m + 9mn - 4$ and

$-11m - 8mn - 5$

Now,

$(m - 7mn + 6)+( 2m + 9mn - 4)+(-11m - 8mn - 5)$

$= m - 7mn + 6+2m + 9mn - 4-11m - 8mn - 5$

$= 6-5-4+m+2m-11m-7mn+9mn-8mn$

$= -3-8m - 6mn$

Hence, this is the answer.

Subtract :
$3xy+5yz-7zx$ from $4xy-yz+zx-9xyz$

$\dfrac { 4xy-{ y }^{ 2 }+zx-{ 9xy }^{ 2 }\\ -3xy+{ 5y }^{ 2 }-72x+{ 0xy }^{ 2 }\\ -\quad \quad +\quad \quad - }{ xy-{ 6y }^{ 2 }+82x-{ 9xy }^{ 2 } }$

$xy-6y^2+82x-9xy^2$

Subtract $24ab-10b-18a$ from $30ab+12b+14a$

$\Rightarrow (30ab+12b+14a)-(24ab-10b-18a)$

$\Rightarrow (30-24)ab+(12+10)b+(14+18)a$

$6ab+22b+32a$

The value of $\left(\dfrac{8}{5}x^3 - \dfrac{3}{2}x^2 - 1\right) - \left(\dfrac{3}{5}x^3 + \dfrac{1}{2}x^2 - 3x + 2\right)$ is

We have,

$\left(\dfrac{8}{5}x^3 - \dfrac{3}{2}x^2 - 1\right) - \left(\dfrac{3}{5}x^3 + \dfrac{1}{2}x^2 - 3x + 2\right)$

$\Rightarrow \dfrac{8}{5}x^3 - \dfrac{3}{2}x^2 - 1 - \dfrac{3}{5}x^3 - \dfrac{1}{2}x^2 + 3x - 2$

$\Rightarrow \dfrac{8-3}{5}x^3 - \dfrac{3+1}{2}x^2 + 3x - 2-1$

$\Rightarrow \dfrac{5}{5}x^3 - \dfrac{4}{2}x^2 + 3x -3$

$\Rightarrow x^3 - 2x^2 + 3x -3$

Hence, this is the answer.

Subtract :
$-3x^{3}-4x+1$ from $-5x^{3}-x-9$

$\dfrac { \quad -{ 5x }^{ 3 }-x-9\\ -\quad -{ 3x }^{ 3 }-4x+1\\ \quad +\quad \quad +\quad \quad - }{ -2{ x }^{ 3 }+3x-10 }$

$-2x^3+3x-10$

Subtract :
$4x^{2}-5x+9$ from $7x^{2}+x-3$

$\dfrac { \quad 7{ x }^{ 2 }+x-3\\ -\quad 4{ x }^{ 2 }-5x+9 }{ 3{ x }^{ 2 }+6x-12 }$

$3x^2+6x-12$

Subtract: $2x^2-x+14,5x-3x^2+8$

1146967_1176218_ans_8ad7ef826fcd42158a12955e684087f6.jpg

Simplify:
$2p^3 - 3p^2 + 4p - 5 - 6p^3 + 2p^2 - 8p - 2 + 6p + 8$

$2p^3-3p^2+4p-5-6p^3+2p^2-8p-2+6p+8$

Reaaranging and collecting the like terms,

$\Rightarrow$

$(2p^3-6p^3)+(-3p^2+2p^2)+(4p-8p+6p)+(-5-2+8)$

$\Rightarrow$

$-4p^3-p^2+2p+1$

From the sum of $3 x ^ { 2 } - 5 x + 2$ and $- 5 x ^ { 2 } - 8 x + 9$ subtract $4 x ^ { 2 } - 7 x + 9.$

Adding:

$(3x^{2}-5x+2)+(-5x^{2}-8x+6)$

Rearranging and collecting the like terms:

$(3-5)x^{2}+(-5-8)x+2+6$

$= -2x^{2}-13x+8$

Subtract

$4x^{2}-9x+7$ from

$-2x^{2}-13x+8$

Change the sign of each term of the expression that is to be subtracted and then add.

Term to be subtracted

$= 4x^{2}-9x+7$

Changing the sign of each term of the expression gives

$-4x^{2}+9x-7$

On adding:

$(-2x^{2}-13x+8)+(-4x^{2}+9x-7) = -2x^{2}-13+8-4x^{2}+9x-7$

$= (-2-4)x^{2}+(-13+9)x+8-7$

$= -6x^{2}-4x+1$

1088029_1180306_ans_8274a04c1f1e489bbf03a831a1d39f47.png

Solve:
Add $3a(a - b + c),2b(a - b + c)$

$3a (a-b+c)+ 2b (a-b+c)$

$=(3a+2b)(a-b+c)$

$=3a^{2}-3ab+3ac+2ab-2b^{2}+2bc$

$= 3a^{2}-ab-2b^{2}+3ac+2bc$

Subtract: $3x^2+3x-5,2x^2-8x-5$

1146890_1176215_ans_b8eb98fc81d1443bbae67737b8708c1e.jpg

Subtract: $-5xy-8x-9,7xy-7x+6$

1146883_1176208_ans_913f5e41bafc41449e5f4275679416ed.jpg

Subtract: $x^2+2xy+y^2,10x^2$

1146863_1176202_ans_3a86503c221548c7bfd70449b91e0ced.jpg

Enter 1 if it is true else enter 0.
$(3x+4y)+(5x+7y)$ $=8x+11y$

$(3x+4y)$ +

$(5x+7y)$

$=(3x+5x)$ +

$(4y+7y)$

$=8x+11y$

Subtract:
(i) $-5y^2$ from $y^2$
(ii) $6xy$ from $-12xy$
(iii) $(a-b)$ from $(a+b)$
(iv) $a(b-5)$ from $b(5-a)$
(v) $-m^2+5mn$ from $4m^2-3mn+8$
(vi) $-x^2+10x-5$ from $5x-10$
(vii) $5a^2-7ab+5b^2$ from $3ab-2a^2-2b^2$
(viii) $4pq-5q^2-3p^2$ from $5p^2+3q^2-pq$

$i)\ y^2-(-5y^2)=6y^2$

$ii)\ -12xy-6xy=-18xy$

$iii)\ a+b-a+b=2b$

$iv)\ 5b-ab-ab+5a=5a+5b-2ab$

$v)\ 4m^2-3mn+8+m^2-5mn=5m^2-8mn+8$

$vi)\ 5x-10+x^2-10x+5=x^2-5x-5$

$vii)\ 3ab-2a^2-2b^2-5a^2+7ab-5b^2=-7a^2-7b^2+10ab$

$viii)\ 5p^2+2q^2-pq-4pq+5q^2+3p^2=8p^2+8q^2-5pq$

Subtract: $a^2+b^2-7ab,3b^2$

1146876_1176204_ans_9b62260eaa20444ba9505d3de319bdc6.jpg

Simplify the given expression:
$4\left( {a + b} \right) - 6\left( {a + b} \right)$

Solution:

$4(a+b)-6(a+b)$

$=4a+4b-6a-6b$

$-2a-2b$

$=2(a+b)$

Solve : $3$ x, - $3$ x, $5$ y, $7$ y

$\begin{matrix} \Rightarrow 3x+\left( { -3x } \right) +5y+7y \\ \Rightarrow \left( { 3x-3x } \right) +12y \\ \Rightarrow 0+12y \\ \Rightarrow 12y \\ \end{matrix}$

Solve:
$- [3{a^2} - 2{a^2} + 9{a^2} - (6{a^2} - ( - 2{a^2} + 3{a^2})]]$

$-\left[ { 3a }^{ 2 }-{ 2a }^{ 2 }+{ 9a }^{ 2 }-\left( { 6a }^{ 2 }-\left( { -2a }^{ 2 }+{ 3a }^{ 2 } \right) \right) \right]$

$=-\left[ { a }^{ 2 }+{ 9a }^{ 2 }-\left( { 6a }^{ 2 }-\left( { a }^{ 2 } \right) \right) \right]$

$=-\left[ { 10a }^{ 2 }-{ 5a }^{ 2 } \right]$

$=-\left[ { 5a }^{ 2 } \right]$

$=-{ 5a }^{ 2 }$

Simplify $10a(2x+y)^3-15b(2x+y)^2+35(2x+y)$

$10a{\left(2x+y\right)}^{3}-15b{\left(2x+y\right)}^{2}+35\left(2x+y\right)$

$=5\left(2x+y\right)\left[2a{\left(2x+y\right)}^{2}-3b\left(2x+y\right)+7\right]$

$=5\left(2x+y\right)\left[2a\left(4{x}^{2}+{y}^{2}-4xy\right)-3b\left(2x+y\right)+7\right]$

$=5\left(2x+y\right)\left[8a{x}^{2}+2a{y}^{2}-8axy-6bx-3by+7\right]$

$=5\left(2x+y\right)\left(8a{x}^{2}+2a{y}^{2}-8axy-6bx-3by+7\right)$

If A= $10W^{3}+20W^{2}-55W+60,$
$B=-25W^{2}+15W-10$ and $C=5W^{2}-10W+20,$ then A+B-C is equal to -

$A=10w^3+20w^2-55w+60$

$B=-25w^2+15w-10$

$C=5w^2-10w+20$

$A+B-C=10w^3-5w^2-40w+50-5w^2+100w-20$

$=10w^3-10w^2-30w+30$

$=10w^2(w-1)-30(w-1)$

$A+B-C=10(w^2-3)(w-1)$

Subtract the sum of $2x^2-4x+11$ and $7x^2-4x+1$ from the sum of $8x^2-7x-7$ and $6x^2+4x-6.$

Let

$S_1=(2x^2-4x+11)+(7x^2-4x+1)=(2+7)x^2+(-4-4)x+(11+1)=9x^2-8x+12$

and

$S_2=(8x^2-7x-7)+(6x^2+4x-6)=(14)x^2+(-7+4)x+(-6-7)=14x^2-3x-13$

Now,

$S_2-S_1=(14x^2-3x-13)-(9x^2-8x+12)=14x^2-3x-13-9x^2+8x-12=5x^2+5x-25$

Simplify :
$(x^{2}-x)-\dfrac{1}{2}(x-3+3x^{2})$

Given:

$\left( { { x^{ 2 } }-x } \right) -\dfrac { 1 }{ 2 } \left( { x-3+3{ x^{ 2 } } } \right)$

$\Rightarrow { x^{ 2 } }-x-\dfrac { x }{ 2 } +\dfrac { 3 }{ 2 } -\dfrac { 3 }{ 2 } { x^{ 2 } }$

$\Rightarrow { x^{ 2 }(1-\dfrac{3}{2}) }-x(1+\dfrac{1}{2})+\dfrac { 3 }{ 2 }$

$\Rightarrow -\dfrac { { { x^{ 2 } } } }{ 2 } -\dfrac { { 3x } }{ 2 } +\dfrac { 3 }{ 2 }$

Subtract by horizontal method:
$2x-4y$ from $-4x+3y$

Given polynomial

$2x-4y$ and

$-4x+3y$

As per question

$-4x+3y-(2x-4y)$

$=-4x+3y-2x+4y$

$=-4x-2x+3y+4y$

$=-6x+7y$

This is the required answer.

Subtract $2x-5y$ from the sum of the binomials $x+7y$ and $3x-6y$ .

$\left( x+7y \right)+\left( 3x-6y \right)$

$=4x+y$

$\left( 4x+y \right)-\left( 2x-5y \right)$

$=4x+y-2x+5y$

$=2x+6y$

Subtract $4 p ^ { 2 } q - 3 p q + 5 p q ^ { 2 } - 8 p + 7 q - 10$ from
$18 - 3 p - 11 q + 5 p q - 2 p q ^ { 2 } + 5 p ^ { 2 } q$

$\Rightarrow 4p^{2}q-3pq+5pq^{2}-8p+7q-10 = x$

$18-3p-11q+5pq-2pq^{2}+5p^{2}q = y$

Now

$x-y =$

$4p^{2}q-3pq+5pq^{2}-8p+7q-10$

$5p^{2}q+5pq-2pq^{2}-3p-11q+18$

______________________________

$-p^{2}q-8pq+7pq^{2}-5p+18q-28$

$y{\left( {x - 2y} \right)^2} - 13 + 4\left( {x - 1} \right)$

Given:

$y(x-2y)^2-13+4(x-1)$

$=y(x^2-4xy+4y^2)-13+4x-4$

$=x^2y-4xy^2+4y^3-13+4x-4$

$=4y^3-4xy^2+x^2y+4x-17$ .

$9{\left( {x - 2y} \right)^2} - 13 + 4\left( {x - 2y} \right)$

1142422_1224935_ans_ebebb3f25178457ca505d0ccb9fbaaa4.jpg

$6 p - 12 q$

$\textbf{Step - 1: Define like terms}$

$\text{Like terms are terms that have the same variables and powers.}$

$\text{ The coefficients do not need to match.}$

$\textbf{Step - 2: Solve the algebraic Expression}$

$6p - 12q = 3\times 2 \times p - 3 \times 2 \times 2 \times q$

$= 3\times 2 (p - 2 \times q)$

$= 6(p - 2q)$

$\textbf{Hence, } \mathbf{ (6p - 12q)= 6(p - 2q)}$

Add $2x^2-8xy+7y^2-8xy^2, 2y^2, 6xy^2-y^2+3x^2, 4y^2-xy-x^2+xy^2$ .

$(2x^2-8xy+7y^2-8xy^2)+(2y^2)+(6xy^2-y^2+3x^2)+(4y^2-xy-x^2+xy^2)$

$\Rightarrow$

$2x^2-8xy+7y^2-8xy^2+2y^2+6xy^2-y^2+3x^2+4y^2-xy-x^2+xy^2$

Rearranging and combining the like terms,

$\Rightarrow$

$(2x^2+3x^2-x^2)+(-8xy-xy)+(7y^2+2y^2-y^2+4y^2)+(-8xy^2+6xy^2+xy^2)$

$\Rightarrow$

$4x^2-9xy+12y^2-xy^2$

$\Rightarrow$

$4x^2+12y^2-9xy-xy^2$

Solve :
$(x^2+4xy)+(4y^2-9z^2)$

Consider the given expression,

$\Rightarrow \left( {{x}^{2}}+4xy \right)+\left( 4{{y}^{2}}-9{{z}^{2}} \right)$

$\Rightarrow {{x}^{2}}+4xy+4{{y}^{2}}-9{{z}^{2}}$

$\Rightarrow {{x}^{2}}+4{{y}^{2}}-9{{z}^{2}}+4xy$

Hence, this is the answer.

$\begin{array} { l } { \text { Add: } x ( x + y ) ; y ( y + z ) : z ( z + x ) } \\ { \text { Add } : 3 m ( 1 - m - n ) \text { and } 2 n ( m - n - 1 ) } \\ { \text { Subtract: } 2 x ( x - 3 y + 4 z ) \text { from } x ( 7 x - y + 9 z ) } \\ { \text { Subtract: } ( a - b + c ) - 2 ( - a + b - c ) \text { from } 5 ( a + 2 b - c ) } \end{array}$

(1)

$x(x+y)+y(y+z)+z(z+x)$

$={x}^{2}+xy+{y}^{2}+yz+{z}^{2}+xz$

$={x}^{2}+{y}^{2}+{z}^{2}+xy+yz+zx$

(2)

$3m(1-m-n)+2n(m-n-1)$

$=3m-3{m}^{2}-3mn+2mn-2{n}^{2}-2n$

$=3m-2n-3{m}^{2}-2{n}^{2}-mn$

(3)

$x(7x-y+9z)-2x(x-3y+4z)$

$=7{x}^{2}-xy+9xz-2{x}^{2}+6xy-8xz$

$=5{x}^{2}+5xy+xz$

(4)

$5(a+2b+c)-[a-b+c-2(-a+b+c)]$

$=5a+10b-5c-[3a-3b+3c]$

$=2a+13b-8c$

$-7p+20p-3$ is a binomial or trinomial.

Since there are two independent terms, so we can say that

$-7p+20p-3$ is a binomial.

Evaluate:
$( x + 4 y + 7 x ) - ( x + 2 y )$

${\textbf{Step - 1: Reduce the given polynomial in 2 variables}}$

${\text{Given}}$

$(x+4y+7x) - (x+2y)$

$= (x+4y+7x) - x-2y$

$= x+4y+7x - x-2y$

$= 4y+7x -2y$

$= 2y+7x$

$= 7x+2y$

${\textbf{Hence ,}}$

$\mathbf{ (x+4y+7x) - (x+2y) = 7x+2y }$

Simplify : $({a^3} - 2{a^2} + 4a - 5) - ( - {a^3} - 8a + 2{a^2} + 5)$

$(a^3-2a^2+4a-5)-(-a^3-8a+2a^2+5)$

$2a^3-4a^2+12a-10$

Add the following expressions:
$a + b - 3, b - a + 3, a - b + 3$

$a+b-3+b-a+3+a-b+3$

$=\left(a-a+a\right)+\left(b+b-b\right)+\left(-3+3+3\right)$ by adding the like terms.

$=a+b+6$

Subtract $4{y^4} - 2{y^3} - 6y - y + 5$ from $5{y^4} - 3{y^3} + 2{y^2} + y - 1$

$\Rightarrow (5y^4-3y^3+2y^2+y-1)-(4y^4-2y^3-6y-y+5)$

$=5y^4-3y^3+2y^2+y-1-4y^4+2y^3+6y+y-5$

$=y^4-y^3+2y^2+8y-6$ .

Add: $6a^3, -4a^3, 10a^3, -8a^3$ .

$6a^3+(-4a^3)+(10a^3)+(-8a^3)$

$\Rightarrow$

$6a^3-4a^3+10a^3-8a^3$

$\Rightarrow$

$4a^3$

Addition
$4x^{2}+4x+\dfrac{3}{2}$ and $2x^{2}+3x+\dfrac{1}{2}$

Addition

$4x^{2}+4x+\frac{3}{2}$ &

$2x^{2}+3x+\frac{1}{2}$

$= 4x^{2}+4x+\frac{3}{2}+2x^{2}+3x+\frac{1}{2}$

$= 6x^{2}+7x+(\frac{3+1}{2})$

$= 6x^{2}+7x+2$

1131956_1250046_ans_58412d39718c4daa8146aa62899b5ad8.jpg

If $A=4x^2+y^2-6xy,\, B=3y^2+12x^2+8xy$ and $C=6x^2+y^2+6xy$ , find $A+B+C$ .

We have

$\begin{array}{l} A=4{ x^{ 2 } }+{ y^{ 2 } }-6xy,\, \, \, \\ B=3{ y^{ 2 } }+12{ x^{ 2 } }+18xy\, \\ and\, C=6{ x^{ 2 } }+{ y^{ 2 } }+6xy \\ Now, \\ A+B+C=4{ x^{ 2 } }+{ y^{ 2 } }-6xy+3{ y^{ 2 } }+12{ x^{ 2 } }+18xy+6{ x^{ 2 } }+{ y^{ 2 } }+6xy \\ =22{ x^{ 2 } }+5{ y^{ 2 } }+8xy \end{array}$

Hence,

$A+B+C=22{ x^{ 2 } }+5{ y^{ 2 } }+8xy$ .

From the sum of $6x^{4}-3x^{3}+7x^{2}-5x+1$ and $-3x^{4}+5x^{3}-9x^{2}+7x-2$ subtract $2x^{4}-5x^{3}+2x^{2}-6x-8$

1292977_1254403_ans_4fccb7b966ee4c12b98bd5c43ac516ef.jpg

$\sqrt { 2 x - 3 } + \sqrt { 7 - 3 x }$

$For\>\sqrt {2x-3},2x-3\>\geq\>0\\x\>\geq(\dfrac{3}{2})\\and\>for\>\sqrt {7-3x}\\7-3x\>\geq 0\\7\geq 3x\\or\>x\leq\>(\dfrac{7}{3})\\hence\>x\in[\>\dfrac{3}{2},\>\dfrac{7}{3}]$

Solve : $(5a^2 - 7ab + 5b^2) - (3ab - 2a^2 -2b^2)$

1302544_1253598_ans_bdd89f39853a47bfa434f182361203b7.jpeg

What would be subtracted from $x^{2}-xy+y^{2}-x+y+3$ to obtain $-x^{2}+3y^{2}-4xy+1$ ?

Let m should be subtracted from

$x^{2}-xy+y^{2}-x+y+3$ to obtain

$-x^{2}+3y^{2}-4xy+1$

$x^{2}-xy+y^{2}-x+y+3-m = -x^{2}+3y^{2}-4xy+1$

$m = x^{2}-xy+y^{2}-x+y+3+x^{2}-3y^{2}+4xy-1$

$m = 2x^{2}-2y^{2}+3xy-x+y+2$

1180222_1284043_ans_8a1a9137873d46af93ce14b3b0cee036.jpg

If $A=2y^{2}+3x-x^{2},\ B=3x^{2}-y^{2}$ and $C=5x^{2}-3xy$ then find
$A-B$

$A = 2y^{2} + 3x - x^{2}$

$B = 3x^{2} - y^{2}$

$A-B = 2y^{2} + 3x - x^{2} - 3x^{2} + y^{2}$

$= -4x^{2} + 3y^{2} + 3x$

[a + {a (a+2)} + 2]

Given expression

$=[a+\{a(a+2)\}+2]$

$=[a+\{a^2+2a\}+2]$

$=[a+a^2+2a+2]$

$=a^2+3a+2$ .

1172822_1269675_ans_83e2390ed5804aeb8fc201f3972196cc.jpeg

If $A = 4 x - 6 y$
$B = 5 x + 4 y$
$C = 3 y - 3 c$ ,
then find the value of
i) $A + B - C$
ii) $2 A - 3 B + 4 C$

$A=4x-6y,B=5x+4y,C=3y-3c\\(i)A+B-C=4x-6y+5x+4y-3y+3c=9x-5y+3c\\(ii)2A-3B+4C=8x-12y-15x-12y+12y-12c=-7x-12y-12c$

How much smaller is $2x^2-2x+5$ than $7+5x^2+3x$

1306695_1266324_ans_35d3b589a3164dee93faf36ad0c86fbe.jpeg

What should be added to $7x^{2}-5x+1$ to get $3x^{2}+4x+7$ .

Let the value y be added to

$7x^2 - 5x + 1$ to give

$3x^2 + 4x + 7$

$y = 3x^2 + 4x + 7 - ( 7x^2 - 5x + 1)$

$= 3x^2 + 4x + 7 - 7x^2 + 5x - 1$

$= - 4x^2 + 9x + 6$

What should be taken away from $3x^2 - 4y^2 + 5xy + 20 \ get \ -x^2-y^2 + 6xy + 20$ .

Given

$\Rightarrow$

$3x^2-4y^2+5xy+20$

Required

$\Rightarrow$

$-x^2-y^2+6xy+20$

Let the necessary cutoff expression be

$a$ .

Then,

$\begin{aligned}{}3{x^2} - 4{y^2} + 5xy + 20 - a& = - {x^2} - {y^2} + 6xy + 20\\a &= 3{x^2} - 4{y^2} + 5xy + 20 + {x^2} + {y^2} - 6xy - 20\\ &= {x^2}\left( {3 + 1} \right) + {y^2}\left( { - 4 + 1} \right) + xy\left( {5 - 6} \right) + 20 - 20\\ &= 4{x^2} - 3{y^2} - xy\end{aligned}$

If $\Delta = 2 x ^ { 2 } + 3 y$
$\square = 8 y ^ { 2 } - 3 x ^ { 4 } + 2 x + 3 y$
$O = 5 x ^ { 2 } + 3 x$
Then $\Delta - 0 - \square$ represents

$\triangle =2x^2+3y$

$\square =8y^2-3x^4+2x+3y$

$\bigcirc =5x^2+3x$

$\triangle -\bigcirc -\square$

$=2x^2+3y-(5x^2+3x)-(8y^2-3x^4+2x+3y)$

$=2x^2+3y-5x^2-3x-8y^2+3x^4-2x-3y$

$=-3x^2-5x-8y^2+3x^4$

$=3x^4-3x^2-8y^2-5x$ .

Add the following algebraic expression using both horizontal and vertical methods. Did you get the same answer with both methods.
$2x^{2}-6x+3;\ -3x^{2}-x-4;\ 1+2x-3x^{2}$

By vertical method

$\begin{array}{l} 2{ x^{ 2 } }-6x+3 \\ -3{ x^{ 2 } }-x-4 \\ \underline { -3{ x^{ 2 } }+2x+1 } \\ \underline { -4{ x^{ 2 } }-5x+0 } \end{array}$

By horizontal method

$\begin{array}{l} 2{ x^{ 2 } }-6x+3-3{ x^{ 2 } }-x-4-3{ x^{ 2 } }+2x+1 \\ =-4{ x^{ 2 } }-5x+0 \end{array}$

$1=2$ both answer are same

Substract second expression from the first expression $x+2y+z,-x-y-33$

According to question

$\begin{array}{l} =\left( { -x-y-33 } \right) from\left( { x+2y+z } \right) \\ =x+2y+z+x+y+33 \\ =2x+3y+z+33 \end{array}$

Subtract:-
$2a\left( a-3b+5c \right)$ from $5a\ \left( 2a-3c+b \right)$

Let

$X=2a(a-3b+5c)$

$Y=5a(2a-3c+b)$

According to the question,

$Y-X$

$\Rightarrow Y-X=5a(2a-3c+b)-2a(a-3b+5c)$

$=10a^2-15ac+5ab-2a^2+6ab-10ac$

$= 8a^2-25ac+11ab$

Subtract $a\left( {b - 5} \right)from\;b\;\left( {5 - a} \right)$

According to the question,

$b(5-a)-a\left( {b - 5} \right)$

$=5b-ab-ab+5a$

$=5(a+b)-2ab$

Hence, this is the answer.

Add the following algebraic expression using both horizontal and vertical methods. Did you get the same answer with both methods
$2x+9y-7z;3y+z+3x;2x-4y-z$

Given expressions are:

$2x+9y-7z, \ 3y+z+3x\text{ and }2x-4y-z$

Adding the given expressions:

$(1). Vertical\, \, method \\ \ \ \ \ \ 2x+9y-7z \\ +\ 3x+3y+z \\ \underline { +\ 2x-4y-z } \\ \underline { \ \ \ \ 7x+8y-7z }\quad .....(i)$

$(2). Horizontal\, \, method \\ 2x+9y-7z+3x+3y+z+2x-4y-z \\ \Rightarrow 7x+8y-7z\quad .....(ii)$

We can see that,

$(i)=(ii)$

Hence, both horizontal and vertical methods give the same answer.

Add
$3a-2b+5c, 2a+5b-7c, -a-b+c$

$\begin{array}{l} =3a+2a-a-2b+5b-b+5c-7c+c \\ we\quad get, \\ =4a+2b-c \end{array}$

What must be added to $5x^{3}-3x+1$ to get $3x^{3}-7x^{2}+8?$

$\begin{array}{l} To\, get\; the\, required\, number\, \\ \left( { 3{ x^{ 3 } }-7{ x^{ 2 } }+8 } \right) -\left( { 5{ x^{ 3 } }-3x+1 } \right) \\ =3{ x^{ 3 } }-7{ x^{ 2 } }+8-5{ x^{ 3 } }+3x-1 \\ =3{ x^{ 3 } }-5{ x^{ 3 } }-7{ x^{ 2 } }+3x+8-1 \\ =-2{ x^{ 3 } }-7{ x^{ 2 } }+3x+8-1 \\ which\, is\, the\, required\, number. \end{array}$

Add
(a) 7x+3y and -4x-2y
(b) 3a-2b+c+3 and a+3b-2c-4

$\begin{array}{l} \left( a \right) 7x+3y\, \, and\, -4x-2y \\ =\left( { 7x+3y } \right) +\left( { -4x-2y} \right) \\ =7x+3y-4x-2y \\ =3x+y \\ \left( b \right) 3a-2b+c+3\, \, and\, \, a+3b-2c-4 \\ =\left( { 3a-2b+c+3 } \right) +\left( { a+3b-2c-4 } \right) \\ =3a-2b+c+3+a+3b-2c-4 \\ =4a+b-c-1 \end{array}$

Subtract : $4pq-3pq+5pq^{2}-8p+7q-10$ from $18-3p-11q+5pq-2pq^{2}+5p^{2}q$

Now,

$\begin{array}{l} =\left( { 18-3p-11q+5pq-2p{ q^{ 2 } }+5{ p^{ 2 } }q } \right) -\left( { 4{ p^{ 2 } }q-3pq+5p{ q^{ 2 } }-8p+7q-10 } \right) \\ =18-3p-11q+5pq-2p{ q^{ 2 } }+5{ p^{ 2 } }q-4{ p^{ 2 } }q+3pq-5p{ q^{ 2 } }+8p-7q+10 \\ =28+5p-18q-2pq-7p{ q^{ 2 } }+{ p^{ 2 } }q \end{array}$

If the land owner gives one forth of his yield of rice of charity, how much yield is left with him. What value is exhibited by the owner?

Let the total rice with the owner be x.

Given for charity =

$\dfrac{x}{4}$

Left with him =

$x-\dfrac{x}{4} = \dfrac{3x}{4}$

Value shown by the owner is helping nature.

Simplify ${\left( {a + 2b} \right)^2}\, - {\left( {a - 2b} \right)^2}$ .

We have,

${\left( {a + 2b} \right)^2}\, - {\left( {a - 2b} \right)^2}$

$\Rightarrow a^2+4b^2+4ab- (a^2 +4b^2-4ab)$

$\Rightarrow a^2+4b^2+4ab- a^2 -4b^2+4ab$

$\Rightarrow 4ab+4ab=8ab$

Hence, this is the answer.

Solve:
$(24m^{4}-16m^{2})+8m^{2}$

$\begin{array}{l} \left( { 24{ m^{ 2 } }-16{ m^{ 2 } } } \right) +8{ m^{ 2 } } \\ \Rightarrow 24{ m^{ 4 } }-16{ m^{ 2 } }+8{ m^{ 2 } } \\ \Rightarrow 24{ m^{ 2 } }-8{ m^{ 2 } } \end{array}$

Add:
$2x\ (y-2x+z)$ and $3y\ (2z+x+x^{2})$

$2x \left( y - 2x + z \right) + 3y \left( 2z + x + {x}^{2} \right)$

$= 2xy - 4 {x}^{2} + 2zx + 6yz + 3xy + 3 {x}^{2}y$

$= 3 {x}^{2} y - 4 {x}^{2} + 5xy + 6yz + 2zx$

Solve
$xy-[yz-zx-yx-(3y-xz)-(x-y-2y)].$

$xy-[yz-zx-yx-(3y-xz)-(x-y-2y)]$

$=xy-[yz-zx-yx-3y+xz-x+y+2y]$

$=xy-[yz-yx-x]$

$=xy-yx+xy+x$

$=2xy-yx+x$

Subtract $4x^2+5y^2$ from $3x - 8{y^2}$ .

According to the question,

$3x-8y^2-(4x^2+5y^2)$

$\Rightarrow 3x-8y^2-4x^2-5y^2$

$\Rightarrow 3x-13y^2-4x^2$

$\Rightarrow 3x-4x^2-13y^2$

Hence, this is the answer.

Evaluate:
$8\left( {{x^3}{y^2}{z^2} + {x^2}{y^3}{z^2} + {x^2}{y^2}{z^3}}+x^2y^2z^2 \right) + 4{x^2}{y^2}{z^2}$

We have,

$8\left( {{x^3}{y^2}{z^2} + {x^2}{y^3}{z^2} + {x^2}{y^2}{z^3}}+x^2y^2z^2 \right) + 4{x^2}{y^2}{z^2}$

$\Rightarrow 8 {{x^3}{y^2}{z^2} + 8{x^2}{y^3}{z^2} + 8{x^2}{y^2}{z^3}}+8x^2y^2z^2 + 4{x^2}{y^2}{z^2}$

$\Rightarrow 8 {{x^3}{y^2}{z^2} + 8{x^2}{y^3}{z^2} + 8{x^2}{y^2}{z^3}}+12x^2y^2z^2$

$\Rightarrow 4(x^2y^2z^2)(2x + 2y + 2z+3)$

Hence, this is the answer.

Simplify
$20x - \left[ {15{x^3} + 5{x^2} - \{ 8{x^2} - (4 - 2x - {x^3}) - 5{x^3}\} - 2x} \right]$

We have,

$20x - \left[ {15{x^3} + 5{x^2} - \{ 8{x^2} - (4 - 2x - {x^3}) - 5{x^3}\} - 2x} \right]$

$\Rightarrow 20x - \left[ {15{x^3} + 5{x^2} - \{ 8{x^2} - 4 +2x+ x^3 - 5{x^3}\} - 2x} \right]$

$\Rightarrow 20x - \left[ {15{x^3} + 5{x^2} - \{ 8{x^2} - 4 +2x- 4x^3\} - 2x} \right]$

$\Rightarrow 20x - \left[ {15{x^3} + 5{x^2} - 8{x^2} + 4 -2x+4x^3 - 2x} \right]$

$\Rightarrow 20x - \left[ {19x^3 - 3x^2+ 4 -4x} \right]$

$\Rightarrow 20x -19x^3 +3x^2- 4 +4x$

$\Rightarrow -19x^3 +3x^2+24x- 4$

Hence, this is the answer.

Add: $a,\,\,c,\,\,-a+b-2c$

Given,

$a+c+\left(-a+b-2c\right)$

$=a+c-a+b-2c$

$=b-c$

subtract: $3a(a+b+c)-2b(a-b+c)$ from $4c(-a+b+c)$

$4c\left(-a+b+c\right)-\left[3a\left(a+b+c\right)-2b\left(a-b+c\right)\right]$

$=-4ac+4bc+4{c}^{2}-\left[3{a}^{2}+3ab+3ac-2ab+2{b}^{2}-2bc\right]$

$=-4ac+4bc+4{c}^{2}-3{a}^{2}-3ab-3ac+2ab-2{b}^{2}+2bc$

$=-3{a}^{2}-2{b}^{2}+4{c}^{2}-ab+6bc-7ac$

Solve: $\dfrac{x+20}{9}+\dfrac{3x}{7}=6$

$\dfrac{x+20}{9}+\dfrac{3x}{7}=6$

$\Rightarrow \dfrac{7x+140+27x}{63}=6$

$\Rightarrow 34x+140=6\times 63$

$\Rightarrow 34x=378‬-140=238$

$\Rightarrow x=\dfrac{238}{34}=7$

Let $P={a}^{2}-{b}^{2}-2ab,\,Q={a}^{2}+4{b}^{2}-6ab,\,R={b}^{2}+6,\,S={a}^{2}-4ab$ and $T=-2{a}^{2}+{b}^{2}-ab+a$ .Find the value of $P+Q+R+S+T$ ?

$P+Q={a}^{2}-{b}^{2}-2ab+{a}^{2}+4{b}^{2}-6ab=2{a}^{2}-8ab+3{b}^{2}$

$P+Q+R=2{a}^{2}-8ab+3{b}^{2}+{b}^{2}+6=2{a}^{2}-8ab+4{b}^{2}+6$

$P+Q+R+S=2{a}^{2}-8ab+4{b}^{2}+6+{a}^{2}-4ab=3{a}^{2}-12ab+4{b}^{2}+6$

$P+Q+R+S+T=3{a}^{2}-12ab+4{b}^{2}+6-2{a}^{2}+{b}^{2}-ab+a={a}^{2}-12ab+5{b}^{2}+6-ab+a$

$\therefore \,P+Q+R+S+T={a}^{2}+5{b}^{2}-12ab-ab+a+6$

Solve: $\dfrac{4\left(x+2\right)}{5}=\dfrac{7+5x}{13}$

given,

$\dfrac{4\left(x+2\right)}{5}=\dfrac{7+5x}{13}$

$\Rightarrow \dfrac{4x+8}{5}=\dfrac{7+5x}{13}$

$\Rightarrow 13\left(4x+8\right)=5\left(7+5x\right)$

$\Rightarrow 52x+104=35+25x$

$\Rightarrow 52x-25x=35-104$

$\Rightarrow 27x=-69$

$\Rightarrow x=\dfrac{-69}{27}=\dfrac{-23}{9}$

Add: $x-3y-2z,\,\,5x+7y-z,\,\,-7x-2y+4z$

Sum

$=\left(x-3y-2z)+(5x+7y-z\right)+\left(-7x-2y+4z\right)$

$=(x+5x-7x)+(-3y+7y-2y)+(-2z-z+4z)$

$=-x+2y+z$

Add: $3x+4y,\,\,x-2y$

$3x+4y+x-2y=4x+2y$

If $A=5{p}^{2}-3{q}^{2}+{r}^{2},\,\,B=-2{q}^{2}+3{p}^{2}-4{r}^{2}$ and $C=-7{r}^{2}+3{p}^{2}+2{q}^{2}$ .Find $A+B-C$

$A+B=5{p}^{2}-3{q}^{2}+{r}^{2}+\left(-2{q}^{2}+3{p}^{2}-4{r}^{2}\right)$

$=5{p}^{2}-3{q}^{2}+{r}^{2}-2{q}^{2}+3{p}^{2}-4{r}^{2}$

$=8{p}^{2}-5{q}^{2}-3{r}^{2}$

$A+B-C=8{p}^{2}-5{q}^{2}-3{r}^{2}-\left(-7{r}^{2}+3{p}^{2}+2{q}^{2}\right)$

$=8{p}^{2}-5{q}^{2}-3{r}^{2}+7{r}^{2}-3{p}^{2}-2{q}^{2}$

$=5{p}^{2}-7{q}^{2}+4{r}^{2}$

Simplify: $\left({x}^{2}-{y}^{2}+2xy+1\right)-\left({x}^{2}+{y}^{2}+4xy-5\right)$

Given,

$\left({x}^{2}-{y}^{2}+2xy+1\right)-\left({x}^{2}+{y}^{2}+4xy-5\right)$

$={x}^{2}-{y}^{2}+2xy+1-{x}^{2}-{y}^{2}-4xy+5$

$=-2{y}^{2}-2xy+6$

Find the sum of $-xyz,\,-2xyz,\,7xyz,\,-xyz$

Given,

$-xyz+\left(-2xyz\right)+7xyz+\left(-xyz\right)$

$=(-xyz-2xyz)+(7xyz-xyz)$

$=-3xyz+6xyz$

$=3xyz$

what must be added to $2{x}^{2}{y}^{2} - 5x +7$ to get the sum $-5{x}^{2}{y}^{2} +7x-9$ ?

Let

$a$ be added to

$2{x}^{2}{y}^{2}-5x+7$ to get the sum

$-5{x}^{2}{y}^{2}+7x-9$

$\Rightarrow a+\left(2{x}^{2}{y}^{2}-5x+7\right)=-5{x}^{2}{y}^{2}+7x-9$

$\Rightarrow a=-5{x}^{2}{y}^{2}+7x-9-\left(2{x}^{2}{y}^{2}-5x+7\right)$

$\Rightarrow a=-5{x}^{2}{y}^{2}+7x-9-2{x}^{2}{y}^{2}+5x-7$

$\Rightarrow a=-7{x}^{2}{y}^{2}+12x-16$

Thus,

$-7{x}^{2}{y}^{2}+12x-16$ is to be added to get the sum.

Subtract $1-{p}^{3}$ from $2{p}^{3}-3p{q}^{2}-2{q}^{3}$

We have,

$2{p}^{3}-3p{q}^{2}-2{q}^{3}-\left(1-{p}^{3}\right)$

$=2{p}^{3}-3p{q}^{2}-2{q}^{3}-1+{p}^{3}$

$=2{p}^{3}+{p}^{3}-3p{q}^{2}-2{q}^{3}-1$

$=3{p}^{3}-3p{q}^{2}-2{q}^{3}-1$

Divide and Simplify:
$4a^{2}+12ab+9b^{2}-25c^{2}$ by $2a+3b+5c$ .

$4{a}^{2}+12ab+9{b}^{2}-25{c}^{2}$

$={\left(2a\right)}^{2}+2\times 2a\times 3b+{\left(3b\right)}^{2}-{\left(5c\right)}^{2}$

$={\left(2a+3b\right)}^{2}-{\left(5c\right)}^{2}$

$=\left(2a+3b+5c\right)\left(2a+3b-5c\right)$

$4{a}^{2}+12ab+9{b}^{2}-25{c}^{2}\div \left(2a+3b+5c\right)$

$=\dfrac{\left(2a+3b+5c\right)\left(2a+3b-5c\right)}{\left(2a+3b+5c\right)}=2a+3b-5c$

Solve: $5\left(x-3\right)-7\left(6-x\right)+3=24-3\left(8-x\right)$

given,

$5\left(x-3\right)-7\left(6-x\right)+3=24-3\left(8-x\right)$

$\Rightarrow \,5x-15-42+7x+3=24-24+3x$

$\Rightarrow\,12x-54=3x$

$\Rightarrow\,12x-3x=54$

$\Rightarrow 9x=54$

$\Rightarrow x=\dfrac{54}{9}=6$

$\therefore\,x=6$

Find the value of $13pqx-5xpq-19qpx$

given,

$13pqx-5xpq-19qpx$

$=13pqx-5pqx-19pqx$

$=\left(13-5-19\right)pqx$

$=-11pqx$

Solve: $\left(x+1\right)\left(2x-1\right)-5x=\left(2x-3\right)\left(x-5\right)+47$

Given,

$\left(x+1\right)\left(2x-1\right)-5x=\left(2x-3\right)\left(x-5\right)+47$

$\Rightarrow x\left(2x-1\right)+1\left(2x-1\right)-5x=2x\left(x-5\right)-3\left(x-5\right)+47$

$\Rightarrow 2{x}^{2}-x+2x-1-5x=2{x}^{2}-10x-3x+15+47$

$\Rightarrow 2{x}^{2}-4x-1=2{x}^{2}-13x+62$

$\Rightarrow -4x-1=-13x+62$

$\Rightarrow -4x+13x=62+1$

$\Rightarrow 9x=63$

$\Rightarrow x=\dfrac{63}{9}=7$

$\therefore\,x=7$

From the sum of $5{a}^{2} - 3a +2$ and $-8{a}^{2} + 5a +6$ , subtract ${a}^{2} + 7a -4$

Sum of

$5{a}^{2}-3a+2+\left(-8{a}^{2}+5a+6\right)$

$=5{a}^{2}-3a+2-8{a}^{2}+5a+6$

Sum

$=-3{a}^{2}+2a+8$

Subtract

${a}^{2}+7a-4$

$=-3{a}^{2}+2a+8-\left({a}^{2}+7a-4\right)$

$=-3{a}^{2}+2a+8-{a}^{2}-7a+4$

$=-4{a}^{2}-5a+12$

By how much does ${x}^{2}-2ax+5$ exceed $3{x}^{2} - 5x+6$ ?

Required solution

$=3{x}^{2}-5x+6-\left({x}^{2}-2ax+5\right)$

$=2{x}^{2}-\left(5+2a\right)x+1$

$\therefore\,\,{x}^{2}-2ax+5$ exceeds

$3{x}^{2}-5x+6$ by

$2{x}^{2}-\left(5+2a\right)x+1$

From the sum ${ 2y }^{ 2 }+3yz,-{ y }^{ 2 }yz-{ z }^{ 2 }$ and $yz+{ 2z }^{ 2 }$ subtract the sum $3y^{ 2 }+{ z }^{ 2 }$ and $y^{ 2 }+yz+{ z }^{ 2 }.$

Sum of

$2y^2+3yz,-y^2-yz-z^2$ and

$yz-2z^2$

$=2y^2+3yz-y^2-yz-z^2+yz+2z^2$

$=y^2+z^2+3yz$

And sum

$=3y^2-z^2,(-y^2+yz+z^2)$

$=3y^2-z^2-y^2+yz+z^2$

$=2y^2+yz$

We need to find:

$(y^2+z^2+3yz)-(2y^2+yz)$

$=y^2+z^2+3yz-2y^2-yz$

$=-y^2+z^2+2yz$

$=-y^2+2yz+z^2$

Add the following
(i) $x-3y-2z$
$5x+7y-8z$
$3x-2y+5z$

Given

$x-3y-2z, 5x+7y-8z, 3x-2y+5z$

$\Rightarrow$ On adding given data

$\Rightarrow x-3y-2z+5x+7y-8z+3x-2y+5z$

$\Rightarrow x+5x+3x-3y+7y-2y-2z-8z+5z$

$\Rightarrow 9x+2y-5z$ .

1204169_1393397_ans_ca1fc8b83e01468188838583b2e00c15.JPG

If $A=4x^ {2}+y^ {2}-6xy;B=3y^ {2}+12x^ {2}+8xy$ $C=6x^ {2}+8y^ {2}+6xy$ find $A+B-C$

$A+B-C=4{x}^{2}+{y}^{2}-6xy+3{y}^{2}+12{x}^{2}+8xy-\left(6{x}^{2}+8{y}^{2}+6xy\right)$

$=\left(4{x}^{2}+12{x}^{2}\right)+\left({y}^{2}+3{y}^{2}\right)+\left(8xy-6xy\right)-\left(6{x}^{2}+8{y}^{2}+6xy\right)$

$=16{x}^{2}+4{y}^{2}+2xy-\left(6{x}^{2}+8{y}^{2}+6xy\right)$

$=16{x}^{2}+4{y}^{2}+2xy-6{x}^{2}-8{y}^{2}-6xy$

$=10{x}^{2}-4{y}^{2}-4xy$

$=2\left(5{x}^{2}-2xy-2{y}^{2}\right)$

Add : $-7mn+5,12mn+2,9mn-8,-2mn-3$

If we add

$-7mn+5,12mn+2,9mn-8,-2mn-3$ then we get,

$(-7mn+5)+(12mn+2)+(9mn-8)+(-2mn-3)$

$=12mn-4$ .

Subtract:
$(x^{2}-2xy+3y^{2})$ from $(4x^{2}-5xy+8y^{2})$

$4x^2-5xy+8y^2-(x^2-2xy+3y^2)\\4x^2-5xy+8y^2-x^2+2xy-3y^2\\3x^2-3xy+5y^2$

Add :
$2x(z-x-y)$ and $2y(z-y-x)$

$2x(z-x-y) = 2xz-2x^{2}-2xy$

$2y(z-y-x) = 2yz-2y^{2}-2xy$

So,

$2x(z-x-y) + 2y(z-y-z) = 2xz-2x^{2}-4xy+2yz-2y^{2}$

Solve the following
$15a(2p-3q)-10b(2p-3q)$

$15a (2p-3q)-10b(2p-3q)$
Take out common in both terms,
The,

$5(2p-3q)[3a-2b]$
Therefore,HCF of

$15a(2p-3q) \quad and \quad 10b(2p-3q) \quad is \quad 5(2p-3q)$ .

If $A=x+2,$ and $B=2x$ then,find $A+2B=$

$A+2B=x+2+2\left(2x\right)$

$\Rightarrow A+2B=x+2+4x$

$\therefore A+2B=5x+2$

What should be added to $1+2x-3x^ {2}$ to get $x^ {2}+x-1$

1306843_1387329_ans_33920ed522504e6c8d8711161bfaa2f9.JPG

Subtract: $(a-b)$ from $(a+b)$

$\left(a+b\right)-\left(a-b\right)$

$=a+b-a+b=2b$

If $\dfrac{a}{b} = \dfrac{5}{3},$ then find the value of $\dfrac{{{a^3} - {b^3}}}{{{b^3}}}.$

Given,

$\dfrac{a}{b} = \dfrac{5}{3},$ .....(1).

Then,

$\dfrac{{{a^3} - {b^3}}}{{{b^3}}}$

$=\left(\dfrac{a}{b}\right)^3-1$

$=\dfrac{5^3}{3^3}-1$ [ Using (1)]

$=\dfrac{125-27}{27}$

$=\dfrac{98}{27}$ .

Simplify :-
$\left( m+n \right) ^{ 2 }-\left( m-n \right) ^{ 2 }$

${\left(m+n\right)}^{2}-{\left(m-n\right)}^{2}$ is of the form

${a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)$

where

$a=m+n,\,\,b=m-n$

$=\left(m+n+m-n\right)\left(m+n-m+n\right)$

$=2m\times 2n=4mn$

Subtract: $5{x^2} - 9xy - 3{y^2}$ from $3{x^2} - 5xy - 2{y^2}$

If we subtract

$5{x^2} - 9xy - 3{y^2}$ from

$3{x^2} - 5xy - 2{y^2}$ then we get,

$(3x^2-5xy-2y^2)-(5x^2-9xy-3y^2)$

$=3x^2-5xy-2y^2-5x^2+9xy+3y^2$

$=-2x^2+4xy+y^2$ .

Subtract the second polynominal from the first polynomial:
${ 5x }^{ 2 }-2y+9;{ 3x }^{ 2 }+5y-7$

$=3{x}^{2}+5y-7-\left(5{x}^{2}-2y+9\right)$

$=3{x}^{2}+5y-7-5{x}^{2}+2y-9$

$=-2{x}^{2}+7y-16$

Simplify
Subtract 3 kg (p - q) from 2 kg (p + q)

$2kg\left(p+q\right)-3kg\left(p-q\right)$

$=kg\left(2p+2q-2p+3q\right)$

$=5qkg$

Simplify : Subtract $3pq ( p-q )$ from $2pq ( p+q )$

$2pq\left(p+q\right)-3pq\left(p-q\right)$

$=pq\left(2p+2q-3p+3q\right)$

$=pq\left(5q-p\right)$

$=5p{q}^{2}-{p}^{2}q$

Write the degree of the following polynomial:
$3xy+7{ xy }^{ 2 }-8{ xy }^{ 3 }+7{ y }^{ 2 }{ x }^{ 2 }$

The degree of a polynomial is the highest of the degrees of its monomials with non-zero coefficients.

Degree of

first term

$=xy=2$

Second term

$=x{y}^{2}=1+2=3$

Third term

$x{y}^{3}=1+3=4$

Fourth term

$={x}^{2}{y}^{2}=2+2=4$

$\therefore$ the highest power is

$4$

$\therefore$ Degree of the polynomial

$=4$

Subtract $\dfrac { 7x+14 }{ 3 } -6x$ from $\dfrac { 17-3x }{ 5 } -\dfrac { 4x+2 }{ 3 }$

$\dfrac{17-3x}{5}-\dfrac{4x+2}{3}-\left(\dfrac{7x+14}{3}-6x\right)$

$=\dfrac{17-3x}{5}-\dfrac{4x+2}{3}-\dfrac{7x+14}{3}+6x$

$=\dfrac{17-3x}{5}-\left(\dfrac{4x+2}{3}+\dfrac{7x+14}{3}\right)+6x$

$=\dfrac{17-3x}{5}-\dfrac{4x+2+7x+14}{3}+6x$

$=\dfrac{17-3x}{5}-\dfrac{11x+16}{3}+6x$

$=\dfrac{51x-15-55x-80}{3}+6x$

$=\dfrac{-4x-95}{3}+6x$

$=\dfrac{-4x-95+18x}{3}=\dfrac{14x-95}{3}$

Subract $5 a^{2}-7 a b+5 b^{2}$ from $3 a b-2 a^{2}-2 b^{2}$

Subtracting

$5 a^2-7 a b+5 b^2$ from

$3 a b-2 a^2-2 b^2$ gives

$3 ab -2 a^2-2 b^2-(5 a^2-7 a b+5 b^2)=3 a b+7 a b-2 a^2-5 a^2-2 b^2-5 b^2=10 a b-7 a^2-7 b^2$

Solve : $\left( \dfrac { 1 }{ 4{ ab }^{ 2 }c } \right) ^{ 2 }\div \left( \dfrac { 3 }{ 2{ a }^{ 2 }{ bc }^{ 2 } } \right) ^{ 4 }$

${\left(\dfrac{1}{4a{b}^{2}c}\right)}^{2}\div{\left(\dfrac{3}{2{a}^{2}b{c}^{2}}\right)}^{4}$

$=\dfrac{1}{16{a}^{2}{b}^{4}{c}^{2}}\times \dfrac{16{a}^{4}{b}^{2}{c}^{4}}{81}$

$=\dfrac{1}{81}{a}^{4-2}{b}^{2-4}{c}^{4-2}$

$=\dfrac{1}{81}{a}^{2}{b}^{-2}{c}^{2}$

$=\dfrac{{a}^{2}{c}^{2}}{81{b}^{2}}$

From the sum of 4 + 3x and 5 - 4x + ${ 2x }^{ 2 },$ subtract the sum of ${ 3x }^{ 2 }-5x$ and ${ 2x }^{ 2 }+2x+5.$

1317472_1489801_ans_75c9ab7d1ea340a585fee28e93a3b33c.JPG

Subtract the second expression from the first expression
$6m^{ 3 } + 4m^{ 2 } + 7m+ 3, 3m^{ 3 } + 4$

Given First expression is

$6 m^3+4 m^2+7 m+3$

Second expression is

$3 m^3+4$

Subtracting second expression from first one will give

$6 m^3+4 m^2+7 m+3-3 m^3-4=3 m^3+4 m^2+7 m-1$

What should be added to $1+2x-{ 3 }^{ x }\quad to\quad get\quad { x }^{ 2 }-x-1\quad$

1314627_1488823_ans_0c7b25ee568f4a818f84a420a3d8bf00.JPG

Solve : $(ax+by)^2+(bx-ay)^2$

$(ax+by)^2+(bx-ay)^2$
By expanding the give question,we get,

$(ax)^2+(by)^2+2axby+(bx)^2+(ay)^2-2bxay$

$a^2x^2+b^2y^2+b^2x^2+a^2y^2$
Re-arranging the above we get,

$a^2x^2+a^2y^2+b^2y^2+b^2x^2$
Take out common in all terms,

$a^2(x^2+y^2)+b^2(x^2+y^2)$

$(x^2+y^2)(a^2+b^2)$

Subtract the second expression from the first expression
$x + 2y + z, -x -y -3z$

Given First expression is

$x+2 y+z$

Second expression is

$-x-y-3 z$

Subtracting second expression from first one will give

$=x+2 y+z-(-x-y-3 z)$

$=x+2 y+z+x+y+3 z$

$=2 x+3 y+4 z$

Subtract $a \left(b-5\right)$ from $b \left(5-a\right)$

$b\left(5-a\right)-a\left(b-5\right)$

$=5b-ba-ab+5a$

$=5a+5b-2ab$

Subtract the second expression from the first expression
$x + 2y + z, -x-y-3z$

Given First expression is

$x+2 y+z$

Second expression is

$-x-y-3 z$

Subtracting second expression from first one will give

$x+2 y+z+x+y+3 z=2 x+3 y+4 z$

What should be taken away from ${ 3x }^{ 2 }-{ 4y }^{ 2 }+5xy+20$ to obtain $-{ x }^{ 2 }-{ y }^{ 2 }+6xy+20?$

1317473_1489790_ans_64bab68a42114109b99ac1e660f1060b.JPG

What should be added to ${ x }^{ 2 }-{ y }^{ 2 }+2xy\quad to\quad obtain\quad { x }^{ 2 }+{ y }^{ 2 }+5xy?$

Let the term added=Z term
i.e. Z term

$+(x^2-y^2+2xy)=x^2+y^2+5xy$
Z term=

$(x^2+y^2+5xy)-(x^2-y^2+2xy)$

$=x^2+y^2+5xy-x^2+y^2-2xy$

$=2y^2+3xy$
The required term is

$2y^2+3xy$

Solve : $2x^2-xy+6y-4y+5xy-4x+6x^2+3y$ .

$2x^2-xy+6y-4y+5xy-4x+6x^2+3y$

Rearranging the like terms together,

$\Rightarrow$

$(2x^2+6x^2)+(-xy+5xy)+(6y-4y+3y)-4x$

$\Rightarrow$

$8x^2+4xy+5y-4x$

Evaluate: $\left( 4{ x }^{ 2 }-\dfrac { 1 }{ 5 } x+7 \right) -\left( -2{ x }^{ 2 }-\dfrac { 1 }{ 2 } x+\dfrac { 1 }{ 3 } \right)$

$= 4x^{2} - \dfrac{1}{5} x + 7 + 2x^{2} + \dfrac{1}{2} x- \dfrac{1}{3}$

$= 6x^{2} + \dfrac{3}{10} x + \dfrac{20}{3}$

For $x=\dfrac { 1 }{ 5 }$ and $y=\dfrac { 3 }{ 7 }$ , verify that $\left( x+y \right) =\left( -x \right) +\left( -y \right) .$

$x=\dfrac{1}{5}$ and

$y=\dfrac{3}{7}$

$\Rightarrow$

$-(x+y)=(-x)+(-y)$

$\Rightarrow$

$-\left(\dfrac{1}{5}+\dfrac{3}{7}\right)=\left(-\dfrac{1}{5}\right)+\left(-\dfrac{3}{7}\right)$

$\Rightarrow$

$-\left(\dfrac{7+15}{35}\right)=-\dfrac{1}{5}-\dfrac{3}{7}$

$\Rightarrow$

$-\dfrac{22}{35}=-\dfrac{22}{35}$

$LHS=RHS$

Add : $\displaystyle \left( 3 x ^ { 2 } - \frac { y } { 3 } - 4 \right) + \left( 2 y ^ { 2 } - 4 x ^ { 2 } + \frac { y } { 4 } + 8 \right)$

Given

$(3x^{2}-\dfrac{y}{3}-4)+(2y^{2}-4x^{2}+\dfrac{y}{4}+8)$

$=3x^{2}-\dfrac{y}{3}-4+2y^{2}-4x^{2}+\dfrac{y}{4}+8$

$=-x^{2}+2y^{2}+\dfrac{-y}{12}+4$

$=-x^{2}+2y^{2}-\dfrac{y}{12}+4$

Add $-5p^2, 3p^2, 2p^2$ .

Now, if we add

$-5p^2, 3p^2, 2p^2$ then we get,

$-5p^2+3p^2+2p^2=0$ .

Subtract $-x^{ 2 }+10x-5 \,\,from \,\,5x-10$

Given algebraic expressions are

$-x^2+10x-5$ and

$5x-10$ .

To find out: The result when we subtract

$-x^2+10x-5$ from

$5x-10$

Hence, the difference will be:

$(5x-10)-(-x^2+10x-5)$

$\Rightarrow 5x-10+x^2-10x+5$

$\Rightarrow x^2+5x-10x-10+5$

Subtracting the like terms, we get:

$\Rightarrow x^2-5x-5$

Hence, the result on subtracting

$-x^2+10x-5$ from

$5x-10$ is

$x^2-5x-5$ .

What should be subtracted from $(6a+7b-10)$ to get $(2a-5b)?$

Let us take

$x$ should be subracted

Subtracting

$x$ from

$(6 a+7 b-10)$ gives

$(2 a-5 b)$

$\implies (6 a+7 b-10)- x=2 a-5 b$

$\implies x=(6 a+7 b-10)-(2 a-5 b)$

$=(6 a-2 a)+(7 b+5 b)-10$

$=3 a+12 b-10$

Add the following equations: $4 a ^ { 2 } + 5 b ^ { 2 } + 6 a b ; 3 a b ; 6 a ^ { 2 } - 2 b ^ { 2 } ; 4 b ^ { 2 } - 5 a b$

Adding all the equations

$4 a^2+5 b^2+6 a b,3 a b,6 a^2-2 b^2,4 b^2-5 a b$ gives

$4 a^2+5 b^2+6 a b+3 a b+6 a^2-2 b^2+4 b^2-5 a b =10 a^2+7 b^2+a b$

Solve: From the sum the of $3x^{2}-5x+26$ and $-5x^{2}-8x+1$ subtract $4x^{2}+7x+1$ .

$3{ x }^{ 2 }-5x+26+(-5{ x }^{ 2 }-8x+1)$

$=3{ x }^{ 2 }-5x+26-5{ x }^{ 2 }-8x+1$

$=-2{ x }^{ 2 }-13x+27$

Subtracting

$4{ x }^{ 2 }+7x+1$ from the obtained expression

$\Rightarrow -2{ x }^{ 2 }-13x+27-4{ x }^{ 2 }-7x-1$

$\Rightarrow -6{ x }^{ 2 }-20x+26$

Simplify: $(m+l)+m+1$ .

$\textbf{Step - 1: Open the bracket and combine like terms.}$

$(m+l)+m+1$

$= m+l+m+1$

$=(m+m)+l+1$

$=2m+l+1$

$\textbf{Hence, }\mathbf{\left(m+l\right)+m+1=2m+l+1}$

Subtract : $4pq-5q^{ 2 }-3p^{ 2 }$ from $5p^{ 2 }+3q^{ 2 }-pq$

$4pq-5q^{ 2 }-3p^{ 2 }$ from

$5p^{ 2 }+3q^{ 2 }-pq$

$=(5p^2+3q^2-pq)-(4pq-5q^2-3p^2)$

$=5p^2+3q^2-pq-4pq+5q^2+3p^2$

$=5p^2+3p^2+3q^2+5q^2-pq-4pq$

$=8p^2+8q^2-5pq$

$(a^{2}+b^{2}+2ab)-(a^{2}+b^{2}-2ab)$

$(a^2+b^2+2ab)-(a^2+b^2-2ab)$

$\Rightarrow$

$a^2+b^2+2ab-a^2-b^2+2ab$

$\Rightarrow$

$a^2-a^2+b^2-b^2+2ab+2ab$

$\Rightarrow$

$4ab$

Add:
$(a+b-3),(b-a+3),(a-b+3)$

Adding

$a+b-3,b-a+3,a-b+3$ gives

$a+b-3+b-a+3+a-b+3$

$=(a-a+a)+(b+b-b)+(3-3+3)$

$=a+b+3$

If both x + 1 and x - 1 are factors of ${ ax }^{ 3 }+{ x }^{ 2 }-2x+b$ , find the values of a and b.

Let

$f(x)=ax^3+x^2-2x+b$

$g(x)=x+1$ ,

$q(x)=x-1$

$g(x)$ is a factor of

$f(x)$ , zero of

$g(x)$ will satisfy

$f(x)$

zero of

$g(x):- x+1=0$ ,

$x=-1$

$\Rightarrow f(-1)=0$

$a(-1)^3+(-1)^2-2(-1)+b=0$

$-a+1+2+b=0$

$-a+3+b=0$

$a-b=3$ .......

$(1)$

Also if

$q(x)$ is a factor of

$f(x)$ , zero of

$q(x)$ will satisfy

$f(x)$

zero of

$q(x):- x-1=0$ ,

$x=1$

$\Rightarrow f(1)=0$

$a(1)^3+(1)^2-2(1)+b=0$

$a+1-2+b=0$

$a+b-1=0$

$a+b=1$ ......

$(2)$

Adding equations

$(1)$ and

$(2)$

$a-b+a+b=3+1$

$2a=4$

$a=2$

$\therefore a+b=1$

$\Rightarrow b=-1$

Hence

$a=2$ and

$b=-1$

Give expressions in the following cases.
$y$ is multiplied by $-8$ and then $5$ is added to the result.

$\Rightarrow y\times (-8)=-8y$

$\Rightarrow -8y + 5$

Give expressions in the following cases.
$y$ is multiplied by $5$ and results is subtracted from $16$ .

$\Rightarrow y\times 5=5y$

$\Rightarrow 16 - 5y$ .

Give expressions in the following cases.
$11$ added to $2m$ .

$11$ added to

$2m\Rightarrow 11 + 2m$ .

Give expressions in the following cases.
$y$ is multiplied by $-8$ .

$y$ is multiplied by

$−8\Rightarrow -8y$

Give expressions in the following cases.
$5$ times $y$ from which $3$ is subtracted.

$5$ times

$y$ from which

$3$ is subtracted

$\Rightarrow 5\times y -3\Rightarrow 5y - 3$ .

Give expressions in the following cases.
$5$ times $y$ to which $3$ is added.

$5$ times

$y$ to which

$3$ is added

$\Rightarrow 5\times y -3\Rightarrow 5y - 3$ ..

Give expressions in the following cases.
$y$ is multiplied by $-5$ and results is added from $16$ .

$y$ is multiplied by

$-5$ and results is added from

$16\Rightarrow (-5)\times y+16\Rightarrow -5y + 16$ .

Give expressions in the following cases.
$11$ subtracted from $2m$ .

$11$ subtracted from

$2m\Rightarrow 11-(2\times m)\Rightarrow 11 - 2m$ .

Identify the terms and factors in the expression given.
$1.2ab-2.4b+3.6a$ .

$1.2ab-2.4b+3.6a$
Terms:

$1.2ab,-2.4b, 3.6a$
Factors:

$1.2, a, b; -2.4,b; 3.6, a$ .

Show that
$\left ( \dfrac{x^{a^{2}+b^{2}}}{x^{ab}} \right )^{a+b}\left ( \dfrac{x^{b^{2}+c^{2}}}{x^{bc}} \right )^{b+c}\left ( \dfrac{x^{c^{2}+a^{2}}}{x^{ac}} \right )^{a+c} = x^{2(a^{3}+b^{3}+c^{3})}$

Given,

$\left(\dfrac{x^{a^2+b^2}}{x^{ab}}\right)^{a+b}\left(\dfrac{x^{b^2+c^2}}{x^{bc}}\right)^{b+c}\left(\dfrac{x^{c^2+a^2}}{x^{ca}}\right)^{c+a}$

$=x^{\left(a^2-ab+b^2\right)\left(a+b\right)}x^{\left(b^2-bc+c^2\right)\left(b+c\right)}x^{\left(a^2-ac+c^2\right)\left(a+c\right)}$

$=x^{\left(a^2+b^2-ab\right)\left(a+b\right)+\left(b^2+c^2-bc\right)\left(b+c\right)+\left(c^2+a^2-ac\right)\left(c+a\right)}$

$=x^{2a^3+2b^3+2c^3}$

$=x^{2(a^3+b^3+c^3)}$

Subtract $4pq-5q^2-3p^2$ from $5p^2+3q^2-pq$ .

$5p^2+3q^2-pq-(4pq-5q^2-3p^2)=5p^2+3q^2-pq-4pq+5q^2+3p^2$

$=5p^2+3p^2+3q^2+5q^2-pq-4pq$

$=8p^2+8q^2-5pq$ .

Subtract $-m^2+5mn$ from $4m^2-3mn+8$ .

$4m^2-3mn+8-(-m^2+5mn)=4m^2-3mn+8+m^2-5mn$

$=4m^2+m^2-3mn-5mn+8$

$=5m^2-8mn+8$ .

Subtract $a(b-5)$ from $b(5-a)$ .

$b(5-a)-a(b-5)=5b-ab-ab+5a=5b-2ab+5a$ .

Classify the following polynomials as polynomials in one-variable, two variables etc.:
$x^{2} - xy + 7y^{2}$ .

Given polynomial

$x^{2} - xy + 7y^{2}$ has two variables

$x$ and

$y$ , hence it is a polynomial in two variable.

Add the following algebraic expressions.
$3a^2b, -4a^2b, 9a^2b$ .

If we add

$3a^2, -4a^2b, 9a^2b$ then we get,

$3a^2b-4a^2b+9a^2b$

$=-a^2b+9a^2b$

$=8a^2b$ .

What should be added to $x^2+xy+y^2$ to obtain $2x^2+3xy$ ?

Let p be added to

$x^2+xy+y^2$ to get

$2x^2+3xy$
i.e.,

$x^2+xy+y^2+p=2x^2+3xy$

taking all the terms to RHS except p

$\Rightarrow p=2x^2+3xy-(x^2+xy+y^2)$

$\Rightarrow p=2x^2+3xy-x^2-xy-y^2$

grouping like terms

$\Rightarrow p=2x^2-x^2-y^2+3xy-xy$

on simplifying we get

$\Rightarrow p=x^2-y^2+2xy$
Hence,

$x^2-y^2+2xy$ should be added.

Subtract $-x^2+10x-5$ from $5x-10$ .

$5x-10-(-x^2+10x-5)=5x-10+x^2-10x+5$

$=x^2+5x-10x-10+5=x^2-5x-5$ .

What should be subtracted from $2a+8b+10$ to get $-3a+7b+16$ ?

Let q should be subtracted.
Then, according to question,

$2a+8b+10-q=-3a+7b+16$

$\Rightarrow -q=-3a+7b+16-(2a+8b+10)$

$\Rightarrow -q=-3a+7b+16-2a-8b-10$

$\Rightarrow -q=-3a-2a+7b-8b+16-10$

$\Rightarrow -q=-5a-b+6$

$\Rightarrow q=-(-5a+b+6)$

$\Rightarrow q=5a+b-6$ .

Subtract $-5xy$ from $12xy$ .

If we subtract

$-5xy$ from

$12xy$ we get,

$12xy-(-5xy)$

$=17xy$ .

Subtract $2a^2$ from $-7a^2$ .

1509728_1672669_ans_45f415b9a3964f68ba9d67cd387cb819.jpg

Subtract $2a-b$ from $3a-5b$ .

1510146_1672670_ans_d06ad6f5fc5f4df58ae296a58a0f0f1a.jpg

Write all the terms of the following algebraic expressions: $3x$

Given,

$3x$

We observe that the number of terms used in the expression

$3x$ . The terms is

$3x$

Add the following: $2x^{2}y, -4x^{2}y, 6x^{2}y, -5x^{2}y$

If we add

$2x^2y,-4x^2y, 6x^2y,-5x^2y$ then we get,

$2x^2y-4x^2y+6x^2y-5x^2y$

$=-2x^2y+6x^2y-5x^2y$

$=4x^2y-5x^2y$

$=-x^2y$

Add the following: $-5xy$ and $9xy$

1516163_1676146_ans_2d68624ce2344992b45d45fff759c165.jpg

Add the following: $3x$ and $7x$

$3x+7x=(3+7)x$

$=10x$

Simplify each of the following: $7x^{3}y+9yx^{3}$

$7x^3y+9yx^3=7x^3y+9x^3y=16x^3y$

Add the following: $7abc, -5abc, 9abc, -8abc$

$7abc-5abc+9abc-8abc$

$=2abc+9abc-8abc$

$=11abc-8abc$

$=3abc$

Add $2x^{2}-3x+1$ to the sum of $3x^{2}-2x$ and $3x+7$ .

Sum of

$3x^2 - 2x$ and

$3x+7$

$\Rightarrow 3x^2-2x+3x+7$

$\Rightarrow 3x^2+x+7$
Now add

$2x^2-3x+1$ in the above sum, we get

$(3x^2+x+7)+(2x^2-3x+1)$

$=3x^2+2x^2+x-3x+7+1$

$=5x^2-2x+8$

Subtract: $4xy$ from $-3xy$

If we subtract

$4xy$ from

$-3xy$ the obtained expression will be

$(-3xy-4xy)$

Thus after the subtraction, the required result is

$-7xy$ .

$\implies -3xy-4xy=-7xy$

Add the following expressions: $5x^{3}+7+6x-5x^{2},2x^{2}-8-9x, 4x-2x^{2}+3x^{3},3x^{3}-9x-x^{2}$ and $x-x^{2}-x^{3}-4$

Adding the given expression as follows:

$(5x^3+7+6x-5x^2)+(2x^2-8-9x)+(4x-2x^2+3x^3)+(3x^3-9x-x^2)+(x-x^2-x^3-4)$

$=x^3(5+3+3-1)+x^2(-5+2-2-1-1)+x(6-9+4-9+1)+(7-8-4)$

$=10x^3-7x^2-7x-5$

Subtract: $7a^{2}b$ from $3a^{2}b$

The required subtraction is done as follows,

$(3a^2b)-(7a^2b)$

$=3a^2b-7a^2b$

$=(3-7)a^2b\qquad$ [ Since they are like terms ]

$=-4a^2b$

Hence, the required answer is

$-4a^2b$

Add the following expressions: $a^{4}-2a^{3}b+3ab^{3}+4a^{2}b^{2}+3b^{4}\ , \ -2a^{4}-5ab^{3}+7a^{3}b-6a^{2}b^{2}+b^{4}$

Given expressions are :

$a^{4}-2a^{3}b+3ab^{3}+4a^{2}b^{2}+3b^{4}$ and

$-2a^{4}-5ab^{3}+7a^{3}b-6a^{2}b^{2}+b^{4}$

Required sum

$= (a^{4}-2a^{3}b+3ab^{3}+4a^{2}b^{2}+3b^{4})+(-2a^{4}-5ab^{3}+7a^{3}b-6a^{2}b^{2}+b^{4})$

$= a^{4}-2a^{3}b+3ab^{3}+4a^{2}b^{2}+3b^{4}-2a^{4}-5ab^{3}+7a^{3}b-6a^{2}b^{2}+b^{4}$

Rearranging and collecting like terms,

$=a^4-2a^4-2a^3b+7a^3b+3ab^3-5ab^3+4a^2b^2-6a^2b^2+3b^4+b^4$

$=(1-2)a^4+(-2+7)a^3b+(3-5)ab^3+(4-6)a^2b^2+(3+1)b^4$

$=-a^{4}+52a^{3}b-2ab^{3}-2a^{2}b^{2}+4b^{4}$

Add the following expressions: $x^{3}-2x^{2}y+3xy^{2}-y^{3}\ ,\ 2x^{3}-5xy^{2}+3x^{2}y-4y^{3}$

Given expressions are:

$x^{3}-2x^{2}y+3xy^{2}-y^{3}$ and

$2x^{3}-5xy^{2}+3x^{2}y-4y^{3}$

Required sum

$=(x^{3}-2x^{2}y+3xy^{2}-y^{3})+(2x^{3}-5xy^{2}+3x^{2}y-4y^{3})$

$=x^{3}-2x^{2}y+3xy^{2}-y^{3}+2x^{3}-5xy^{2}+3x^{2}y-4y^{3}$

Combining like terms,

$=x^{3}+2x^{3}-2x^{2}y+3x^{2}y+3xy^{2}-5xy^{2}-y^{3}-4y^{3}$

$=(1+2)x^{3}+(-2+3)x^{2}y+(3-5)xy^{2}-(1+4)y^{3}$

$=3x^{3}+x^{2}y-2xy^{2}-5y^{3}$

Subtract: $-2x$ from $-5y$

$-5y-(-2x)$

$=-5y+2x$

$=2x-5y$

Add $x^{2}+2xy+y^{2}$ to the sum of $x^{2}-3y^{2}$ and $2x^{2}-y^{2}+9$ .

$\textbf{Step 1: Find sum of }\boldsymbol{x^2-3y^2\ \&\ 2x^2-y^2+9}\textbf{ by adding corresponding terms.}$

$\text{Given expressions are,}$

$x^2+2xy+y^2$

$.....(1)$

$x^2-3y^2$

$.....(2)$

$2x^2-y^2+9$

$.....(3)$

$\text{Now, Adding (2) and (3), we get}$

$(x^2-3y^2)+(2x^2-y^2+9)=(1+2)x^2-(3+1)y^2+9$

$=3x^2-4y^2+9$

$\textbf{Step 2: Find sum of }\boldsymbol{3x^2-4y^2+9\ \&\ x^2+2xy+y^2}\textbf{ by adding corresponding terms.}$

$(3x^2-4y^2+9)+(x^2+2xy+y^2)=(3+1)x^2-(4-1)y^2+2xy+9$

$=4x^2-3y^2+2xy+9$

$\textbf{Hence, Addition of all three linear expressions are }\boldsymbol{4x^2-3y^2+2xy+9.}$

Add the following:
$4ab-5bc+7ca$
$-3ab+2bc-3ca$
$5ab-3bc+4ca$

1518136_1676752_ans_af708dc9b49344428fde86aafb2eda2a.jpg

Add the following expressions: $8a-6ab+5b, -6a-ab-8b$ and $-4a+2ab+3b$

Given expressions are:

$8a-6ab+5b, -6a-ab-8b$ and

$-4a+2ab+3b$

Sum

$=(8a-6ab+5b)+( -6a-ab-8b)+(-4a+2ab+3b)$

$=8a-6ab+5b -6a-ab-8b-4a+2ab+3b$

$=(8a-6a-4a)+( -6ab-ab+2ab)+(5b-8b+3b)$

$=-2a-5ab+0b$

$=-2a-5ab$

What should be added to $xy-3yz+4zx$ to get $4xy-3zx+4yz+7$ ?

Let, the polynomial that should be added be

$P$ .

Then according to question,

$(xy-3yz+4zx)+P=(4xy-3zx+4yz+7)\\ \Rightarrow P=(4xy-3zx+4yz+7)-(xy-3yz+4zx)\\\Rightarrow P=4xy-3zx+4yz+7-xy+3yz-4zx\\$

Rearranging and collecting like terms,

$P=4xy-xy-3zx-4zx+4yz+3yz+7\\ \Rightarrow P=3xy-7zx+7yz+7$

Hence,

$3xy-7zx+7yz+7$ should be added to get the required result.

From $p^{3}-4+3p^{2}$ , subtract $5p^{2}-3p^{3}+p-6$

1517937_1676792_ans_4e3a54f31d3e4e79a19d36a40236b1b4.jpg

From $1-5y^{2}$ , take away $y^{3}+7y^{2}+y+1$

1515031_1676799_ans_a8fcc1b21c584fd98cde6fcd36816ff8.jpg

Subtract: $x^{3}+2x^{2}y+6xy^{2}-y^{3}$ from $y^{3}-3xy^{2}-4x^{2}y$

The subtraction is carried out as follows,

$(y^3-3xy^2-4x^2y)-(x^3+2x^2y+6xy^2-y^3)$

$=y^3-3xy^2-4x^2y-x^3-2x^2y-6xy^2+y^3$

Rearranging the like terms together,

$=(y^3+y^3)+(-3xy^2-6xy^2)+(-4x^2y-2x^2y)-x^3$

$=2y^3-9xy^2-6x^2y-x^3$

What should be subtracted from $x^{2}-xy+y^{2}-x+y+3$ to obtain $-x^{2}+3y^{2}-4xy+1$ ?

Let

$'\text{A}'$ be the required expression.

Then, we have

$(x^2-xy+y^2-x+y+3)-\text{A}=-x^2+3y^2-4xy+1$

$\Rightarrow \text{A}=(x^2-xy+y^2-x+y+3)-(-x^2+3y^2-4xy+1)$

$\Rightarrow \text{A}=x^2-xy+y^2-x+y+3+x^2-3y^2+4xy-1$

$\Rightarrow \text{A}=(x^2+x^2)+(-xy+4xy)+(y^2-3y^2)-x+y+2$

$\Rightarrow \text{A}=2x^2+3xy-2y^2-x+y+2$

From $x^{3}-5x^{2}+3x+1$ , take away $6x^{2}-4x^{3}+5+3x$

1516976_1676802_ans_7a9cbb2fc92d4a25bce5b0c366da8daa.jpg

From the sum of $3x^{2} - 5x + 2$ and $-5x^{2} - 8x + 9$ subtract $4x^{2} - 7x + 9$ .

Let us add

$3x^2-5x+2$ ad

$-5x^2-8x+9$ first,

$\begin{array}{l}\left( {3{x^2} - 5x + 2} \right) + \left( { - 5{x^2} - 8x + 9} \right)\\ \Rightarrow \left( {3{x^2} - 5{x^2}} \right) + \left( { - 5x - 8x} \right) + \left( {2 + 9} \right)\\ \Rightarrow - 2{x^2} - 13x + 11\end{array}$

Now, we will subtract

$4x^2-7x+9$ from the above-obtained result,

$\begin{array}{l}\left( { - 2{x^2} - 13x + 11} \right) - \left( {4{x^2} - 7x + 9} \right)\\ \Rightarrow - 2{x^2} - 13x + 11 - 4{x^2} + 7x - 9\\ \Rightarrow \left( { - 2{x^2} - 4{x^2}} \right) + \left( { - 13x + 7x} \right) + \left( {11 - 9} \right)\\ \Rightarrow - 6{x^2} - 6x + 2\end{array}$

Subtract the sum of $13x-4y+7z$ and $-6z+6x+3y$ from the sum of $6x-4y-4z$ and $2x+4y-7$

Subtract: $6x^{3}-7x^{2}+5x-3$ from $4-5x+6x^{2}-8x^{3}$

The required subtraction is done as follows,

$(4-5x+6x^2-8x^3)-(6x^3-7x^2+5x-3)$

$=4-5x+6x^2-8x^3-6x^3+7x^2-5x+3$

Rearrnaging and collecting like terms,

$=4+3-5x-5x+6x^2+7x^2-8x^3-6x^3$

$=(4+3)-(5+5)x+(6+7)x^2-(8+6)x^3$

$=7-10x+13x^2-14x^3$

From $7+x-x^{2}$ , take away $9+x+3x^{2}+7x^{3}$

Here "take away" means subtraction of second polynomial from the first polynomial,

$\left( {7 + x - {x^2}} \right) - \left( {9 + x + 3{x^2} + 7{x^3}} \right) = 7 + x - {x^2} - 9 - x - 3{x^2} - 7{x^3}$

$= - 7{x^3} + \left( { - 3{x^2} - {x^2}} \right) + \left( {x - x} \right) + \left( {7 - 9} \right)$

$= - 7{x^3} - 4{x^2} - 2$

Add:
$4x^2-4xy+4y^2-3, 5+6y^2-8xy+x^2$ and $6-2xy+2x^2-5y^2$

Given:

$4x^2-4xy+4y^2-3, 5+6y^2-8xy+x^2$ and

$6-2xy+2x^2-5y^2$

To add the given terms we have to add coefficient of like terms:

$\therefore (4x^2-4xy+4y^2-3)+( 5+6y^2-8xy+x^2)+(6-2xy+2x^2-5y^2)\\$

$=(4+1+2)x^2+(-4-8-2)xy+(4+6-5)y^2+(-3+5+6)\\$

$=7x^2-14xy+5y^2+8\\$

Write the constant term in $\dfrac{\pi}{2}x^2+7x-\dfrac{2}{5}\pi$ .

$\textbf{Step 1: Identify the term with no variable in equation}$

$\text{Given- Equation is } p(x)=\dfrac{\pi}{2} x^2 + 7x -\dfrac{2\pi}{5}$

$\text{Constant term is the term which no variable with it.}$

$\text{By definition,}$

$\Rightarrow \text{Constant term is } \dfrac{-2\pi}{5}$

$\textbf{Therefore,from given quadratic equation,constant term is } \mathbf{\dfrac{-2\pi}{5} }$

Add:
$3a-3b+4c, 2a+3b-8c, a+6b+c$

Given

$3a-3b+4c, 2a+3b-8c, a6b+c$

To add the given terms we have to add coefficient of like terms:

$\therefore (3a-3b+4c)+( 2a+3b-8c)+( a+6b+c)\\$

$=(3+2+1)a+(-3+3+6)b+(4-8+1)c\\$

$=6a+6b-3c$

Add:
$2x^3-9x^2+8, 3x^2-6x-5, 7x^3-10x+1$ and $3+2x-5x^2-4x^3$

Given:

$2x^3-9x^2+8, 3x^2-6x-5, 7x^3-10x+1$ and

$3+2x-5x^2-4x^3$

To add the given terms we have to add coefficient of like terms:

$\therefore (2x^3-9x^2+8)+( 3x^2-6x-5)+( 7x^3-10x+1)+(3+2x-5x^2-4x^3)\\$

$=(2+0+7-4)x^3+(-9+3+0-5)x^2+(0-6-10+2)x+(8-5+1+3)\\$

$=5x^3-11x^2-14x+7\\$

Add:
$5x-8y+2z, 3z-4y-2x, 6y-z-x$ , and $3x-2z-3y$ .

Given

$5x-8y+2z, 3z-4y-2x, 6y-z-x$ and

$3x-2z-3y$

To add the given terms we have to add coefficient of like terms:

$\therefore (5x-8y+2z)+( 3z-4y-2x)+( 6y-z-x)+(3x-2z-3y)\\$

$=(5-2-1+3)x+(-8-4+6-3)y+(2+3-1-2)z\\$

$=5x-9y+2z\\$

If $P=a^{2}-b^{2}+2ab, Q=a^{2}+4b^{2}-6ab,\ R=b^{2}+b,\ S=a^{2}-4ab$ and $T=-2a^{2}+b^{2}-ab+a$ . Find $P+Q+R+S-T$ .

Given

$P=a^{2}-b^{2}+2ab$

$Q=a^{2}+4b^{2}-6ab$

$R=b^{2}+b$

$S=a^{2}-4ab$ and

$T=-2a^{2}+b^{2}-ab+a$

$\therefore P+Q+R+S-T$

$=(a^{2}-b^{2}+2ab)+(a^{2}+4b^{2}-6ab)+(b^{2}+b)+(a^{2}-4ab)-(-2a^{2}+b^{2}-ab+a)$

$=a^{2}-b^{2}+2ab+a^{2}+4b^{2}-6ab+b^{2}+b+a^{2}-4ab+2a^{2}-b^{2}+ab-a$

Rearranging and collecting like terms,

$=a^{2}+a^{2}+a^{2}+2a^{2}-b^2+4b^2+b^2-b^2+2ab-6ab-4ab+ab+b-a$

$=(1+1+1+2)a^2+(-1+4+1-1)b^2+(2-6-4+1)ab+b-a$

$=5a^2+3b^2-7ab+b-a$

Hence,

$P+Q+R+S-T=5a^2+3b^2-7ab+b-a$

How much is $x-2y+3z$ greater than $3x+5y-7$ ?

Let say,

$x−2y+3z$ is greater than

$3x+5y−7$ by

$P$ .

Then according to question,

$( x−2y+3z)=P+(3x+5y−7)\\ \Rightarrow P=( x−2y+3z)-(3x+5y−7)\\\Rightarrow P=x−2y+3z-3x-5y+7\\$

Rearranging and collecting like terms,

$\Rightarrow P=x-3x-2y-5y+3z+7\\ \Rightarrow P=-2x-7y+3z+7$

Hence,

$x−2y+3z$ is greater than

$3x+5y−7$ by

$-2x-7y+3z+7$ .

If $P=7x^{2}+5xy-9y^{2}\ ,\ Q=4y^{2}-3x^{2}-6xy$ and $R=-4x^{2}+xy+5y^{2}$ , show that $P+Q+R=0$ .

Given,

$P=7x^{2}+5xy-9y^{2}$

$Q=4y^{2}-3x^{2}-6xy$ and

$R=-4x^{2}+xy+5y^{2}$

To prove :

$P+Q+R=0$

$\therefore P+Q+R=(7x^{2}+5xy-9y^{2})+(4y^{2}-3x^{2}-6xy)+ (-4x^{2}+xy+5y^{2})\\=7x^{2}+5xy-9y^{2}+4y^{2}-3x^{2}-6xy-4x^{2}+xy+5y^{2}$

Rearranging and collecting like terms,

$=7x^{2}-3x^{2}-4x^{2}+5xy-6xy+xy-9y^{2}+4y^{2}+5y^{2}\\=(7-3-4)x^2+(5-6+1)xy+(-9+4+5)y^2\\=(0)x^2+(0)xy+(0)y^2=0$

Hence, proved.

How much is $x^{2}-2xy+3y^{2}$ less than $2x^{2}-3y^{2}+xy$ ?

Required expression

$=(2x^2-3y^2+xy)-(x^2-2xy+3y^2)$

$=2x^2-3y^2+xy-x^2+2xy-3y^2$

$=(2x^2-x^2)+(-3y^2-3y^2)+(xy+2xy)$

$=x^2-6y^2+3xy$

Add:
$7x, -3x, 5x, -x, -3x$

Given terms are,

$7x, -3x,5x,-x,-3x$

All the terms have same variable

$x$ .

So, while adding their numeric coefficients will be added.

To add the given terms :

$7x+( -3x)+5x+(-x)+(-3x)\\$

$=7x -3x+5x-x-3x\\$

$=(7-3+5-1-3)\ x\\$

$=5x$

Hence, the sum is

$5x$

Add:
$8ab, -5ab, 3ab, -ab$

Given

$8ab, -5ab, 3ab, -ab$

All terms have same variable

$ab$ .

So, simply their numeric coefficients will be added.

To add the given terms :

$8ab+(-5ab)+3ab+(-ab)\\$

$=8\ ab$

$-5\ ab$

$+3\ ab-ab$

$=(8-5+3-1)\ ab\\$

$=5\ ab$

Add the following:
$2x^2 - 3xy + y^2$
$-7x^2 -5xy - 2y^2$
$4x^2 + xy -6y^2$
-------------------------------

$2x^2 - 3xy + y^2$

$-7x^2 -5xy - 2y^2$

$4x^2 + xy -6y^2$

-------------------------------

$-x^2 -7xy -7y^2$

--------------------------------

Subtract:
$2a-5b+2c-9$ from $3a-4b-c+6$

We have to subtract

$2a-5b+2c-9$ from

$3a-4b-c+6$

So, simply numeric coefficients of same varaibles will be subtracted.

$\therefore (3a-4b-c+6)-(2a-5b+2c-9)\\$

$= (3-2)a+(-4-(-5))b+(-1-2)c+(6-(9))\\$

$= a+b-3c+15\\$

Add the following:
$x 3y 2z$
$5x + 7y z$
$-7x 2y + 4z$
--------------------------

$x – 3y – 2z$

$5x + 7y – z$

$-7x – 2y + 4z$

--------------------------

$-x + 2y + z$

---------------------------

Add the following:
$\ \ 4xy - 5yz - 7z$
$– 5xy +2yz + zx$
$– 2xy -3yz +3zx$
------------------------------

$\ \ 4xy - 5yz - 7z$

$– 5xy +2yz + zx$

$– 2xy -3yz +3zx$

------------------------------

$-3xy -6yz - 3zx$

---------------------------------

Add the following:
$m^2 -4m +5$
$-2m^2 + 6m -6$
$-m^2 - 2m -7$
--------------------------

$m^2 -4m +5$

$-2m^2 + 6m -6$

$-m^2 - 2m -7$

--------------------------

$-2m^2 + 0m -8$

------------------------------

$\Rightarrow$

$-2m^2 - 8$

From the sum of $3x^2 - 5x + 2$ and $-5x^2 -8x + 6,$ subtract $4x^2 -9x + 7.$

Sum of

$3x^2 - 5x + 2$ and

$-5x^2 -8x + 6$

$=(3x^2 - 5 + 2) + (-5x^2 - 8x +6)$

Rearranging and collecting the like terms, we get:

$=3x^2 -5x^2 -5x -8x +2 + 6$

$= (3 - 5)x^2 - (5 - 8) x + 2 + 6$

$= - 2x^2 - 13x + 8$

Now Subtract

$4x^2 -9x + 7$ from

$-2x^2 - 13x +8$

$(-2x^2 - 13x +8 )-(4x^2 -9x + 7)$

$= -2x^2 -13x + 8 - 4x^2 + 9x - 7$

$= -2x^2 -13x + 8 -4x^2 - 9x - 7$

Rearranging and collecting the like terms, we get:

$= -2x^2 - 4x^2 - 13x - 9x + 8 - 7$

$= (-2 - 4)x^2 - (13 + 9)x + 8 - 7$

$=-6x^2 - 4x + 1$

Add:
$x^3+y^3-z^3+3xyz,\ \ -x^3+y^3+z^3 -6xyz,\ \ x^3-y^3-z^3-8xyz$

The sum of the given expressions

$=(x^3+y^3-z^3+3xyz)+(-x^3+y^3+z^3 -6xyz)+(x^3-y^3-z^3-8xyz)$

Rearranging and collecting like terms

$=(x^3 - x^3 +x^3)+(y^3 + y^3 -y^3)+(-z^3+z^3-z^3)+(3xyz-6xyz-8xyz)$

$=x^3+y^3-z^3-11xyz$

Add the following expressions:
$5x, 7x, -6x$

We add the like terms,

$=5x+7x+(−6x)$

$=12x+(-6x)$

$=6x$

Add the following expressions:
$x-3y+4z, y-2x-8z, 5x-2y-3z$

Required sum,

$=(x-3y+4z)+(y-2x-8z)+(5x-2y-3z)$
Collecting like terms,

$-x-2x+5x-3y+y-2y+4z-8z-3z$
Adding like terms,

$=(1-2+5)x+(-3+1-2)y+(4-8-3)z$

$=4x-4y-7z$

If $A = 7x^2 + 5xy -9y^2, B = -4x^2 + xy + 5y^2$ and $C = 4y^2 - 3x^2 - 6xy$ then show that $A +B+C= 0.$

Given:

$A$

$= 7x^2 + 5xy - 9y^2$

$B$

$= 4x^2 + xy + 5y^2$

$C$

$= 4y^2 - 3x^2 - 6xy$

Substitute the value of

$A,\ B$ and

$C$ in

$A + B + C,$

$A + B + C =$

$(7x^2 + 5xy - 9y^2) + (-4x^2 + xy + 5y^2) + (4y^2 - 3x^2 - 6xy)$

$= 7x^2 + 5xy -9y^2 - 4x^2 + xy +5y^2 + 4y^2 -3x^2-6xy$

$= 7x^2 - 4x^2 - 3x^2 + 5xy + xy - 6xy -9y^2 +5y^2 +4y^2$

$= (7-4-3)x^2 + (5 +1 - 6) xy - (9 - 5 - 4) y^2$

$= 0 x^2 + 0xy - 0y^2$

$= 0$

Hence, proved

$A + B + C = 0 .$

Add the following expressions:
$\dfrac34 x^{2}, \ 5x^{2}, \ -3x^{2}, -\dfrac14 x^{2}$

Given terms are

$\dfrac34 x^{2}, \ 5x^{2}, \ -3x^{2}, -\dfrac14 x^{2}$

All the above terms have the same literal factors.

Hence, they are all like terms.

Now, adding all the terms, we get:

$\dfrac34 x^{2}+ 5x^{2}+ (-3x^{2})+\left(-\dfrac14 x^{2}\right)$

$=\dfrac34 x^{2}-\dfrac14 x^{2}+ 5x^{2}-3x^{2}$

$=\left(\dfrac{3-1}{4}\right)x^{2}+2x^{2}$

$=\dfrac{2}{4}x^{2}+2x^{2}$

$=\dfrac{1}{2}x^{2}+2x^{2}$

$=\left(\dfrac{1+4}{2}\right)x^{2}$

$=\dfrac{5}{2}x^{2}$

Hence, the sum of the given terms is

$\dfrac{5}{2}x^{2}$ .

Add the following expressions:
$2x^{2}-3y^{2}, 5x^{2}+6y^{2}, -3x^{2}-4y^{2}$

Required sum,

$=(2x^{2}-3y^{2})+(5x^{2}+6y^{2})+(-3x^{2}-4y^{2})$
Collecting like terms,

$=2x^{2}+5x^{2}-3x^{2}-3y^{2}+6y^{2}-4y^{2}$
Adding like terms,

$=(2+5-3)x^{2}+(-3+6-4)y^{2}$

$=4x^{2}-y^{2}$

Add the following expressions:
$5a^{2}b, -8a^{2}b, 7a^{2}b$

In the above questions terms having the same literal factors are like terms.

Now add the like terms

$=5a^{2}b-8a^{2}b+7a^{2}b$

$=12a^{2}b-8a^{2}b$

$=4a^{2}b$

Add the following expressions:
$\dfrac32x^{3}-\dfrac14x^{2}+\dfrac53,\hspace{2mm} \dfrac{-5}4x^{3}+ \dfrac35x^{2}-x+\dfrac15,\hspace{2mm} -x^{2}+\dfrac38x-\dfrac8{15}$

The required sum of the given expressions

$=\left(\dfrac32x^{3}-\dfrac14x^{2}+\dfrac53 \right)+\left( \dfrac{-5}4x^{3}+\dfrac35x^{2}-x+\dfrac15\right)+\left( -x^{2}+\dfrac38x-\dfrac8{15}\right)$

Rearranging and collecting like terms,

$=\left(\dfrac32x^{3}-\dfrac54x^{3}\right)+\left(\dfrac14x^{2}+\dfrac35x^{2}-x^{2}\right)+ \left( -x+\dfrac38x \right)+\left(\dfrac53+\dfrac15-\dfrac8{15}\right)$

$=\left(\dfrac32-\dfrac54\right)x^{3}+\left(-\dfrac14+\dfrac35-1\right)x^{2}+ \left(-1+\dfrac38 \right)x+\left(\dfrac53+\dfrac15-\dfrac8{15}\right)$

$=\dfrac14x^{3}-\dfrac{13}{20}x^{2}-\dfrac58x+\dfrac43$

Add the following expressions:
$\dfrac85x+\dfrac{11}7y+\dfrac94xy,\hspace{2mm} \dfrac{-3}2x-\dfrac53y-\dfrac95xy$

The required sum of the given expressions

$=\left(\dfrac85x+\dfrac{11}7y+\dfrac94xy\right)+\left(\dfrac{-3}2x-\dfrac53y-\dfrac95xy\right)$
Rearranging and collecting like terms,

$=\left(\dfrac85x-\dfrac32x\right)+\left(\dfrac{11}7y-\dfrac53y\right)+\left(\dfrac94xy-\dfrac95xy\right)$

$=\left(\dfrac85-\dfrac32\right)x+\left(\dfrac{11}7-\dfrac53\right)y+\left(\dfrac94-\dfrac95\right)xy$

$=\dfrac1{10}x-\dfrac2{21}y+\dfrac9{20}xy$

Add the following expressions:
$5x-2x^{2}-8, 8x^{2}-7x-9, 3+7x^{2}-2x$

Required sum,

$=(5x-2x^{2}-8)+(8x^{2}-7x-9)+(3+7x^{2}-2x)$
Collecting like terms,

$=2x^{2}+8x^{2}+7x^{2}+5x-7x-2x-8-9+3$

$=(-2+8+7)x^{2}+(5-7-2)x+(-8-9+3)$

$=13x^{2}-4x-14$

Add the following expressions:
$(2/3)a-(4/5)v+(3/5)c, -(3/4)a-(5/2)b+(2/3)c, (5/2)a+(7/4)b-(5/6)c$

Required sum,

$=[(2/3)a-(4/5b)+(3/5)c]+[-(3/4)a-(5/2)b+(2/3)c]+[(5/2)a+(7/4)b-(5/6)c]$
Collecting like terms,

$=(2/3)a-(3/4)a+(5/2)a-(4/5)b-(5/2)b+(7/4)b+(3/5)c+(2/3)c-(5/6)c$

$=[(2/3)-(3/4)+(5/2)]a+[(-4/5)-(5/2)+(7/4)]b+[(3/5)+(2/3)-(5/6)]c$

$=[(8-9+30)/12)a]+[(16-15+30)/20)b]+[(18+20-25)/30)c]$

$=(29/12)a+(31/20)b+(13/30)c$

Identify the terms and their factors in the algebraic expressions given below: $xy+2x^2y^2$

$xy+2x^2y^2$
term is

$xy,2x^2y^2$
Factor of

$xy=x,y$
Factor of

$2x^2y^2=2,x,x,y,y$

Subtract:
$(x^{2}-y^{2})$ from $(2x^{2}-3y^{2}+6xy)$ .

We know that,

The difference of two like terms is a like term whose coefficient is the difference of the numerical coefficient of the two like terms.

$\therefore$ The required difference

$=(2x^{2}-3y^{2}+6xy)-(x^{2}-y^{2})$

$=2x^{2}-3y^{2}+6xy-x^{2}+y^{2}$
Rearranging and collecting like terms

$=(2x^{2}-x^{2})+(-3y^{2}+y^{2})+6xy$

$=(2-1)x^{2}+(-3+1)y^{2}+6xy$

$=(1)x^{2}+(-2)y^{2}+6xy$

$=x^{2}-2y^{2}+6xy$

Subtract:
$(a^{2}+b^{2}-2ab)$ from $(a^{2}+b^{2}+2ab)$

We know that,

The difference of two like terms is a like term whose coefficient is the difference of the numerical coefficient of the two like terms.

$\therefore$ The required difference

$=(a^{2}+b^{2}+2ab)-(a^{2}+b^{2}-2ab)$

$=a^{2}+b^{2}+2ab-a^{2}-b^{2}+2ab$

Rearranging and collecting like terms

$=(1-1)a^{2}+(1-1)b^{2}+(2+2)ab$

$=(0)a^{2}+(0)b^{2}+(4)ab$

$=4ab$

Subtract:
$(x-y)$ from $(4y-5x)$

The required difference

$=(4y-5x)-(x-y)$

$=4y-5x-x+y$
Rearranging and collecting like terms

$=(-5x-x)+(4y+y)$

$=(-5-1)x+(4+1)y$

$=-6x+5y$

Subtract $(2a-3b+4c)$ from the sum of $(a+3b-4c), \hspace{1mm}(4a-b+9c)$ and $(-2b+3c-a)$ .

First we find the sum of

$(a+3b-4c), (4a-b+9c)$ and

$(-2b+3c-a)$

$\therefore \text{The required sum}=(a+3b-4c)+(4a-b+9c)+(-2b+3c-a)$

$=(a+3b-4c+4a-b+9c-2b+3c-a)$
Rearranging and collecting like terms

$=(a+4a-a)+(3b-b-2b)+(-4c+9c+3c)$

$=(1+4-1)a+(3-1-2)b+(-4+9+3)c$

$=4a+(0)b+8c$

$=4a+8c$

Then subtract

$(2a-3b+4c)$ from the above sum

$\therefore\text{The required difference }=(4a+8c)-(2a-3b+4c)$

$=4a+8c-2a+3b-4c$
Rearranging and collecting like terms

$=(4a-2a)+(3b)+(8c-4c)$

$=2a+3b+4c$

Subtract:
$(x-y+3z)$ from $(2z-x-3y)$

We know that,

The difference of two like terms is a like term whose coefficient is the difference of the numerical coefficient of the two like terms.

$\therefore$ The required difference

$=(2z-x-3y)-(x-y+3z)$

$=2z-x-3y-x+y-3z$
Rearranging and collecting like terms

$=(2z-3z)+(-x-x)+(-3y+y)$

$=(2-3)z+(-1-1)x+(-3+1)y$

$=-z-2x-2y$

Subtract:
$x^{2}$ from $-3x^{2}$

We know that the difference of two like terms is a like term whose coefficient is the difference of the numerical coefficient of the two like terms.

$\therefore$ The required difference

$=(-3x^2)-(x^2)$

$=-3x^2-x^2$

$=(-3-1)x^2$

$=-4x^2$

Simplify the following by combining the like terms and then write whether the expression is a monomial, a binomial or a trinomial.
${x}^{4}+3{x}^{3}y+3{x}^{2}{y}^{2}-3{x}^{3}y-3x{y}^{3}+{y}^{4}-3{x}^{2}{y}^{2}$

$\textbf{Step 1:Combine the like terms and simplify}$

$\begin{aligned}&x^{4}+3 x^{3} y+3 x^{2} y^{2}-3 x^{3} y-3 x y^{3}+y^{4}-3 x^{2} y^{2} \\&=x^{4}+y^{4}-3 x y^{3}+\left(3 x^{3} y-3 x^{3} y\right)+\left(3 x^{2} y^{2}-3 x^{2} y^{2}\right) \\&=x^{4}+y^{4}-3 x y^{3}\end{aligned}$

$\textbf{Step 2:Identify the number of terms in the expression}$

$x^{4}, y^{4}$ and

$-3 x y^{3}$ are three unlike terms in the simplified expression.
So, the given expression is a trinomial.

$\textbf{Hence, } {x}^{4}+3{x}^{3}y+3{x}^{2}{y}^{2}-3{x}^{3}y-3x{y}^{3}+{y}^{4}-3{x}^{2}{y}^{2} \textbf{ will result in a trinomial.}$

Sum or difference of two like terms is _____

Consider two like terms

$10 x^{2} y$ and

$16 x^{2} y$

$\textbf{Part 1: Finding sum of two like terms.}$

$\bullet$ The sum of

$10 x^{2} y$ and

$16 x^{2} y=10 x^{2} y+16 x^{2} y=26 x^{2} y$

$\bullet \ 10 x^{2} y, 16 x^{2} y$ , and

$26 x^{2} y$ have same algebraic factors

$x^{2} y$ so they are like terms.Therefore, the sum of two like terms gives a like term.

$\textbf{Part 2: Finding difference of two like terms}$

$\bullet$ The difference of

$10 x^{2} y$ and

$16 x^{2} y=16 x^{2} y-10 x^{2} y=6 x^{2} y$

$\bullet \ 10 x^{2} y, 16 x^{2} y$ , and

$6 x^{2} y$ have same algebraic factors

$x^{2} y$ so they are like terms. Therefore, the difference of two like terms gives a like term.

$\textbf{Hence, the sum or difference of two like terms is a } \underline{\textbf{like term}}$

If $({ x }^{ 2 }{ y }+{y}^{2}+3)$ is subtracted from $(3{ x }^{ 2 }{ y }+2{y}^{2}+5)$ , then coefficient of $y$ in the result is _____

$\textbf{Step 1: Subtracting the given expressions}$

$\begin{aligned}\left(3 x^{2} y+\right.&\left.2 y^{2}+5\right)-\left(x^{2} y+y^{2}+3\right)=3 x^{2} y+2 y^{2}+5-x^{2} y-y^{2}-3 \\&=\left(3 x^{2} y-x^{2} y\right)+\left(2 y^{2}-y^{2}\right)+(5-3) \text { (Grouping like terms) } \\&=2 x^{2} y+y^{2}+2\end{aligned}$

$\textbf{Step 2: Identify the coefficient of the variable y}$
In the expression

$2 x^{2} y+y^{2}+2$ , the term

$2 x^{2} y$ , has variable

$y$ .
We can see that

$2 x^{2}$ is the coefficient of

$y$ , in the term

$2 x^{2} y$ .
Therefore,

$2 x^{2}$ is the coefficient of

$y$ , in

$2 x^{2} y+y^{2}+2$

$\textbf{Hence, the coefficient of y in the given result is } 2x^2$

How much is $3a^2-5ab+7b^2+3$ greater than $2a^2+2ab+5$ ?

Required expression

1772624_1802271_ans_37f2f3bb2d364ff6973ca7592496a3fe.PNG

Simplify the following by combining the like terms and then write whether the expression is a monomial, a binomial or a trinomial.
$3{ x }^{ 2 }{ y }{z}^{2}-3x{ y }^{ 2 }{ z }+{ x }^{ 2 }{ y }{ z }^{ 2 }+7x^{ 2 }{ y }^{ 2 }{ z }$

$\textbf{Step 1:Combine the like terms and simplify}$

$\begin{aligned}&3 x^{2} y z^{2}-3 x y^{2} z+x^{2} y z^{2}+7 x y^{2} z=\left(3 x^{2} y z^{2}+x^{2} y z^{2}\right)+\left(7 x y^{2} z-3 x y^{2} z\right) \\&3 x^{2} y z^{2}-3 x y^{2} z+x^{2} y z^{2}+7 x y^{2} z=4 x^{2} y z^{2}+4 x y^{2} z\end{aligned}$

$\textbf{Step 2:Identify the number of terms in the expression}$

$4 x^{2} y z^{2}$ and

$4 x y^{2} z$ are the two unlike terms in the simplified expression.
So, the given expression is a binomial.

$\textbf{Hence, } 3x^2yz^2 - 3xy^2z + x^2yz^2+ 7xy^2z \textbf{ will result in a binomial.}$

Add the following expressions:
${p}^{2}-7pq-{q}^{2}$ and $-3{p}^{2}-2pq+7{q}^{2}$

$\textbf{Combine the like terms and simplify.}$

$\begin{aligned}&\left(p^{2}-7 p q-q^{2}\right)+\left(-3 p^{2}-2 p q+7 q^{2}\right)=\left(p^{2}-3 p^{2}\right)+(-7 p q-2 p q)+\left(-q^{2}+7 q^{2}\right) \\&=>\left(p^{2}-7 p q-q^{2}\right)+\left(-3 p^{2}-2 p q+7 q^{2}\right)=-2 p^{2}-9 p q+6 q^{2}\end{aligned}$

$\textbf{Hence, the required expression is } -2 p^{2}-9 p q+6 q^{2}$

Simplify the following by combining the like terms and then write whether the expression is a monomial, a binomial or a trinomial.
$50{x}^{3}-21x+107+41{x}^{3}-x+1-93+71x-31{x}^{3}$

$\textbf{Step 1:Combine the like terms and simplify}$

$\begin{aligned}&50 x^{3}-21 x+107+41 x^{3}-x+1-93+71 x-31 x^{3} \\&=\left(50 x^{3}+41 x^{3}-31 x^{3}\right)+(71 x-21 x-x)+(107+1-93) \\&50 x^{3}-21 x+107+41 x^{3}-x+1-93+71 x-31 x^{3}=60 x^{3}+49 x+15\end{aligned}$

$\textbf{Step 2:Identify the number of terms in the expression}$

$60x^3$ ,

$49x$ and

$15$ are the three unlike terms in the simplified expression.
So, the given expression is a trinomial.

$\textbf{Hence, } 50{x}^{3}-21x+107+41{x}^{3}-x+1-93+71x-31{x}^{3} \textbf{ will result in a trinomial.}$

Simplify the following by combining the like terms and then write whether the expression is a monomial, a binomial or a trinomial.
${p}^{3}{q}^{2}r+p{q}^{2}{r}^{3}+3{p}^{2}q{r}^{2}-9{p}^{2}q{r}^{2}$

$\textbf{Step 1:Combine the like terms and simplify}$

$p^{3} q^{2} r+p q^{2} r^{3}+\left(3 p^{2} q r^{2}-9 p^{2} q r^{2}\right)=p^{3} q^{2} r+p q^{2} r^{3}-6 p^{2} q r^{2}$

$\textbf{Step 2:Identify the number of terms in the expression}$

$p^{3} q^{2} r, p q^{2} r^{3}$ and

$-6 p^{2} q r^{2}$ are three unlike terms in the simplified expression.
So, the given expression is a trinomial.

$\textbf{Hence, } {p}^{3}{q}^{2}r+p{q}^{2}{r}^{3}+3{p}^{2}q{r}^{2}-9{p}^{2}q{r}^{2} \textbf{ will result in a trinomial.}$

Add the following expressions:
${a}^{2}+3ab-bc,{b}^{2}+3bc-ca$ and ${c}^{2}+3ca-ab$

Required sum

$={a}^{2}+3ab-bc+{b}^{2}+3bc-ca+{c}^{2}+3ca-ab$

Collecting like terms,

$={a}^{2}+{b}^{2}+{c}^{2}+3ab-ab+3bc-bc+3ca-ca$

$={a}^{2}+{b}^{2}+{c}^{2}+2ab+2bc+2ca$

Subtract :
$-{a}^{2}-ab$ from ${b}^{2}+ab$

${b}^{2}+ab-(-{a}^{2}-ab)$

$\\={b}^{2}+ab+{a}^{2}+ab\\={b}^{2}+2ab+{a}^{2}$

Add the following expressions:
${p}^{2}qr+p{q}^{2}r+pq{r}^{2}$ and $-3p{q}^{2}r-2pq{r}^{2}$

$\textbf{Combine the like terms and simplify.}$

$\begin{aligned}&\left(p^{2} q r+p q^{2} r+p q r^{2}\right)+\left(-3 p q^{2} r-2 p q r^{2}\right) \\&=p^{2} q r+\left(p q^{2} r-3 p q^{2} r\right)+\left(p q r^{2}-2 p q r^{2}\right) \\&=>\left(p^{2} q r+p q^{2} r+p q r^{2}\right)+\left(-3 p q^{2} r-2 p q r^{2}\right) \\&=p^{2} q r-2 p q^{2} r-p q r^{2}\end{aligned}$

$\textbf{Hence, the required expression is } p^{2} q r-2 p q^{2} r-p q r^{2}$

Add the following expressions:
$ab+bc+ca$ and $-bc-ca-ab$

$\textbf{Combine the like terms and simplify.}$

$\begin{aligned}&(a b+b c+c a)+(-b c-c a-a b) \\&=(a b-a b)+(b c-b c)+(c a-c a) \\&=>(a b+b c+c a)+(-b c-c a-a b)=0+0+0=0\end{aligned}$

$\textbf{Hence, the required expression is }0$

Add the following expressions:
${x}^{3}-{x}^{2}y-x{y}^{2}-{y}^{3}$ and ${x}^{3}-2{x}^{2}y+3x{y}^{2}+4y$

$\textbf{Combine the like terms and simplify.}$

$\begin{aligned}&\left(x^{3}-x^{2} y-x y^{2}-y^{3}\right)+\left(x^{3}-2 x^{2} y+3 x y^{2}+4 y\right) \\&=\left(x^{3}+x^{3}\right)+\left(-x^{2} y-2 x^{2} y\right)+\left(-x y^{2}+3 x y^{2}\right)-y^{3}+4 y \\&=>\left(x^{3}-x^{2} y-x y^{2}-y^{3}\right)+\left(x^{3}-2 x^{2} y+3 x y^{2}+4 y\right)=2 x^{3}-3 x^{2} y+2 x y^{2}-y^{3}+4 y\end{aligned}$

$\textbf{Hence, the required expression is } 2 x^{3}-3 x^{2} y+2 x y^{2}-y^{3}+4 y$

Add the following expressions:
${x}^{3}{y}^{2}+{x}^{2}{y}^{3}+3{y}^{4}$ and ${x}^{4}+3{x}^{2}{y}^{3}+4{y}^{4}$

$\textbf{Combine the like terms and simplify.}$

$\begin{aligned}&\left(x^{3} y^{2}+x^{2} y^{3}+3 y^{4}\right)+\left(x^{4}+3 x^{2} y^{3}+4 y^{4}\right) \\&=x^{4}+\left(3 y^{4}+4 y^{4}\right)+x^{3} y^{2}+\left(x^{2} y^{3}+3 x^{2} y^{3}\right) \\&=>\left(x^{3} y^{2}+x^{2} y^{3}+3 y^{4}\right)+\left(x^{4}+3 x^{2} y^{3}+4 y^{4}\right) \\&=x^{4}+7 y^{4}+x^{3} y^{2}+4 x^{2} y^{3}\end{aligned}$

$\textbf{Hence, the required expression is } x^{4}+7 y^{4}+x^{3} y^{2}+4 x^{2} y^{3}$

Add the following expressions:
$uv-vw,vw-wu$ and $wu-uv$

Required sum

$=uv-vw+vw-wu+wu-uv$

Collecting like terms,

$=uv-uv-vw+vw-wu+wu$

$=0$

Add the following expressions:
$t-{t}^{2}-{t}^{3}-14$ , $15{t}^{3}+13+9t-8{t}^{2}$ , $12{t}^{2}-19-24t$ and $4t-9{t}^{2}+19{t}^{3}$

$\textbf{Combine the like terms and simplify.}$

$\begin{aligned}&\left(t-t^{2}-t^{3}-14\right)+\left(15 t^{3}+13+9 t-8 t^{2}\right)+\left(12 t^{2}-19-24 t\right)+\left(4 t-9 t^{2}+19 t^{3}\right) \\&=\left(-t^{3}+15 t^{3}+19 t^{3}\right)+\left(-t^{2}-8 t^{2}+12 t^{2}-9 t^{2}\right)+(t+9 t-24 t+4 t)+(-14+13-19) \\&=33 t^{3}-6 t^{2}-10 t-20\end{aligned}$

$\textbf{Hence, the required expression is } 33 t^{3}-6 t^{2}-10 t-20$

How much should $5x^3+3x^2-2x+1$ be increased to get $6x^2+7$ ?

Required expression

$=6x^2+7-(5x^3+3x^2-2x+1)$

$=6x^2+7-5x^3-3x^2+2x-1$

$=-5x^3+3x^2+2x+6$

Add the following expressions:
${p}^{2}-q+r$ , ${q}^{2}-r+p$ and ${r}^{2}-p+q$

$\textbf{Combine the like terms and simplify.}$

$\begin{aligned}&\left(p^{2}-q+r\right)+\left(q^{2}-r+p\right)+\left(r^{2}-p+q\right) \\&=p^{2}+q^{2}+r^{2}+(r-r)+(-q+q)+(r-r) \\&=>\left(p^{2}-q+r\right)+\left(q^{2}-r+p\right)+\left(r^{2}-p+q\right)=p^{2}+q^{2}+r^{2}+0+0+0 \\&=p^{2}+q^{2}+r^{2}\end{aligned}$

$\textbf{Hence, the required expression is } p^{2}+q^{2}+r^{2}$

Subtract:
$-7{p}^{2}qr$ from $-3{p}^{2}qr$

Required difference

$=-3{p}^{2}qr-(-7{p}^{2}qr)$

$=-3{p}^{2}qr+(7{p}^{2}qr)$

$=4{p}^{2}qr$

Subtract :
$-2{a}^{2}-2{b}^{2}$ from $-{a}^{2}-{b}^{2}+2ab$ .

$\textbf{Combine the like terms and simplify.}$

$\begin{aligned}&-a^{2}-b^{2}+2 a b-\left(-2 a^{2}-2 b^{2}\right)=-a^{2}-b^{2}+2 a b+2 a^{2}+2 b^{2}=\left(-a^{2}+2 a^{2}\right)+\left(-b^{2}+2 b^{2}\right)+2 a b \\&=>-a^{2}-b^{2}+2 a b-\left(-2 a^{2}-2 b^{2}\right)=a^{2}+b^{2}+2 a b\end{aligned}$

$\textbf{Hence, the required expression is } a^{2}+b^{2}+2 a b$ .

Subtract:
$ab-bc-ca$ from $-ab+bc+ca$

$(-ab+bc+ca)-(ab-bc-ca)$

Collecting like terms,

$=-ab-ab+bc+bc+ca+ca$

$=-2ab+2bc+2ca$

Subtract:
$-4{x}^{2}y-{y}^{3}$ from ${x}^{3}+3x{y}^{2}-{x}^{2}y$

$\textbf{Combine the like terms and simplify.}$

$\begin{aligned}&\left(x^{3}+3 x y^{2}-x^{2} y\right)-\left(-4 x^{2} y-y^{3}\right)=x^{3}+3 x y^{2}-x^{2} y+4 x^{2} y+y^{3}=x^{3}+y^{3}+3 x y^{2}+\left(-x^{2} y+4 x^{2} y\right) \\&=>\left(x^{3}+3 x y^{2}-x^{2} y\right)-\left(-4 x^{2} y-y^{3}\right)=x^{3}+y^{3}+3 x y^{2}+3 x^{2} y\end{aligned}$

$\textbf{Hence, the required expression is } x^{3}+y^{3}+3 x y^{2}+3 x^{2} y$ .

Subtract:
${x}^{4}+3{x}^{3}{y}^{3}+5{y}^{4}$ from $2{x}^{4}-{x}^{3}{y}^{3}+7{y}^{4}$

$\textbf{Combine the like terms and simplify.}$

$\begin{aligned}&\left(2 x^{4}-x^{3} y^{3}+7 y^{4}\right)-\left(x^{4}+3 x^{3} y^{3}+5 y^{4}\right)=2 x^{4}-x^{3} y^{3}+7 y^{4}-x^{4}-3 x^{3} y^{3}-5 y^{4} \\&=\left(2 x^{4}-x^{4}\right)+\left(-x^{3} y^{3}-3 x^{3} y^{3}\right)+\left(7 y^{4}-5 y^{4}\right) \\&=>\left(2 x^{4}-x^{3} y^{3}+7 y^{4}\right)-\left(x^{4}+3 x^{3} y^{3}+5 y^{4}\right)=x^{4}-4 x^{3} y^{3}+2 y^{4}\end{aligned}$

$\textbf{Hence, the required expression is } x^{4}-4 x^{3} y^{3}+2 y^{4}$ .

What should be added to ${x}^{3}+3{x}^{2}y+3x{y}^{2}+{y}^{3}$ to get ${x}^{3}+{y}^{3}$

$\textbf{Step 1: Subtract } {x}^{3}+3{x}^{2}y+3x{y}^{2}+{y}^{3} \textbf{ from } {x}^{3}+{y}^{3}$

$\begin{aligned}&x^{3}+y^{3}-\left(x^{3}+3 x^{2} y+3 x y^{2}+y^{3}\right) \\&\quad=x^{3}+y^{3}-x^{3}-3 x^{2} y-3 x y^{2}-y^{3}\end{aligned}$

$\textbf{Step 2:Combine the like terms and simplify}$

$\begin{aligned}&=\left(x^{3}-x^{3}\right)+\left(y^{3}-y^{3}\right)-3 x^{2} y-3 x y^{2} \\&=>x^{3}+y^{3}-\left(x^{3}+3 x^{2} y+3 x y^{2}+y^{3}\right)=-3 x^{2} y-3 x y^{2}\end{aligned}$

$\textbf{Hence, } -3 x^{2} y-3 x y^{2} \textbf{ should be added to } x^{3}+3 x^{2} y+3 x y^{2}+y^{3} \textbf{ to get } x^{3}+y^{3}$

Subtract:
$2(ab+bc+ca)$ from $-ab-bc-ca$

$(-ab-bc-ca)-\{2(ab+bc+ca)\}$

$=-ab-bc-ca-2ab-2bc-2ca$

$=-3ab-3bc-3ca$

What should be added to $3pq+5{p}^{2}{q}^{2}+{p}^{3}$ to get ${p}^{3}+2{p}^{2}{q}^{2}+4pq$

$\textbf{Step 1: Subtract } 3pq+5{p}^{2}{q}^{2}+{p}^{3} \textbf{ from } {p}^{3}+2{p}^{2}{q}^{2}+4pq$

$\begin{aligned}&p^{3}+2 p^{2} q^{2}+4 p q-\left(3 p q+5 p^{2} q^{2}+p^{3}\right) \\&=p^{3}+2 p^{2} q^{2}+4 p q-3 p q-5 p^{2} q^{2}-p^{3}\end{aligned}$

$\textbf{Step 2:Combine the like terms and simplify}$

$\begin{aligned}&=\left(p^{3}-p^{3}\right)+\left(2 p^{2} q^{2}-5 p^{2} q^{2}\right)+(4 p q-3 p q) \\&=>p^{3}+2 p^{2} q^{2}+4 p q-\left(3 p q+5 p^{2} q^{2}+p^{3}\right)=-3 p^{2} q^{2}+p q\end{aligned}$

$\textbf{Hence, } -3p^2q^2 + pq \textbf{ should be added to }3pq+5{p}^{2}{q}^{2}+{p}^{3} \textbf{ to get } {p}^{3}+2{p}^{2}{q}^{2}+4pq$

Subtract:
${x}^{3}{y}^{2}+3{x}^{2}{y}^{2}-7x{y}^{3}$ from ${x}^{4}+{y}^{4}+3{x}^{2}{y}^{2}-x{y}^{3}$

$\textbf{Combine the like terms and simplify.}$

$\begin{aligned}&x^{4}+y^{4}+3 x^{2} y^{2}-x y^{3}-\left(x^{3} y^{2}+3 x^{2} y^{2}-7 x y^{3}\right) \\&=x^{4}+y^{4}+3 x^{2} y^{2}-x y^{3}-x^{3} y^{2}-3 x^{2} y^{2}+7 x y^{3} \\&=x^{4}+y^{4}+\left(3 x^{2} y^{2}-3 x^{2} y^{2}\right)+\left(-x y^{3}+7 x y^{3}\right)-x^{3} y^{2} \\&=>x^{4}+y^{4}+3 x^{2} y^{2}-x y^{3}-\left(x^{3} y^{2}+3 x^{2} y^{2}-7 x y^{3}\right) \\&=x^{4}+y^{4}+6 x y^{3}-x^{3} y^{2}\end{aligned}$

$\textbf{Hence, the required expression is } x^{4}+y^{4}+6 x y^{3}-x^{3} y^{2}$ .

Subtract:
$4.5{x}^{5}-3.4{x}^{2}+5.7$ from $5{x}^{4}-3.2{x}^{2}-7.3x$

$\textbf{Combine the like terms and simplify.}$

$\begin{aligned}&5 x^{4}-3.2 x^{2}-7.3 x-\left(4.5 x^{5}-3.4 x^{2}+5.7\right) \\&=5 x^{4}-3.2 x^{2}-7.3 x-4.5 x^{5}+3.4 x^{2}-5.7 \\&=5 x^{4}-4.5 x^{5}+\left(3.4 x^{2}-3.2 x^{2}\right)-7.3 x-5.7 \\&=>5 x^{4}-3.2 x^{2}-7.3 x-\left(4.5 x^{5}-3.4 x^{2}+5.7\right) \\&=-4.5 x^{5}+5 x^{4}+0.2 x^{2}-7.3 x-5.7\end{aligned}$

$\textbf{Hence, the required expression is }-4.5 x^{5}+5 x^{4}+0.2 x^{2}-7.3 x-5.7$ .

Subtract:
$11-15{y}^{2}$ from ${y}^{3}-15{y}^{2}-y-11$

$\begin{aligned}&y^{3}-15 y^{2}-y-11-\left(11-15 y^{2}\right)=y^{3}-15 y^{2}-y-11-11+15 y^{2} \\&=y^{3}+\left(-15 y^{2}+15 y^{2}\right)-y+(-11-11) \\&=>y^{3}-15 y^{2}-y-11-\left(11-15 y^{2}\right)=y^{3}-y-22\end{aligned}$

$\textbf{Hence, the required expression is }y^{3}-y-22$ .

What should be subtracted from $-7mn+2{m}^{2}+3{n}^{2}$ to get ${m}^{2}+2mn+{n}^{2}$

$\textbf{Step 1:In order to get the result, we subtract } m^2 +2mn+n^2 \textbf{ from } -7mn+2m^2+3n^2$

$-7 m n+2 m^{2}+3 n^{2}-\left(m^{2}+2 m n+n^{2}\right)=-7 m n+2 m^{2}+3 n^{2}-m^{2}-2 m n-n^{2}$

$\textbf{Step 2:Combine the like terms and simplify}$

$\begin{aligned}&=(-7 m n-2 m n)+\left(2 m^{2}-m^{2}\right)+\left(3 n^{2}-n^{2}\right) \\&=>-7 m n+2 m^{2}+3 n^{2}-\left(m^{2}+2 m n+n 2\right)=-9 m n+m^{2}+2 n^{2}\end{aligned}$

$\textbf{Hence, } m^2+2n^2-9mn \textbf{ should be subtracted from } -7mn+2m^2+3n^2 \textbf{ to get } m^2 +2mn+n^2$

Simplify the following expression.
$5x+22-6x=23$ .

$\begin{matrix} 5x+22-6X=23 \\ \Rightarrow \left( { 5x-6x } \right) +\left( { 22-23 } \right) =0 \\ \Rightarrow \left( { -x-1 } \right) =0 \\ \Rightarrow -\left( { x+1 } \right) =0 \\ \Rightarrow x+1=0\, \, \, \, Ans \\ \end{matrix}$

How much is ${y}^{4}-12{y}^{2}+y+14$ greater than $17{y}^{3}+34{y}^{2}-51y+68$ .

$\textbf{Step 1:In order to get the result, we subtract } 17{y}^{3}+34{y}^{2}-51y+68 \textbf{ from } {y}^{4}-12{y}^{2}+y+14$

$\begin{aligned}&y^{4}-12 y^{2}+y+14-\left(17 y^{3}+34 y^{2}-51 y+68\right) \\&=y^{4}-12 y^{2}+y+14-17 y^{3}-34 y^{2}+51 y-68\end{aligned}$

$\textbf{Step 2:Combine the like terms and simplify}$

$\begin{aligned}&=y^{4}-17 y^{3}+\left(-12 y^{2}-34 y^{2}\right)+(y+51 y)+(14-68) \\&=>y^{4}-12 y^{2}+y+14-\left(17 y^{3}+34 y^{2}-51 y+68\right)=y^{4}-17 y^{3}-46 y^{2}+52 y-54\end{aligned}$

Hence,

$y^{4}-12 y^{2}+y+14$ is

$y^{4}-17 y^{3}-46 y^{2}+52 y-54$ greater than

$17 y^{3}+34 y^{2}-51 y+68$

How much is $21{a}^{3}-17{a}^{2}$ less than $89{a}^{3}-64{a}^{2}+6a+16$

$\textbf{Step 1:In order to get the result, we subtract } 21a^3-17a^2 \textbf{ from } 89a^3 - 64a^2 + 6a + 16$

$89 a^{3}-64 a^{2}+6 a+16-\left(21 a^{3}-17 a^{2}\right)=89 a^{3}-64 a^{2}+6 a+16-21 a^{3}+17 a^{2}$

$\textbf{Step 2:Combine the like terms and simplify}$

$\begin{aligned}&=\left(89 a^{3}-21 a^{3}\right)+\left(-64 a^{2}+17 a^{2}\right)+6 a+16 \\&=>89 a^{3}-64 a^{2}+6 a+16-\left(21 a^{3}-17 a^{2}\right)=68 a^{3}-47 a^{2}+6 a+16\end{aligned}$

Hence,

$21 a^{3}-17 a^{2}$ is

$68 a^{3}-47 a^{2}+6 a+16$ less than

$89 a^{3}-64 a^{2}+6 a+16$

What should be subtracted from $2{x}^{3}-3{x}^{2}y+2x{y}^{2}+3{y}^{3}$ to get ${x}^{3}-2{x}^{2}y+3x{y}^{2}+4{y}^{3}$

$\textbf{Step 1:In order to get the result, we subtract } x^3-2x2y+3xy^2+4y^3 \textbf{ from } 2x^3-3x^2y+2xy^2+3y^3$

$\begin{aligned}&2 x^{3}-3 x^{2} y+2 x y^{2}+3 y^{3}-\left(x^{3}-2 x^{2} y+3 x y^{2}+4 y^{3}\right) \\&=2 x^{3}-3 x^{2} y+2 x y^{2}+3 y^{3}-x^{3}+2 x^{2} y-3 x y^{2}-4 y^{3}\end{aligned}$

$\textbf{Step 2:Combine the like terms and simplify}$

$\begin{aligned}&=\left(2 x^{3}-x^{3}\right)+\left(-3 x^{2} y+2 x^{2} y\right)+\left(2 x y^{2}-3 x y^{2}\right)+\left(3 y^{3}-4 y^{3}\right) \\&=>2 x^{3}-3 x^{2} y+2 x y^{2}+3 y^{3}-\left(x^{3}-2 x^{2} y+3 x y^{2}+4 y^{3}\right)=x^{3}-x^{2} y-x y^{2}-y^{3}\end{aligned}$

$\textbf{Hence, } x^{3}-x^{2} y-x y^{2}-y^{3} \textbf{ should be subtracted from } 2 x^{3}-3 x^{2} y+2 x y^{2}+3 y^{3} \textbf{ to get } x^{3}-2 x^{2} y+3 x y^{2}+4 y^{3}$

From the sum of ${x}^{2}-{y}^{2}-1$ , ${y}^{2}-{x}^{2}-1$ and $1-{x}^{2}-{y}^{2}$ subtract $-(1+{y}^{2})$

$\textbf{Step 1: Add } x^{2}-y^{2}-1, y^{2}-x^{2}-1 \textbf{ and } 1-x^{2}-y^{2}$

$\begin{aligned}&\left(x^{2}-y^{2}-1\right)+\left(y^{2}-x^{2}-1\right)+\left(1-x^{2}-y^{2}\right) \\&=\left(x^{2}-x^{2}-x^{2}\right)+\left(-y^{2}+y^{2}-y^{2}\right)+(-1-1+1)=-x^{2}-y^{2}-1\end{aligned}$

$\textbf{Step 2: Subtract } -(1+y^2) \textbf{ from above expression }$

$\begin{aligned}&-x^{2}-y^{2}-1-\left\{-\left(1+y^{2}\right)\right\} \\&=-x^{2}-y^{2}-1+1+y^{2} \\&=-x^{2}\end{aligned}$

$\textbf{Hence, the required expression is } -x^2.$

How much does $93{p}^{2}-55p+4$ exceed $13{p}^{3}-5{p}^{2}+17p-90$

$\textbf{Step 1:In order to get the result, we subtract 13p^3-5p^2+17p-90 from 93p^2-55p+4}$

$93 p^{2}-55 p+4-\left(13 p^{3}-5 p^{2}+17 p-90\right)=93 p^{2}-55 p+4-13 p^{3}+5 p^{2}-17 p+90$

$\textbf{Step 2:Combine the like terms and simplify}$

$\begin{aligned}&=-13 p^{3}+\left(93 p^{2}+5 p^{2}\right)+(-55 p-17 p)+(4+90) \\&=>93 p^{2}-55 p+4-\left(13 p^{3}-5 p^{2}+17 p-90\right)=-13 p^{3}+98 p^{2}-72 p+94\end{aligned}$

$\textbf{Hence, } 93 p^{2}-55 p+4 \textbf{ exceeds } 13 p^{3}-5 p^{2}+17 p-90 \textbf{ by } -13 p^{3}+98 p^{2}-72 p+94$

Each symbol given below represents an algebraic expression: (first figure)
The symbol are then represented in the expression: (second figure)
Find the expression which is represented by the given symbols.

$\textbf{Step1: Add the expression representing triangle and circle}$
Since, the triangle represents the expression

$2x^2+3y$ and the circle represent the expression

$5x^2+3x$ .

$\begin{gathered}\text { Therefore, triangle plus circle }=2 x^{2}+3 y+5 x^{2}+3 x \\ \ \ \ \ \ \ \ \ \ \ \ \quad \ \ \ \quad \ \ \ \quad \ \ \ \quad \ \ \ \ \ =7 x^{2}+3 y+3 x\end{gathered}$

$\textbf{Step 2:Subtract the expression representing square from the sum of the expressions representing}$

$\ \ \ \ \ \ \ \ \ \ \ \ \textbf{ triangle and circle.}$

$\begin{aligned}&=7 x^{2}+3 y+3 x-\left(8 y^{2}-3 x^{2}+2 x+3 y\right) \\&=7 x^{2}+3 y+3 x-8 y^{2}+3 x^{2}-2 x-3 y \\&=\left(7 x^{2}+3 x^{2}\right)-8 y^{2}+(3 x-2 x)+(3 y-3 y) \quad \text { (grouping like terms) } \\&=10 x^{2}-8 y^{2}+x\end{aligned}$

$\textbf{Hence, } 10x^2-8y^2+x \textbf{ is the required expression represented by the above symbols.}$

Subtract the sum of $12ab-10{b}^{2}-18{a}^{2}$ and $9ab+12{b}^{2}+14{a}^{2}$ from the sum of $ab+2{b}^{2}$ and $3{b}^{2}-{a}^{2}$

$\textbf{Step 1:Find the sum of given expression}$

$\begin{aligned}&\left(12 a b-10 b^{2}-18 a^{2}\right)+\left(9 a b+12 b^{2}+14 a^{2}\right) \\&=21 a b+2 b^{2}-4 a^{2}\end{aligned}$
And

$\left(a b+2 b^{2}\right)+\left(3 b^{2}-a^{2}\right)=a b+5 b^{2}-a^{2}$

$\textbf{Step 2: Subtract the above two expressions to get the result.}$

$a b+5 b^{2}-a^{2}-\left(21 a b+2 b^{2}-4 a^{2}\right)=a b+5 b^{2}-a^{2}-21 a b-2 b^{2}+4 a^{2}=3 b^{2}+3 a^{2}-2$

$\textbf{Hence, the required expression is } 3b^2+3a^2-20ab.$

Subtract $9{a}^{2}-15a+3$ from unity

We know that Unity is

$1$ , so the subtraction is

$\bullet \ 1-\left(9 a^{2}-15 a+3\right)=1-9 a^{2}+15 a-3=-9 a^{2}+15 a-2$

$\textbf{Hence, the required expression is } -9a^2+15a-2$

To what expression must $99x^3-33x^2-13x-41$ be added to make the sum zero.

$\bullet$ Let the expression added be

$t$

$\bullet$ Then,

$99 x^{3}-33 x^{2}-13 x-41+t=0$

$\bullet \ t=-99 x^{3}+33 x^{2}+13 x+41$ (transposing terms to right hand side)

$\textbf{Hence, } -99x^3+33x^2+13x+41 \textbf{ is added to } 99x^3-33x^2-13x-41 \textbf{to make the sum zero.}$

Observe the following nutritional chart carefully:
Food Item Carbohydrates
Rajma $60\text{g}$
Cabbage $5\text{g}$
Potato $22\text{g}$
Carrot $11\text{g}$
Tomato $4\text{g}$
Apples $14\text{g}$
Write an algebraic expression for the amount of carbohydrates in $g$ for
(a) $y$ units of potatoes and $2$ units of rajma
(b) $2x$ units tomatoes and $y$ units apples

(a)
One unit of Potato contains

$22g$ of carbohydrates Therefore, y units of potatoes will contain

$22 \times y=22y$ g of carbohydrates.

Also, one unit of Rajma contains

$60g$ of carbohydrates Therefore,

$2$ units of rajma will contain

$2 \times 60=120$ g of carbohydrates.

So, y units of potatoes and

$2$ units of rajma will contain

$22y+120$ g of carbohydrates.

$\textbf{Hence, the required expression for the amount of carbohydrates in g for y units of potatoes and }$

$\textbf{2 units of rajma is 22y+120.}$

(b)
One unit tomato contains

$4g$ of carbohydrates Therefore,

$2x$ units of tomatoes will contain

$4 \times 2x=8x$ g of carbohydrates.

Also, one unit of apples contain

$14g$ of carbohydrates Therefore, y units apples will contain

$14 \times y=14y$ g of carbohydrates.

So,

$2x$ units tomatoes and y units apples will contain

$8x+14y$ g of carbohydrates.

$\textbf{Hence, the required expression for the amount of carbohydrates in g for 2x units of tomatoes}$

$\textbf{and y units of apples is 8x+14y}$ .

Add: $9ax$ , $+ 3by - cz$ , $— 5by + ax + 3cz$

$9ax+3by-cz-5by+ax+3cz$

$= 9ax+ax+3by-5by-cz+3cz$

$= 10ax-2by+2cz$

Number of terms in the expression $a^{2} + bc \times d$ is __________ .

$a^2+bc\times d=a^2+bcd$

So, terms in

$a^2+bcd$ are

$a^2$ and

$bcd$

Thus, number of terms are $2$

Add: $xy^{2}z^{2} + 3x^{2}y^{2}z - 4x^{2}yz^{2}$ , $-9x^{2}y^{2}z + 3xy^{2}z^{2} + x^{2}yz^{2}$

$xy^{2}z^{2} + 3x^{2}y^{2}z - 4x^{2}yz^{2} +(-9x^{2}y^{2}z + 3xy^{2}z^{2} + x^{2}yz^{2})$

$=xy^{2}z^{2} +3xy^2z^{2} + 3x^{2}y^{2}z -9x^{2}y^2z- 4x^{2}yz^{2} +x^{2}yz^2$ (on combining like terms)

$=4xy^{2}z^{2}-6x^{2}y^{2}z-3x^{2}yz^{2}$

Add: $7a^{2}bc\ , -3abc^{2}\ ,\ 3a^{2}bc\ ,\ 2abc^{2}$

Given terms are

$7a^{2}bc\ ,\ -3abc^{2}\ ,\ 3a^{2}bc\ ,\ 2abc^{2}$

Required sum

$=7a^{2}bc+ (-3abc^{2})+(3a^{2}bc)+(2abc^{2})$

Rearranging and collecting like terms,

$= 7a^{2}bc+3a^{2}bc-3abc^{2}+2abc^{2}$

$= (7+3)a^{2}bc+(-3+2)abc^{2}$

$= 10a^{2}bc-abc^{2}$

Add: $5x^{2}-3xy+4y^{2}-9$ , $7y^{2}+5xy-2x^{2}+13$

$5x^{2}-3xy+4y^{2}-9 + 7y^{2}+5xy-2x^{2}+13$

$=5x^{2}-2x^{2}-3xy + 5xy +4y^{2}+7y^{2}-9+13$

$=7x^{2}+2xy+11y^{2}+4$

Subtract $6x^{2} - 4xy + 5y^{2}$ from $8y^{2} + 6xy - 3x^{2}$

$8y^{2} + 6xy — 3x^{2}-(6x^{2} — 4xy + 5y^{2})$

$8y^{2} + 6xy — 3x^{2}-6x^{2} + 4xy - 5y^{2})$

$=-9x^{2}+10xy+3y^{2}$

Subtract $2ab^{2}c^{2} + 4a^{2}b^{2}c - 5a^{2}bc^{2}$ from $-10a^{2}b^{2}c + 4ab^{2}c^{2} +2a^{2}bc^{2}$

$-10a^{2}b^{2}c + 4ab^{2}c^{2} +2a^{2}bc^{2} - (2ab^{2}c^{2} + 4a^{2}b^{2}c - 5a^{2}bc^{2})$

$-10a^{2}b^{2}c + 4ab^{2}c^{2} +2a^{2}bc^{2} - 2ab^{2}c^{2} - 4a^{2}b^{2}c + 5a^{2}bc^{2}$

=

$-10a^{2}b^{2}c - 4a^{2}b^{2}c + 4ab^{2}c^{2} - 2ab^{2}c^{2} + 2a^{2}bc^{2} +5a^{2}bc^{2}$

=

$-14a^{2}b^{2}c+4ab^{2}c^{2} +7a^{2}bc^{2}$

Subtract $3t^{4} - 4t^{3} + 2t^{2}- 6t + 6$ from $-4t^{4} + 8t^{3}-4t^{2} - 2t+ 11$

$-4t^{4} + 8t^{3}-4t^{2} - 2t+ 11 -(3t^{4} - 4t^{3} + 2t^{2}- 6t + 6)$

$-4t^{4} + 8t^{3}-4t^{2} - 2t+ 11 -3t^{4} + 4t^{3} - 2t^{2}+ 6t - 6)$

$= — 4t^{4} -3t^{4}+ 8t^{3}+4t^{3}-4t^2 -2t^{2} - 2t+ 6t+ 11- 6$

$= -7t^{4}+12t^{3}-6t^{2}+4t+5$

Simplify: $(3x + 2y)^{2} - (3x - 2y)^{2}$

We have,

$(3x + 2y)^{2} = (3x)^{2} + 2.3x.2y + (2y)^{2}$ (

$\because (a+b)^2= a^2+2ab+b^2)$ )

$=9x^{2} + 12xy + 4y^{2}$

And,

$(3x-2y)^{2} = (3x)^{2}-2.3x.2y + (2y)^{2}$ (

$\because (a-b)^2= a^2-2ab+b^2)$ )

$= 9x^{2}-12xy + 4y^{2}$

Therefore,

$(3x + 2y)^{2} - (3x - 2y)^{2} = 9x^{2} + 12xy + 4y^{2}-9x^{2} + 12xy-4y^{2}$

$= 24xy$

Simplify: $(3x + 2y)^{2} + (3x - 2y)^{2}$

We have,

$(3x + 2y)^{2} = (3x)^{2} + 2.3x.2y + (2y)^{2}$

$=9x^{2} + 12xy + 4y^{2}$

And,

$(3x-2y)^{2} = (3x)^{2}-2.3x.2y + (2y)^{2}$

$= 9x^{2}-12xy + 4y^{2}$

Therefore,

$(3x + 2y)^{2} + (3x — 2y)^{2} = 9x^{2} + 12xy + 4y^{2} + 9x^{2}- 12xy + 4y^{2}$

$= 18x^{2} + 8y^{2}$

$= 2(9x^{2} + 4y^{2})$

Simplify: $(1.5p + 1.2q)^{2} - (1.5p - 1.2q)^{2}$

We have

$(1.5p + 1.2q)^{2} = (1.5p)^{2} + 2(1.5p)(1.2q) + (1.2q)^{2}$

And

$(1.5p - 1.2q)^{2} = (1.5p)^{2} 2(1.5p)(1.2q) + (1.2q)^{2}$

Therefore,

$(1.5p + 1.2q)^{2} — (1.5p - 1.29)^{2}$

$= (1.5p)^{2} + 2(1.5p)(1.2q) + (1.2q)^{2}-(1.5p)^{2} + 2(1.5p)(1.2q)-(1.2q)$

$= 2(1.5p)(1.2q)$

Find the value of $a$ , if $pqa = (3p + q)^{2} - (3p - q)^{2}$

$pqa = (3p + q)^{2} - (3p - q)^{2}$

We know that,

$a^{2}- b^{2} =(a+b)(a-b)$

By using this formula,

$pqa = [(3p + q) + (3p — q)] [(3p + q) - (3p - q)]$

$pqa = [(3p + q + 3p — q)] [(3p + q -3p + q)]$

$pqa = (6p) \times (2q)$

$a = \dfrac{6p \times 2q}{pq} =\dfrac{(6 \times 2)pq}{pq}$

$a=12$

Identify the terms and their factors in the algebraic expressions given below:
$1.2 ab-2.4 b+3.6 a$

$1.2 ab-2.4b+3.6a$
Term is

$1.2ab,-2.4b,3.6a$
Factor of

$1.2ab=1.2,a,b$
Factor of

$-2.4b=-2.4,b$
Factor of

$3.6 a=3.6,a$

Simplify: $(2.5m + 1.5q)^{2} + (2.5m - 1.5q)^{2}$

We have

$(2.5m+1.5q)^2=(2.5m)^2+2(2.5m)(1.5q)+(1.5q)^2\quad$ [using

$(a+b)^2=a^2+2ab+b^2$ ]

$=6.25m^2+7.5mq+2.25q^2$

$(2.5m-1.5q)^2=(2.5m)^2-2(2.5m)(1.5q)+(1.59)^2\quad$ [using

$(a-b)^2=a^2-2ab+b^2$ ]

$=6.25m^2-7.5mq+2.25q^2$

Therefore,

$(2.5m + 1.59)^{2} + (2.5m - 1.5q)^{2}$

$=(6.25m^2+7.5mq+2.25q^2)+(6.25m^2-7.5mq+2.25q^2)$

$=6.25m^2+7.5mq+2.25q^+6.25m^2-7.5mq+2.25q^2$

Collecting like terms,

$=6.25m^2+6.25m^2+2.25q^2+2.25q^2+7.5mq-7.5mq$

$=12.50m^2+4.50q^2$

Find the sum of the following algebraic expressions: $7a^2-5a +2,3a^2-7,2a+9,1+2a-5a^2$

Find the sum of the following algebraic expressions: $5m-7n,3n-4m+2,2m-3mn-5$

1886469_1801816_ans_3fb0e7578ba0428eba6afcb6db56038a.jpeg

Find the sum of the following algebraic expressions: $4x^2y,-3xy^2,-5xy^2,5x^2y$

$(4x^2y)+(-3xy^2)+(-5xy^2)+(5x^2y)$

$=4x^2y+5x^2y-3xy^2-5xy^2$

$=9x^2y-8xy^2$

Simplify the following: $3xy^2-5x^2y+7xy-8xy^2-4xy+6x^2y$

$3xy^2-5x^2y+7xy-8xy^2-4xy+6x^2y$

$=3xy^2-8xy^2-5x^2y+6x^2y+7xy-4xy$

$=-5xy^2+x^2y+3xy$

Find the sum of the following algebraic expressions: $3x^3-5x^2+2x+1,3x-2x^2-x^3,2x^2-7x+9$

$3x^3-5x^2+2x+1,3x-2x^2-x^3,2x^2-7x+9$

$2x^3-5x^2-2x+10$
1730555_1801817_ans_d0fc5ef709e7423e9dc242a2aaae8f1c.PNG

Simplify the following: $2x^2+3y^2-5xy+5x^2-y^2+6xy-3x^2$

$2x^2+3y^2-5xy+5x^2-y^2+6xy-3x^2$

$2x^2+5x^2-3x^2+3y^2-y^2-5xy+6xy$

$4x^2+2y^2+xy$

Find the sum of the following algebraic expressions: $14x+10y-12xy-13,18-7x-10y+8xy,4xy$

$14x+10y-12xy-13,18-7x-10y+8xy,4xy$

$14x+10y-12xy-13$

$-7x-10y+8xy+18$

$+4xy$

-----------------------------------------

$7x$

$+5$

__________________________

$7x+5$

Find the sum of the following algebraic expressions: $-7mn + 5, 12 mn+2,8mn-8,-2mn-3$

$-7mn + 5, 12 mn+2,8mn-8,-2mn-3$

$=-7mn+5$

$+12mm+2$

$+8mn-8$

$+(-2mn-3)$

$=-7mn+5+12mn+2+8mn-8-2mn-3$

$=11mn-4$

Find the sum of the following algebraic expressions: $a+b-3,b-a+3,a-b+3$

Given terms :

$a+b-3,b-a+3,a-b+3$

$a+b-3$

$-a+b+3$

$a-b+3$

__________

$a+b+3$

__________

Sum

$=a+b+3$

Subtract:-7xy from -2xy

By Subtracting

$-7xy$ from

$-2xy$ we get,

$=-2xy-(-7xy)$

$=-2xy+7xy$

$=5xy$

Using the horizontal method:
Add $x^2+y^2-2xy,-2x^2-y^2-2xy$ and $3x^2+y^2+xy$

$=x^2+y^2-2xy+(-2x^2-y^2-2xy)+(3x^2+y^2+xy)$

$=x^2-2x^2+3x^2+y^2-y^2+y^2-2xy-2xy+xy$

$=2x^2+y^2-3xy$

Using column method, add $ab+2bc-ca$ and $2ab-bc-ca$ and subtract $4ab+5bc-3ca$

1730522_1802258_ans_aac8e396f7a04225831e11c4b8f99dd9.png

Subtract: $2x^4-7x^2+5x+3$ from $x^4-3x^3-2x^2+3$

$2x^4-7x^2+5x+3$ from

$x^4-3x^3-2x^2+3$

$=(x^4-3x^3-2x^2+3)-(2x^4-7x^2+5x+3)$

$=x^4-3x^3-2x^2+3-2x^4+7x^2-5x-3$

$=x^4-2x^4-3x^3-2x^2+7x^2-5x+3-3$

$-x^4-3x^3+5x^2-5x$

Subtract p-2q+r from the sum of 10p-r and 5p+2q.

Subtract p-2q+r from the sum of 10p-r and 5p+2q .By adding 10p-4+5p+2q and 5p+2q, we get

$=10p-r+5p+2q$

$=15p+2q-r$
Now,

$(15p+2q-r)-(p-2q+r)$

$=15p+2q-r-p+2q-r$

$14p+4q-2r$

Simplify the following: $5x^4-7x^2+8x-1+3x^3-9x^2+7-3x^4+11x-2+8x^2$

$5x^4-7x^2+8x-1+3x^3-9x^2+7-3x^4+11x-2+8x^2$

$=5x^4-3x^4+3x^3-7x^2-9x^2+8x^2+8x+11x-1+7-2$

$=2x^4+3x^3-8x^2+19x+4$

Subtract the sum of $3x^2 + 5xy + 7y^2 + 3$ and $2x^2 - 4xy - 3y^2 + 7$ from $9x^2 - 8xy + 11y^2$

First, adding

$3x^2 + 5xy + 7y^2 + 3$ and

$2x^2 - 4xy - 3y^2 + 7$ , we have
=

$3x^2 + 5xy + 7y^2 + 3 + 2x^2 - 4xy - 3y^2 + 7$
=

$5x^2 + xy + 4y^2 + 10$
Now,
Subtracting

$5x^2 + xy + 4y^2 + 10$ from

$9x^2 - 8xy + 11y^2$
=

$(9x^2 - 8xy + 11y^2) - (5x^2 + xy + 4y^2 + 10)$
=

$9x^2 - 8xy + 11y^2 - 5x^2 -xy - 4y^2 -10$
=

$4x^2 - 9xy + 7y^2 - 10$

What must be added to $5x^3-2x^2+3x+7$ to get $7x^3+7x-5$ ?

Required expression

$=7x^3+7x-5-(5x^3-2x^2+3x+7)$

$=7x^3+7x-5-5x^3+2x^2-3x-7$

$=2x^3+2x^2+4x-12$

What must be subtracted from $3a^2 - 5ab - 2b^2 - 3$ to get $5a^2 - 7ab - 3b^2 + 3a$ ?

From the question, its understood that we have to subtract

$5a^2 - 7ab - 3b^2 + 3a$ from

$3a^2 - 5ab - 2b^2 - 3$ .
=

$3a^2 - 5ab - 2b^2 - 3 - (5a^2 - 7ab - 3b^2 + 3a)$
=

$3a^2 - 5ab - 2b^2 - 3 - 5a^2 + 7ab + 3b^2 - 3a$
=

$-2a^2 + 2ab + b^2 - 3a - 3$

Subtract the sum of $3x^2+2xy-2y^2$ and $5y^2-7xy$ from $5x^2+2y^2-3xy$

Sum of

$3x^2+2xy-2y^2$ and

$5y^2-7xy$

$=3x^2+2xy-2y^2+5y^2-7xy$

$=3x^2-5xy+3y^2$
Now

$5x^2+2y^2-3xy$

$3x^2+3y^2-5xy$

$- \ \ \ - \ \ \ \ \ \ +$
--------------------------

$2x^2-y^2+2xy$
------------------------

$=2x^2+2xy-y^2$

Simplify:
$(x + 5) (x + 6) (x + 7)$

Add the following expressions:
$5p^2q^2 + 4pq + 7, 3 + 9pq - 2p^2q$

$5p^2q^2 + 4pq + 7,3 + 9pq - 2p^2q^2$
On adding the expressions, we have
=

$5p^2q^2 4 + 4pq + 7 + 3 + 9pq - 2p^2q^2$
=

$5p^2q^2 - 2p^2q^2 + 4pq + 9pq + 7 + 3$
=

$3p^2q^2 + 13pq + 10$

Subtract:
$8xy + 4yz + 5zx$ from $12xy - 3yz - 4zx + 5xyz$

Subtracting

$8xy + 4yz + 5zx$ from

$12xy - 3yz - 4zx + 5xyz$ , we have

$(12xy - 3yz - 4zx + 5xyz) - (8xy + 4yz + 5zx)$
=

$12xy - 3yz - 4zx + 5xyz - 8xy - 4yz - 5zx$
=

$4xy - 7yz - 9zx + 5xyz$

Add the following:
$4p(2 - p^2)$ and $8p^3 - 3p$

Adding,

$4p(2 - p^2)$ and

$8p^3 - 3p$
=

$8p - 4p^3 + 8p^3 - 3p$
=

$5p + 4p^3$
=

$4p^3 + 5p$

Add the following expressions:
$I^2 + m^2 + n^2, Im + mn, mn + nI + Im$

$1^2 + m^2 + n^2, 1m + mn, mn + nl, n1 + 1m$
On adding the expressions, we have
=

$1^2 + m^2 + n^2 + 1m + mn + mn + n1 + n1 + 1m$
=

$1^2 + m^2 + n^2 + 21m + 2mn + 2n1$

Subtract:
$6x(x - y + z) - 3y(x + y - z)$ from $2z(-x + y + z)$

Subtracting,

$6x(x - y + z) - 3y(x + y - z)$ from

$2z(-x + y + z)$

$\Rightarrow \ \ 6x^2 - 6xy + 6xz - 3xy - 3y^2 + 3yz$ from

$- 2xz + 2yz + 2z^2$
=

$(-2xz + 2yz + 2z^2) - (6x^2 - 6xy + 6xz - 3xy - 3y^2 + 3yz)$
=

$- 2xz + 2yz + 2z^2 - 6x^2 + 6xy - 6xz + 3xy + 3y^2 - 3yz$
=

$9xy - yz - 8zx - 6x^2 + 3y^2 + 2z^2$

Subtract:
$4p^2q - 3pq + 5pq^2 - 8p + 7q - 10$ from $18 - 3p - 11q + 5pq - 2pq^2 + 5p^2q$

Subtracting (

$4p^2q - 3pq + 5pq^2 - 8p + 7q - 10$ ) from (

$18 - 3p - 11q + 5pq - 2pq^2 + 5p^2q$ )

we have,
=

$(18 - 3p - 11q + 5pq - 2pq^2 + 5p^2q) - (4p^2q - 3pq + 5pq^2 - 8p + 7q - 10)$
=

$18 - 3p - 11q + 5pq - 2pq^2 + 5p^2q - 7p^2q + 3pq - 5pq^2 + 8p - 7q + 10$
=

$28 + 5p - 78q + 8pq - 7pq^2 + p^2q$

Add the following expressions:
$4x^3 - 7x^2 + 9, 3x^2 - 5x + 4, 7x^3 - 11x + 1, 6x^2 - 13x$

$4x^3 - 7x^2 + 9, 3x^2 - 5x + 4, 7x^3 - 11x + 1, 6x^2 - 13x$
On adding the expressions, we have
=

$4x^3 - 7x^2 + 9 + 3x^2 - 5x + 4 + 7x^3 - 11^2 + 1 + 6x^2 - 13x$
=

$4x^2 + 7x^3 - 7x^2 + 3x^2 + 6x^2 + 5x - 11x - 13x + 9 + 4 + 1$
=

$11x^3 - 2x^2 - 29x + 14$

Add the following:
$7xy(8x + 2y - 3)$ and $4xy^2(3y - 7x + 8)$

By Adding ,

$7xy(8x + 2y - 3)$ and

$4xy^2(3y - 7x + 8)$

we get,

$=56x^2y + 14xy^2 - 21xy + 12xy^3 - 28x^2y^2 + 32xy^2$

$=12xy^3 - 28x^2y^2 + 56x^2y + 46xy^2 - 21xy$

Show that:
$(3p/7 - 7q/6)^2 + pq = 9p^2/49 + 49q^2/36$

Taking LHS, we have
LHS =

$\Big( \frac{3}{7}p - \frac{7}{6}q \Big)^2 + pq$

$= \Big( \frac{3}{7} p \Big)^2 - 2 \times \frac{3}{7} p \times \frac{7}{6} q + \Big( \frac{7}{6} q \Big)^2 + pq \ \ \ \ \ { (a - b)^2 = a^2 - 2ab + b^2 }$

$= \frac{9}{49} p^2 - pq + \frac{49}{36} q^2 + pq$

$= \frac{9}{49} p^2 + \frac{49}{36}q^2 = RHS$

Simplify:
$(p + q) (r + s) + (p - q)(r - s) - 2(pr + qs)$

$(p + q) (r + s) + (p - q) (r - s) - 2(pr + qs)$

$=(pr + ps + qr + qs) + (pr - ps - qr + qs) - 2pr - 2qs$

$=2pr-2pr+2qs-2qs+qr-qr+ps-ps$

$=0$

Show that:
$(p-q)(p+q)+ (q-r)(q+r)+(r-p)(r+p) = 0$

Taking LHS, we have
LHS =

$(p-q)(p+q)+ (q-r)(q+r)+(r-p)(r+p)$

$= p^2 - q^2 + q^2 - r^2 + r^2 - p^2 \ \ \ \ \ [ Using, \ (a + b) (a - b) = a^2 - b^2]$

$= 0 = RHS$

$2a + b c = 2a + ()$

$2a + b – c = 2a + (b – c)$ .

Simplify the following:
$(2x + 5y)^2 + (2x - 5y)^2$

$(2x + 5y)^2 + (2x - 5y)^2$

$[Using, \ (a \pm b)^2 = a^2 \pm 2ab + b^2]$

$= (2x)^2 + 2 \times 2x \times 5y + (5y)^2 + (2x)^2 - 2 \times 2x \times 5y + (5y)^2$

$= 4x^2 + 20xy + 25y^2 + 4x^2 - 20xy + 25y^2$

$= 8x^2 + 50y^2$

Simplify:
$(x + y + z) (x - y + z) + (x + y - z) (- x + y + z) - 4zx$

$(x + y + z) (x - y + z) + (x + y - z) (-x + y + z) - 4zx$
=

$x^2 - xy + xz + xy - y^2 + yz + xz - yz + z^2 - x^2 + xy + xz - xy + x^2 + yx + xz - yz - z^2 - 4zx$
=

$0$

Simplify:
$(p + q - 2r) (2p - q + r) - 4qr$

Given expression is

$(p + q - 2r) (2p - q + r) - 4qr$
We can simplify the given expression by multiplying the given terms as:

$p(2p - q + r) + q(2p - q + r) - 2r(2p - q + r) - 4qr$

$\Rightarrow 2p^2 - pq + pr + 2pq - q^2 + qr - 4pr + 2qr - 2r^2 - 4qr$

$\Rightarrow 2p^2 - q^2 - 2r^2 + pq - 3pr - qr$

Hence,

$(p + q - 2r) (2p - q + r) - 4qr=2p^2 - q^2 - 2r^2 + pq - 3pr - qr$ .

Simplify the following expression.
$6x-20y+7z-8x+25y-11z$ .

$\begin{matrix} 6x-20y+7z-8x+25y-11z \\ \Rightarrow \left( { 6x-8x } \right) +\left( { 25y-20y } \right) +\left( { 7z-11z } \right) \\ \Rightarrow -2x+5y-4z \\ \end{matrix}$

Show that:
$(4x + 7y)^2 - (4x - 7y)^2 = 112 xy$

Taking LHS we have
LHS =

$(4x + 7y)^2 - (4x - 7y)^2$

$[ Using, \ (a \pm b)^2 = a^2 \pm 2ab + b^2]$

$= [ (4x)^2 + 2 \times 4x \times 7y + (7y)^2 ]- [ (4x)^2 - 2 \times 4x + 7y + (7y)^2 ]$

$= ( 16 x^2 + 56 xy + 49 y^2 ) - ( 16 x^2 - 56 xy + 49 y^2)$

$= 16 x^2 + 56 xy + 49y^2 - 16x^2 + 56 xy - 49y^2$

$= 112 xy = RHS$

Evaluate: $(9a 3a) + 4a$ .

$(9a – 3a) + 4a$ ,
The value of the expression

$(9a – 3a) + 4a$ is calculated as follows,

$(9a – 3a) + 4a = 6a + 4a$
We get,

$= 10a$ .
Hence, the value of the expression

$(9a – 3a) + 4a = 10a$ .

Subtract:
$5 a - 3 b + 2 c$ from $a - 4 b - 2 c$

$5 a - 3 b + 2 c$ from

$a - 4 b - 2 c$
The value of the subtraction is calculated as follows

$(a - 4 b - 2 c) - (5 a - 3 b + 2 c)$

$= a - 5 a - 4 b + 3 b - 2 c - 2 c$
We get,

$= - 4 a - b - 4 c$

$a 3b + 5c = a (..)$

$a – 3b + 5c = a – (3b – 5c)$ .

$a + 4b 4c = a + (..)$

$a + 4b – 4c = a + (4b - 4c)$ .

$a + b c + d = a + (..)$

$a + b – c + d = a + (b – c + d)$ .

$5a + 4b + 4x 2c = 4x ()$ .

$5a + 4b + 4x – 2c = 4x – (2c – 5a – 4b)$ .

Subtract:
$4 x - 6 y + 3 z$ from $12 x + 7 y - 21 z$

$4 x - 6 y + 3 z$ from

$12 x + 7 y – 21 z$
The value of the subtraction is calculated as follows

$(12 x + 7 y - 21 z) - (4 x - 6 y + 3 z)$

$= 12 x - 4 x + 7 y + 6 y - 21 z - 3 z$
On further calculation, we get

$= 8 x + 13 y - 24 z$

$m + n p = - (..)$ .

$m + n – p = - (p – m – n)$ .

$x^{2} y^{2} + z^{2} = x^{2} ()$

$x^{2} – y^{2} + z^{2} = x^{2} – (y^{2} – z^{2})$ .

$3x z + y = 3x (...)$

$3x – z + y = 3x – (z – y)$ .

Subtract:
$5 - a - 4 b + 4 c$ from $5 a - 7 b + 2 c$

$5 - a - 4 b + 4 c$ from

$5 a - 7 b + 2 c$
The value of the subtraction is calculated as follows

$(5 a - 7 b + 2 c) - (5 - a - 4 b + 4 c)$

$= 5 a + a - 7 b + 4 b + 2 c - 4 c - 5$
We get,

$=6 a - 3 b - 2 c - 5$

Subtract the sum of $x + y$ and $x - z$ from the sum of $x - 2 z$ and $x + y + z.$

Sum of

$x+y$ and

$x-z$ will be,

$\left( {x + y} \right) + \left( {x - z} \right) = x + y + x - z$

$= x + x + y - z$

$=2x+ y - z$

Sum of

$x - 2z$ and

$x + y + z$ will be,

$\left( {x - 2z} \right) + \left( {x + y + z} \right) = x - 2z + x + y + z$

$= x + x + y - 2z + z$

$= 2x + y - z$

From the above two results, we can conclude that sum of

$x + y,\ x-z$ and

$x - 2z,\ x + y + z$ is equal so, difference of sum of

$x + y,\ x-z$ from sum of

$x - 2z$ and

$x + y + z$ will be equal to

$0.$

Subtract ( $5x + 6y - 3z$ ) from ( $3x + 5y - 4z$ )

The value of the subtraction is calculated as,

$(3x + 5y - 4z) - (5x + 6y - 3z)$

$= 3x - 5x + 5y - 6y - 4z + 3z$

On simplification,

we get,

$= - 2 x - y - z$

Hence,

$(3 x + 5 y - 4 z) - (5 x + 6 y - 3 z) = - 2 x - y - z$

Subtract:
$2 ab + cd - ac - 2 bd$ from $ab - 2 cd + 2 ac + bd$

$2 ab + cd - ac - 2 bd$ from

$ab - 2 cd + 2 ac + bd$
The value of the subtraction is calculated as follows

$(ab - 2 cd + 2 ac + bd) - (2 ab + cd - ac - 2 bd)$

$= ab - 2 ab - 2 cd - cd + 2 ac + ac + bd + 2 bd$
On calculating further, we get

$= - ab - 3 cd + 3 ac + 3 bd$

The sum of two expressions is $5 x^2 - 3 y^2$ . If one of them is $3 x^2 + 4 xy - y^2$ , find the other.

The other expression is calculated as follows

$(5x^2 - 3y^2) - (3x^2 + 4xy - y^2)$

$= 5x^2 - 3x^2 - 4xy - 3y^2 + y^2$
We get,

$= 2 x^2 - 4 xy - 2y^2$

From the sum of $x + y - 2 z$ and $2x - y + z$ , subtract $x + y + z$ .

The given algebraic expressions are

$(x + y - 2 z),\ (2 x - y + z)$ and

$(x + y + z)$ .

We need to subtract the third expression from the sum of the first two.

Hence, we will get:

$(x + y - 2 z) + (2 x - y + z) - (x + y + z)$

$= x + 2x - x + y - y - y - 2 z - z + z$

$= 2x - y - 2 z$

Hence,

$(x + y - 2 z) + (2 x - y + z) - (x + y + z) = 2x - y - 2 z$ .

By how much should $x + 2y - 3 z$ be increased to get $3x$ ?

The terms calculated as per the question is as follows

$3x - (x + 2y - 3z)$

$= 3x - x - 2y + 3z$
We get,

$= 2x - 2y + 3z$

From the sum of $3 a - 2 b + 4 c$ and $3 b - 2 c$ subtract $a - b - c$ .

The value of terms as per the question is calculated as shown below

$(3 a - 2 b + 4 c ) + (3 b - 2 c) - (a - b - c)$

$= 3 a - 2 b + 4 c + 3 b - 2 c - a + b + c$
On further calculation, we get

$= 3 a - a + 3 b + b - 2 b + 4 c + c - 2 c$

$= 2 a + 2 b + 3 c$
Hence,

$(3 a - 2 b + 4 c) + (3 b - 2 c) - (a - b - c) = 2 a + 2 b + 3 c$

Solve: $2x^{2}+5x-1+8x+x^{2}+7-6x+3-3x^{2}$

Consider the given expression

$2x^2+5x-1+8x+x^2+7-6x+3-3x^2$ and solve it by combining like terms as shown below:

$2x^{ 2 }+5x-1+8x+x^{ 2 }+7-6x+3-3x^{ 2 }\\ =(2x^{ 2 }+x^{ 2 }-3x^{ 2 })+(5x+8x-6x)+(-1+7+3)\\ =(3x^{ 2 }-3x^{ 2 })+(13x-6x)+(-1+10)\\ =0+7x+9\\ =7x+9$

Hence,

$2x^{ 2 }+5x-1+8x+x^{ 2 }+7-6x+3-3x^{ 2 }=7x+9$ .

Take $1 - a + a^2\,\, from \,\,a^2 + a + 1$

$\textbf{Step-1: Subtract the first term from the second term.}$

$\text{Take }$

$1 - a + a^2\,\, \text{from} \,\,a^2 + a + 1$

$\text{The value of the subtraction is calculated as,}$

$(a^2 + a + 1) - (1 - a + a^2)$

$= a^2 - a^2 + a + a + 1 - 1$

$\text{We get,}$

$= a + a$

$= 2\,a$

$\textbf{Hence, answer is 2a.}$

Separate constant terms and variable terms from the following :
$8 , x , 6 + x,- 5xy^{2},15az^{2} ,32 z/ xy ,y^{2} / 3x$

The constant term is

$8.$
The variable terms are

$x , 6xy , 6 + x,- 5xy^{2},15az^{2} ,32 z/ xy ,y^{2} / 3x$

Enclose the given terms in brackets as required:
$x - y - z = z - (..........)$

$x - y - z = z - (y -x )$

Simplify :
$9x - (-4x + 5)$

$9x - (-4x + 5)$
We can write is as

$= 9x + 4x - 5$
So we get

$= 13x - 5$

Simplify :
$3a / 8 + 4a / 5 - (a / 2 + 2a / 5)$

$3a / 8 + 4a / 5 - (a / 2 + 2a / 5)$
Here the LCM of

$8, 5$ and

$2$ is

$40$

$= [15a + 32a - ( 20a + 16a)] / 40$
By further calculation

$= (47a - 36a) / 40$

$= 11a / 40$

Simplify :
$- 2x (x + y) + x^{2}$

$- 2x (x + y) + x^{2}$
We can write it as

$-2x^{2} - 2xy + x^{2}$
So we get

$= - x^{2} - 2xy$

Add:
$x^3-x^2y+5xy^2 +y^3; -x^3 -9xy^2 +y^3; 3x^2y+9xy^2$

$\dfrac {\quad x^3-x^2y+5xy^2 +y^3 \\ -x^3+0x^2y-9xy^2 +y^3 \\ +0x^3+3x^2 y+9xy^2+0y^3}{\quad \ \ \ \ \ \ 2x^2y+5xy^2 +2y^3}$

Simplify :
$(p - 2q) - (3q - r)$

$(p - 2q) - (3q - r)$
We can write it as

$= p - 2q - 3q + r$
So we get

$= p - 5q + r$

Simplify :
$p + q - (p - q) + (2p - 3q)$

$p + q - (p - q) + (2p - 3q)$
We can write it as

$= p + q - p + q + 2p - 3q$
So we get

$= 2p - q$

Simplify :
$6a - (-5a - 8b) + (3a + b)$

$6a - (-5a - 8b) + (3a + b)$
We can write it as

$= 6a + 5a + 8b + 3a + b$
So we get

$= 6a + 5a + 3a + 8b + b$

$= 14a + 9b$

Simplify :
$2b / 5 - 7b / 15 + 13b/ 3$

$2b / 5 - 7b / 15 + 13b/ 3$
Here the LCM of

$3, 5$ and

$15$ is

$15$

$= (6b - 7b + 65b) / 15$

$= 64b / 15$

Simplify :
$b (2b - 1/b) - 2b (b - 1/b)$

$b (2b - 1/b) - 2b (b - 1/b)$
We can write it as

$= 2b^{2} - 1 - 2b^{2} + 2$
So we get

$= 1$

Solve x $\dfrac{10x^2+12x+36x}{x}=\dfrac{120+x^2}{x}$

$10x^2 + 12x + 36x – 120 – x^2 = 0$

$9x^2 + 36x – 108 = 0$ divide by

$9$

$x^2 + 4x – 12 = 0$

$x^2 + 6x – 2x – 12 = 0$

$(x + 6) – 2 (x + 6) = 0$

$(x + 6) (x – 2) = 0$

$x = - 6$ (or)

$x = 2$

$\text { Take away } 8 x-7 y-8 p+10 q \text { from } 10 x+10 y-7 p+9 q$

$\begin{array}{lllllll}10x &+&10y&-&7p&+&9q \\8x &-&7y&-&8p&+&10q \\(-)&&(+)&&(+)&&(-) \\ \hline 2x &+&17y&+&p&-&q\end{array}$

What must be added
$x^4 -x^3 +x^2 +x+3$ to obtain $x^4+x^2-1$ ?

$\dfrac {\quad x^4+0x^3+x^2+0x -1\\ +x^4 -x^3 \ \ +x^2 +x\ +3\\ -\quad\ +\quad\ \ -\quad-\quad-}{x^3 \quad -x-4}$

Subtract:
$abc -4cad- bcd$ form $3abc+5bcd -cad$

$\dfrac {\quad 3abc +5bcd -cad \\ +\ abc -bcd -4cad\\ -\quad \ \ +\quad\ \ \ +}{2abc+6dcb +3cad}$

How much more than
$2x^2 +4xy+2y^2$ is $5x^2 +10xy-y^2$ ?

$\dfrac {\quad5x^2+10xy-y^2 \\ +2x^2+4xy +2y^2}{\ \ 3x^2 -6xy-3y^2}$

Take $m^2+m+4$ from $-m^2+3m+6$ and the results from $m^2+m+1$

$\dfrac {-m^2+3m+6\\ + m^2+ m+ 4\\ - \quad -\quad -}{-2m^2+2m+2}$

$\dfrac {\ +m^2+m+1\\ -2m^2+ 2m+ 2\\ +\quad -\quad -}{3m^2-m-1}$

Take away $3a^3 +4x^2-5x+6$ from $3x-4x^2 +5x-6$

$\dfrac {\quad 3x^3-4x^2+5x-6\\ -3x^3+4x^2 -5x+6\\ +\quad \ \ -\quad \ \ \ +\quad -}{\quad 6x^3 -8x^2 +10x-12}$

Subtract the sum of
$5y^2+y-3$ and $y^2-3y+7$ from $6y^2+y-2$

$\dfrac {\quad5y^2+y-3\\+ y^2-3y+7}{\quad 6y^2-2y+4}$

$\dfrac {\quad 6y^2 +y-2\\ \quad 6y^2-2y+4\\ -\quad \ \ + \quad\ \ -}{\quad\quad3y-6}$

How much less $2a^2+1$ is than $3a^2-6$ ?

$\dfrac {\quad3a^2-6\\ \quad2a^2+1\\ -\quad \ \ -}{a^2-7}$

Subtract:
$a^2 +ab+b^2$ from $4a^2-3ab +2b^2$

$\dfrac {\quad4a^2-3ab+2b^2\\ +\ a^2+\ \ ab+\ b^2\\ -\quad \ - \quad \ \ -}{3a^2-4ab+b^2}$

Two adjacent sides of a rectangle are $2 x^{2}-5 x y+3 z^{2}$ and $4 x y-x^{2}-z^{2}$ . Find its perimeter.

Let

$l=2 x^{2}-5 x y+3 z^{2}$ and

$b=4 x y-x^{2}-z^{2}$
Perimeter of a rectangle

$=2(1+b)$

$=2\left[2 x^{2}-5 x y+3 z^{2}+4 x y-x^{2}-z^{2}\right]$

$\left.=4 x^{2}-10 x y+6 z^{2}+8 x y-2 x^{2}-2 z^{2}\right]$

$=2 x^{2}+4 z^{2}-2 x y$

Add :
$5(x+y),\ 3(x+2y),\ 5(2x-y)$

$\begin{array}{l} 5\left( { x+y } \right) ,\, \, 3\left( { x+2y } \right) ,\, \, 5\left( { 2x-y } \right) \\ addition\to \\ \Rightarrow 5x+5y+3x+6y+10x-5y \\ \Rightarrow 18x+6y \\\end{array}$

What should be added to $2xy+3yz+4zx+5xyz$ to get $x+y+z+2xy+3yz+4zx-xyz$ ?

$\begin{array}{l} x+y+z+2xy+3zy+4zx-xyz \\ -\left( { 2xy+3yz+4zx+5xyz } \right) \\ =x+y+z-6xyz \end{array}$

should be added

The perimeter of a triangle is $15 -23 x+9$ and two of its sides are $5 \times 2+8 x-1$ and $6 \times 2-9 x+ 4.$ Find the third side

Let the sides of the triangle be $a, b$ and $c$ Let $a=5 x^{2}+8 x-1$

$b=6 x^{2}-9 x+4$

Perimeter $=15 x^{2}-23 x+9$

$a+b+c=15 x^{2}-23 x+9$

$5 x^{2}+8 x-1+6 x^{2}-9 x+4+c$

$=15 x^{2}-23 x+9$

$11 x^{2}-x+3+c=15 x^{2}-23 x+9$

$c=15 x^{2}-23 x+9-\left(11 x^{2}-x+3\right)$

$=15 x^{2}-23 x+9-11 x^{2}+x-3$

$c=4 x^{2}- 22 x+6$

$\therefore$ The third side is $4 x^{2}-22 x+6$

Add: $5x-8y+2z, 3z-4y-2x, 6y-z-x$ and $3x-2z-3y$ .

Let us add the given expressions

$5x−8y+2z,3z−4y−2x,6y−z−x$ and

$3x−2z−3y$ as shown below:

$(5x−8y+2z)+(3z−4y−2x)+(6y−z−x)+(3x−2z−3y)\\ =5x−8y+2z+3z−4y−2x+6y−z−x+3x−2z−3y\\ =(5x-2x-x+3x)+(-8y-4y+6y-3y)+(2z+3z-z-2z)\quad \quad \quad (Combining\quad like\quad terms)\\ =(5-2-1+3)x+(-8-4+6-3)y+(2+3-1-2)z\\ =5x-9y+2z$

Hence,

$(5x−8y+2z)+(3z−4y−2x)+(6y−z−x)+(3x−2z−3y)=5x-9y+2z$ .

Simpllify $:\ 2x^{2}+5x-1+8x+x^{2}+7-6x+3-3x^{2}$

$2{x}^{2}+5x-1+8x+{x}^{2}+7-6x+3-3{x}^{2}$

$=3{x}^{2}-3{x}^{2}+13x-6x+10-1$

$=7x+9$ Ans.

Add: $(4x^2-7xy+4y^2-3), (5+6y^2-8xy+x^2)$ and $(6-2xy+2x^2-5y^2)$ .

Let us add the given expressions

$4x^2−7xy+4y^2−3,5+6y^2−8xy+x^2$ and

$6−2xy+2x^2−5y^2$ as shown below:

$(4x^{ 2 }−7xy+4y^{ 2 }−3)+(5+6y^{ 2 }−8xy+x^{ 2 })+(6−2xy+2x^{ 2 }−5y^{ 2 })\\ =4x^{ 2 }−7xy+4y^{ 2 }−3+5+6y^{ 2 }−8xy+x^{ 2 }+6−2xy+2x^{ 2 }−5y^{ 2 }\\ =(4x^{ 2 }+x^{ 2 }+2x^{ 2 })+(−7xy−8xy−2xy)+(4y^{ 2 }+6y^{ 2 }-5y^{ 2 })+(-3+5+6)\quad \quad \quad (Combining\quad like\quad terms)\\ =(4+1+2)x^{ 2 }+(-7-8-2)xy+(4+6-5)y^{ 2 }+8\\ =7x^{ 2 }-17xy+5y^{ 2 }+8$

Hence,

$(4x^{ 2 }−7xy+4y^{ 2 }−3)+(5+6y^{ 2 }−8xy+x^{ 2 })+(6−2xy+2x^{ 2 }−5y^{ 2 })=7x^{ 2 }-17xy+5y^{ 2 }+8$ .

Subtract
$4a+3b$ from $2a+2b-4$ .

Let us subtract the given expression

$4a+3b$ from

$2a+2b-4$ as shown below:

$(2a+2b-4)-(4a+3b)\\ =2a+2b-4-4a-3b\\ =(2a-4a)+(2b-3b)-4\quad \quad \quad (Combining\quad like\quad terms)\\ =(2-4)a+(2-3)b-4\\ =-2a-b-4$

Hence,

$(2a+2b-4)-(4a+3b)=-2a-b-4$ .

Simplify $4x-(3y-x+2z)$

$4x- \left(3y-x+2z \right)$

$= 4x-3y+x-2z$

$= \boxed {5x-3y-2z}$ Ans

Add: $2x^3-9x^2+8, 3x^2-6x-5, 7x^3-10x+1$ and $3+2x-5x^2-4x^3$ .

Let us add the given expressions

$2x^3−9x^2+8,3x^2−6x−5,7x^3−10x+1$ and

$3+2x−5x^2−4x^3$ as shown below:

$(2x^{ 3 }−9x^{ 2 }+8)+(3x^{ 2 }−6x−5)+(7x^{ 3 }−10x+1)+(3+2x−5x^{ 2 }−4x^{ 3 })\\ =2x^{ 3 }−9x^{ 2 }+8+3x^{ 2 }−6x−5+7x^{ 3 }−10x+1+3+2x−5x^{ 2 }−4x^{ 3 }\\ =(2x^{ 3 }+7x^{ 3 }-4x^{ 3 })+(−9x^{ 2 }+3x^{ 2 }−5x^{ 2 })+(−6x−10x+2x)+(8-5+1+3)\quad \quad \quad (Combining\quad like\quad terms)\\ =(2+7-4)x^{ 3 }+(-9+3-5)x^{ 2 }+(-6-10+2)x+7\\ =5x^{ 3 }-11x^{ 2 }-14x+7$

Hence,

$(2x^{ 3 }−9x^{ 2 }+8)+(3x^{ 2 }−6x−5)+(7x^{ 3 }−10x+1)+(3+2x−5x^{ 2 }−4x^{ 3 })=5x^{ 3 }-11x^{ 2 }-14x+7$ .

Simplify $2(3b-5a)-7[9-6{2-5(a-6)}]$

$=2(3b−5a)−7[9−62−5(a−6)]\\$

$=6b–10a–(63–434–35(a–6))\\$

$=6b–10a–(63–434–35a–210)\\$

$=6b–10a–63+434+35a+210\\$

$=25a+6b+581.$

Find what must be subtracted from the polynomial $4y^{4} + 12y^{3} + 6y^{2} + 50y + 26$ so that the obtained polynomial is exactly divisible by $y^{2} +4y+2$ .

Polynomial is

$4y^4+12y^3+6y^2+50y+26$

Divisor

$=y^2+4y+2$

Divisor -

So, remainder

$=2y-2$

$2y-2$ must be subtracted from the polynomial so that the obtained polynomial is exactly divisible by

$y^2+4y+2.$

Hence, the answer is

$2y-2.$

Add: $6p+4q-r+2r-5p-6, 11q-7p+2r-1$ and $2q-3r+4$ .

Let us add the given expressions

$6p+4q-r+2r-5p-6, 11q-7p+2r-1$ and

$2q−3r+4$ as shown below:

$(6p+4q-r)+(2r-5p-6)+(11q-7p+2r-1)+(2q−3r+4)\\ =6p+4q-r+2r-5p-6+11q-7p+2r-1+2q−3r+4\\ =(6p-5p-7p)+(4q+11q+2q)+(-r+2r+2r-3r)+(-6-1+4)\quad \quad \quad (Combining\quad like\quad terms)\\ =(6-5-7)p+(4+11+2)q+(-1+2+2-3)r-3\\ =-6p+17q-3$

Hence,

$(6p+4q-r)+(2r-5p-6)+(11q-7p+2r-1)+(2q−3r+4)=-6p+17q-3$ .

$\dfrac{a}{a-c}+\dfrac{b}{b-c}$ .

$\begin{matrix} \frac { a }{ { \left( { a-c } \right) } } +\frac { b }{ { \left( { b-c } \right) } } \\ =a\left( { b-c } \right) +b\left( { a-c } \right) \\ =ab-ac+ab-bc \\ =2ab-c\left( { a+b } \right) \, \, Ans. \\ \end{matrix}$

Find the sum of the following.
$2x+4y, -5y$ .

$\begin{matrix} 2x+4y,-5y \\ \left( { 2x+4y } \right) +\left( { -5y } \right) \\ 2x+4y-5y=2x-y.\, \, \, \, Ans. \\ \end{matrix}$

$-2\left( { x }^{ 2 }-{ y }^{ 2 }+xy \right) -3\left( { x }^{ 2 }+{ y }^{ 2 }-xy \right)$

$-2(x^2-y^2+xy) -3 (x^2-y^2-xy)$

$=-2x^2-3x^2+2y^2-3y^2-2xy+3xy$

$=-5x^2-y^2+xy$

Find the sum of the following.
$3pq-6hk$ , $-3qp+14kh$ .

$3pq-6hk+\left( { -3qp+14kh } \right) \\ \Rightarrow \left( { 3pq-3pq } \right) +\left( { 14kh-6kh } \right) \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left[ { \therefore pq=qp } \right] \, \, and \\ \Rightarrow 0+8kh\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left[ { kh=hk } \right] \\ \therefore 8kh\, \, \, \, Ans$

Simplify the following expression.
$x+3y+6x+4y$ .

$\begin{matrix} x+3y+6x+4y \\ \Rightarrow \left( { x+6x } \right) +\left( { 3y+6y } \right) \\ \Rightarrow \left( { 7x+9y } \right) \, \, Ans. \\ \end{matrix}$

Subtract $(20-3x+4x^2-x^3)$ from $(x^3-x^2+2x-19)$ .

To subtract

$(20-3x+4x^2-x^3)$ from

$(x^3-x^2+2x-19)$ .

$\left( { { x^{ 3 } }-{ x^{ 2 } }+2x-19 } \right) -\left( { 20-3x+4{ x^{ 2 } }-{ x^{ 3 } } } \right) \\ = { x^{ 3 } }-{ x^{ 2 } }+2x-19-20+3x-4{ x^{ 2 } }+{ x^{ 3 } }\\ = { x^{ 3 } }+{ x^{ 3 } }-{ x^{ 2 } }-4{ x^{ 2 } } +2x+3x-19-20\\= 2{ x^{ 3 } }-5{ x^{ 2 } }+5x-39\, \, \,$

Hence, the required answer is

$(2{ x^{ 3 } }-5{ x^{ 2 } }+5x-39)$ .

Add the following algebraic expression using both horizontal and vertical methods. Did you get the same answer with both methods
$2x^ {2}-6xy+3;-3x^ {2}-x-4;1+2x-3x^ {2}$

$\begin{array}{l} Horizontal\, \, method \\ \Rightarrow \left( { 2{ x^{ 2 } }-6xy+3 } \right) +\left( { -3{ x^{ 2 } }-x-4 } \right) +\left( { 1+2x-3{ x^{ 2 } } } \right) \\ \Rightarrow 2{ x^{ 2 } }-6xy+3-3{ x^{ 2 } }-x-4+1+2x-3{ x^{ 2 } } \\ \Rightarrow 2{ x^{ 2 } }-3{ x^{ 2 } }-3{ x^{ 2 } }-6xy+3-4+1+2x-x \\ \Rightarrow { x^{ 2 } }\left( { 2-3-3 } \right) -6xy+x\left( { 2-1 } \right) \\ \Rightarrow -4{ x^{ 2 } }-6xy+x \\ Vertical\, \, method, \\ 2{ x^{ 2 } }\, \, \, \, \, \, \, \, -6xy\, \, \, \, \, \, \, \, 3 \\ -3{ x^{ 2\, \, \, \, \, \, } }\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, -4\, \, -x \\ -3{ x^{ 2 } }\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, 1+2x \\ -4{ x^{ 2 } }-6xy+x \\ \Rightarrow -4{ x^{ 2 } }-6xy+x \\ yes,\, \, got\, \, same\, \, answer\, \, with\, \, both\, \, methods. \end{array}$

Evaluate : $8 ( x + y ) ^ { 2 } - 26 ( x + y ) + 15$ = $0$

$8(x+y)^{2}-2b(x+y)+15=0$ Let

$x+y=t$

$8t^{2}-26t+15=0$

$t=\frac{26\pm \sqrt{676-480}}{2\times 8}=\frac{26\pm \sqrt{196}}{16}=\frac{3}{4},\frac{5}{2}$ .

$\Rightarrow x+y= 3/4$ &

$x+= 5/2$ .

1105737_1181321_ans_4027d4c5d08d4572b72d0de6db4d4ff1.jpg

Solve: $8 ( p - q ) ^ { 2 } - 12 ( p - q ) ^ { 2 }$

$8(p-q)^{2}-12(p-q)^{2}$

$=(p-q)^{2}(8-12)$

$=(p-q)^{2}(-4)$

$=-4(p-q)^{2}$

$=-4(p^{2}+q^{2}-2pq)$

$=8pq-4p^{2}-4q^{2}$

Verify: $(6x^{3}y-3xy-7)-2(-2xy-1)=(9x^{3}y-2xy-13)-(3x^{3}y-3xy-8).$

$6{ x^{ 3 } }y-3xy-4xy-7+2=9{ x^{ 3 } }y-3{ x^{ 3 } }y-2xy+3xy-13+8 \\ 6{ x^{ 3 } }y+xy-5=6{ x^{ 3 } }y+xy-5 \\ Hence\, \, L.H.S.=R.H.S \\$

Add the following algebraic expression using both horizontal and vertical methods. Did you get the same answer with both methods
$x^ {2}-2xy+3y^ {2};5y^ {2}+3xy-6x^ {2}$

$\begin{array}{l} Vertical\, \, method \\ { x^{ 2 } }-2xy+3{ y^{ 2 } } \\ +-\underline { 6{ x^{ 2 } }+3xy+5{ y^{ 2 } } } \\ -5{ x^{ 2 } }+xy+8{ y^{ 2 } } \\ Horizontal\, \, method \\ { x^{ 2 } }-2xy+3{ y^{ 2 } }+5{ y^{ 2 } }+3xy-6{ x^{ 2 } } \\ =-5{ x^{ 2 } }+xy+8{ y^{ 2 } } \\ 1=2 \\ Both\, ans\, \, are\, \, same \end{array}$

Simplify:
$(pq-qr+pr)(pq+qr)-(pr+pq)(p+q-r)$

$\begin{array}{l} \left( { pq-qr+pr } \right) \left( { pq+qr } \right) -\left( { pr+pq } \right) \left( { p+q-r } \right) \\ =\left[ { { p^{ 2 } }{ q^{ 2 } }+p{ q^{ 2 } }r-p{ q^{ 2 } }r-{ q^{ 2 } }{ r^{ 2 } }+{ p^{ 2 } }qr+pq{ r^{ 2 } } } \right] -\left[ { { p^{ 2 } }r+pqr-p{ r^{ 2 } }+{ p^{ 2 } }q+p{ q^{ 2 } }-pqr } \right] \\ ={ p^{ 2 } }{ q^{ 2 } }+p{ q^{ 2 } }r-p{ q^{ 2 } }r-{ q^{ 2 } }{ r^{ 2 } }+{ p^{ 2 } }qr+pq{ r^{ 2 } }-{ p^{ 2 } }r-pqr+p{ r^{ 2 } }-{ p^{ 2 } }q-p{ q^{ 2 } }+pqr \\ ={ p^{ 2 } }{ q^{ 2 } }-{ q^{ 2 } }{ r^{ 2 } }+{ p^{ 2 } }qr+pq{ r^{ 2 } }-{ p^{ 2 } }r+p{ r^{ 2 } }-{ p^{ 2 } }q-p{ q^{ 2 } } \end{array}$

Subtract the sum of ${ x }^{ 2 }-5xy+2{ y }^{ 2 }$ and ${ y }^{ 2 }-2xy-3{ x }^{ 2 }$ from the sum of $6{ x }^{ 2 }-8xy-{ y }^{ 2 }$ and $2xy-2{ y }^{ 2 }-{ x }^{ 2 }$ .

1302244_1506543_ans_c2d58c3b500741d7ac945db9784a3886.jpg

Add : $2x(z-x-y)$ and $2y(z+y-x)$

2045134_1471524_ans_6242cceb697c41c582257a09f995a915.jpg

What should be added to $3x^{2}-2x+8$ to get $4x^{2}-x+2?$

Let the polynomial that should be added be a

Given

$3{ x }^{ 2 }-2x+8+a=4{ x }^{ 2 }-x+2$

$a=\left( 4{ x }^{ 2 }-x+2 \right) -\left( 3{ x }^{ 2 }-2x+8 \right)$

$=4{ x }^{ 2 }-3{ x }^{ 2 }-x+2x+2-8$

$={ x }^{ 2 }+x-6$

${ x }^{ 2 }-2xy+{ 3 }y^{ 2 };{ 5y }^{ 2 }+3xy-{ 6x }^{ 2 }$

2039652_1321913_ans_3522234a90f042dc8e3220908b316ad8.jpg

Add the following : $- 3 m n , - 2 m n , 6 m n$

$\textbf{Step 1: Open the bracket and add the alike terms.}$

$-3mn+(-2mn)+6mn = -3mn-2mn+6mn$

$=mn(-3-2+6)$

$= mn(-5+6)$

$=mn(6-5)$

$= mn(1)$

$= mn$

$\textbf{Hence, the sum is mn. }$

Add the following :
$3 x ^ { 2 } ; - 7 x ^ { 2 } ; 5 x ^ { 2 }$

$\textbf{Step 1: Add the alike terms and then simplify it.}$

$3x^2 + (-7x^2)+5x^2$

$= (3+5)x^2-7x^2$

$= 8x^2-7x^2$

$= x^2$

$\textbf{Hence, the sum is } x^2.$

Add : $x ^ { 2 } + y ^ { 2 } + 3 x y - 6$ and $2 x ^ { 2 } - 4 y ^ { 2 }$

$\textbf{Step 1: Add alike terms.}$

$(x^2+y^2+3xy-6)+(2x^2-4y^2)$

$= (1+2)x^2+(1-4)y^2+3xy-6$

$=3x^2-3y^2+3xy-6$

$=3(x^2-y^2+xy-2)$

$\textbf{Hence, the answer is }3(x^2-y^2+xy-2).$

Solve:
$Add-2p^{2}+7p^{2}$

$\begin{array}{l} Add\, \, -2{ p^{ 2 } }+7{ p^{ 2 } } \\ -2{ p^{ 2 } }+7{ p^{ 2 } } \\ =5{ p^{ 2 } } \end{array}$

$add:p\left( p-q \right) ,q\left( q-r \right) and\quad r\left( r-p \right)$

2040799_1332339_ans_d6176dd4295143a4b485e6227e8c263e.jpg

How much does $a^{2}-3ab+2b^{2}$ exceed $2a^{2}-7ab+9b^{2}$ ?

2039850_1676843_ans_1037cf0e0aa9454f8147754a1f1b2589.png

Simplify $\dfrac{36t^{-3}}{6^{-3}\times t^{-5}}(t\neq 0)$

$\dfrac{36t^{-3}}{6^{-3}\times t^{-5}}=\dfrac{6^{2}t^{-3}}{6^{-3}t^{-5}}=6^{2+3}t^{-3+5}=6^{5}t^{2}$

$=7776\ t^{2}$

Add $a^{3}+b^{3}-3$ to the sum of $2a^{3}-3b^{3}-3ab+7$ and $-a^{3}+b^{3}+3ab-9$ .

1516692_1676765_ans_3b51e9d6f0a9461996f427504514f633.jpg

Subtract $3x-4y-7z$ from the sum of $x-3y+2z$ and $-4x+9y-11z$ .

2039816_1672692_ans_f0f253366cac4ccaa34acfe474cd65ca.jpg

Subtract: $-4x$ from $3y$

1514964_1676773_ans_e97c0607eda3479bab02d6d5ef237601.jpg

From the sum of $x^{2}+3y^{2}-6xy, 2x^{2}-y^{2}+8xy, y^{2}+8$ and $x^{2}-3xy$ subtract $-3x^{2}+4y^{2}-xy+x-y+3$ .

2039849_1676824_ans_215c78823e55431b8b19c9785f669b05.JPG

Simplify combining like terms:
3a-2b-ab-(a-b+ab)+3ab+b-a

1323269_1588430_ans_2070f01ea95e4b8dbf69f5437bfc7dfc.jpg

Add the polynomials $x^{3}-2x^{2}-9$ and $5x^{2}+2x+9$

1323345_1588610_ans_0e1b9a40dd5b4ff78302f5986245a576.jpg

What must be subtracted from $2x^{2}-xy+5y^{2}$ to make it $-5x^{2}-3xy-2y^{2}$ ?

2044629_1567498_ans_2f222212bcb54209a4759e7a256de11b.jpg

Take away $\dfrac{2}{3}ac-\dfrac{5}{7}ab+\dfrac{2}{3}bc$ from $\dfrac{3}{2}ab-\dfrac{7}{4}ac-\dfrac{5}{6}bc$ .

2039815_1672691_ans_6c030d0d9bb94cfdaa0f643b188ae1e6.jpg

Solve the equations:
$ax^{2}+bxy+cy^{2}=bx^{2}+cxy+ay^{2}=d$

1971899_1746135_ans_3840293322c9431d8d84422437240dcf.jpg

What must be added to $12x^{3}-4x^{2}+3x-7$ to make the sum $x^{3}+2x^{2}-3x+2$ ?

2039851_1676846_ans_7a22c51633f94a078c3a4480f2de2a1a.png

Solve the following equations:
$\dfrac{x^3 + 1}{x^2 -1} = x + \sqrt{\dfrac{6}{x}}$ .

1977125_1743553_ans_989a1dae0bfe4048a99c00ec7b9022e6.jpg

Food Item	Carbohydrates
Rajma	$60\text{g}$
Cabbage	$5\text{g}$
Potato	$22\text{g}$
Carrot	$11\text{g}$
Tomato	$4\text{g}$
Apples	$14\text{g}$

Algebraic Expressions - Class 7 Maths - Extra Questions

Add:6ax−2by+3cz,6by−11ax−cz6ax-2by+3cz, 6by-11ax-cz and 10cz−2ax−3by10cz-2ax-3by

Add:6p+4q-r+3, 2r-5p-6, 11q-7p+2r-16p+4q-r+3, 2r-5p-6, 11q-7p+2r-1 and 2q-3r+42q-3r+4

Subtract :-2a + b + 6d-2a + b + 6d from 5a -2b -3c5a -2b -3c

Subtract 5x^2 -3xy + y^2 5x^2 -3xy + y^2 from 7x^2 - 2xy -4y^27x^2 - 2xy -4y^2

Subtract :a - 2b -3c a - 2b -3c from -2a + 5b -4c-2a + 5b -4c

Subtract -11 x^2y^2 + 7xy - 6 -11 x^2y^2 + 7xy - 6 from 9x^2y^2 -6xy +99x^2y^2 -6xy +9

Simplify : x^4 - 6x^3 + 2x - 7 + 7x^3 - x + 5x^2 + 2 - x^4 x^4 - 6x^3 + 2x - 7 + 7x^3 - x + 5x^2 + 2 - x^4

Add :2x^3–3x^2+7x–8,\ \ –5x^3+2x^2–4x+1,\ \ 3–6x+5x2–x^32x^3–3x^2+7x–8,\ \ –5x^3+2x^2–4x+1,\ \ 3–6x+5x2–x^3

Add 2 + x -x^2+6x^3,\ -6-2x+4x^2-3x^3,\ \ 2 +x^2,\ \ 3-x^3+4x-2x^22 + x -x^2+6x^3,\ -6-2x+4x^2-3x^3,\ \ 2 +x^2,\ \ 3-x^3+4x-2x^2

Subtract :6x^3 - 7x^2 + 5x - 36x^3 - 7x^2 + 5x - 3 from 4 - 5x + 6x^2 -8x^34 - 5x + 6x^2 -8x^3

Subtract:-8xy-8xy from 7xy7xy

Subtract x - 2y - zx - 2y - z from the sum of 3x - y + z3x - y + z and x + y - 3 zx + y - 3 z.

Enclose the given terms in brackets as required:x^{2}- y^{2} + z^{2} + 3x - 2y = x^{2} - (..............)x^{2}- y^{2} + z^{2} + 3x - 2y = x^{2} - (..............)

Enclose the given terms in brackets as required:x^{2} - xy^{2} - 2xy - y^{2} = x^{2} - (..........)x^{2} - xy^{2} - 2xy - y^{2} = x^{2} - (..........)

Simplify :k / 2 + k / 3 + 2k / 5k / 2 + k / 3 + 2k / 5

Simplify :2p / 3 - 3p / 52p / 3 - 3p / 5

Simplify :y / 5 + y - 19y / 15y / 5 + y - 19y / 15

Simplify :x + x / 2 + x / 3x + x / 2 + x / 3

Simplify :3y / 4 - y / 53y / 4 - y / 5

Simplify :y / 4 + 3y / 5y / 4 + 3y / 5

Simplify:\dfrac{2x}{5} +\dfrac{3x}{4} - \dfrac{3x}{5}\dfrac{2x}{5} +\dfrac{3x}{4} - \dfrac{3x}{5}

Simplify :6k / 7 - (8k / 9 - k / 3)6k / 7 - (8k / 9 - k / 3)

Simplify :4a / 7 - 2a / 3 + a / 74a / 7 - 2a / 3 + a / 7

Simplify :x / 2 - x / 8x / 2 - x / 8

Simplify :x / 5 + (x + 1)/ 2x / 5 + (x + 1)/ 2

Write the number of terms in following polynomial: ax-by+y \times z ax-by+y \times z

If a = 3x - 5y, b = 6x + 3ya = 3x - 5y, b = 6x + 3y and c = 2y - 4xc = 2y - 4x, then 2a - 3b + 4c. 2a - 3b + 4c. ==

Subtract: 8x - 7y - 8p + 10q8x - 7y - 8p + 10q from 10x + 10y - 7p + 9q10x + 10y - 7p + 9q.

Subtract: a- b - 2ca- b - 2c from 4a + 6b - 2c4a + 6b - 2c.

If a = 3x - 5y, b = 6x + 3ya = 3x - 5y, b = 6x + 3y and c = 2y - 4xc = 2y - 4x, then a + b - c a + b - c ==

Subtract the sum of 2x-{x}^{2}+52x-{x}^{2}+5 and -4x-3+7{x}^{2}-4x-3+7{x}^{2} from 55 is

How much does -5a^{2}+3a-5b^{2}-5a^{2}+3a-5b^{2} exceed 7a^{2}+4a-9b^{2}7a^{2}+4a-9b^{2}?

Simplify:4x+3-(2x-5y)4x+3-(2x-5y)

(n-10)^2 + (10-n)(n-10)^2 + (10-n)

Factorize \left( 1+\dfrac { 1 }{ x } +\dfrac { 1 }{ { x }^{ 2 } } +\dfrac { 1 }{ { x }^{ 3 } } \right)\left( 1+\dfrac { 1 }{ x } +\dfrac { 1 }{ { x }^{ 2 } } +\dfrac { 1 }{ { x }^{ 3 } } \right)

-3{x}^{2}y+4{x}^{2}y-5{x}^{2}{y}^{2}-3{x}^{2}y+4{x}^{2}y-5{x}^{2}{y}^{2}What is a coficient of xx

Subtract the sum of a+2b-3ca+2b-3c and 2c-3b-4a2c-3b-4a from the sum of 5b-4c+a5b-4c+a and 2c-3b-4a2c-3b-4a.

Simplify \left( a+b \right) \left( c-d \right) +\left( a-b \right) \left( c+d \right) \left( a+b \right) \left( c-d \right) +\left( a-b \right) \left( c+d \right)

Simplify : { (4m+5n) }^{ 2 }-{ (4m-5n) }^{ 2 }{ (4m+5n) }^{ 2 }-{ (4m-5n) }^{ 2 }

What is the result when 2x^2+5x-62x^2+5x-6 is subtracted from 4x^2-9x+64x^2-9x+6

Enclose the given terms in brackets as required:4a - 9 + 2b - 6 = 4a - (..........)4a - 9 + 2b - 6 = 4a - (..........)

Simplify :a/10 + 2a/5a/10 + 2a/5

Enclose the given terms in brackets as required:-2a^{2} + 4ab - 6a^{2}b^{2} + 8ab^{2} = -2a (...........)-2a^{2} + 4ab - 6a^{2}b^{2} + 8ab^{2} = -2a (...........)

Simplify :2x - (x + 2y - z)2x - (x + 2y - z)

Simplify :x / 2 + x / 4x / 2 + x / 4

Simplify :8 (2a + 3b - c) - 10 (a + 2b + 3c)8 (2a + 3b - c) - 10 (a + 2b + 3c)

Add the following:2x^3-9x^2+8,3x^2-6x-5,7x^3-10x+12x^3-9x^2+8,3x^2-6x-5,7x^3-10x+1 and 3+2x-5x^2-4x^33+2x-5x^2-4x^3

Add:8ab, -5ab, 3ab, -ab8ab, -5ab, 3ab, -ab

Simplify:p+2(q- \overline { r+p } )p+2(q- \overline { r+p } )

Add: p(p-q),q(q-r)p(p-q),q(q-r) and r(r-p)r(r-p)

Combine the following : ( x - 3 ) ( x + 3 ) \left( x ^ { 2 } + 9 \right)( x - 3 ) ( x + 3 ) \left( x ^ { 2 } + 9 \right)

Simplify: (3.5e-4.5f)(1.5e+4f+ef)-4.5e+10f(3.5e-4.5f)(1.5e+4f+ef)-4.5e+10f.

Identify terms and factors in the expression given.-4x+5y-4x+5y.

Identify the numerical coefficients of terms (other than constants) in the following expression(5-3t)(5-3t)

Simplify :a^{2}+4a^{2}-3a+6-10+5aa^{2}+4a^{2}-3a+6-10+5a

Solve:z+z=?z+z=?

Simplify the following expression.6xy+13x-2yx-5x6xy+13x-2yx-5x.

Subtract: {4(2p - 3q)4(2p - 3q)} - {4(2p - 3q)4(2p - 3q)}

Simplify: 9{\left( {x - 2y} \right)^2} - 13 + 4\left( {x - 2} \right)9{\left( {x - 2y} \right)^2} - 13 + 4\left( {x - 2} \right)

Find and correct errors of the following mathematical expressions:2x+3y =5xy2x+3y =5xy

Find and correct errors of the following mathematical expressions:x+2x+3x = 5xx+2x+3x = 5x

Find and correct errors of the following mathematical expressions:5y+2y+y-7y =05y+2y+y-7y =0

Simplify the polynomial and write it in standard form:2(x+3)-(-3x+7)2(x+3)-(-3x+7)

The length and breadth a rectangular field are ( 2 x + 3 y )( 2 x + 3 y ) units and ( 3 x - y )( 3 x - y ) respectively. Find the perimeter of the field in the form of an algebraic expression .

Solve :2a - 3b - [4a - 3b - \{ a - 2c - (a - 2b - c)\} ]]2a - 3b - [4a - 3b - \{ a - 2c - (a - 2b - c)\} ]]

Add:-(4x^2-12xy+9y^2)+(25x^2+4xy-32y^2)(4x^2-12xy+9y^2)+(25x^2+4xy-32y^2)

Evaluate:(l+m) -4m\quad (l+m) -4m\quad

From the sum of 3x-y+113x-y+11 and -y-11-y-11 subtract 3x-y-113x-y-11

Find the difference between P\left( x \right) ={ x }^{ 4 }-{ 3x }^{ 2 }+4x+5,g\left( x \right) ={ x }^{ 2 }+1-xP\left( x \right) ={ x }^{ 4 }-{ 3x }^{ 2 }+4x+5,g\left( x \right) ={ x }^{ 2 }+1-x

Identify the terms and factors in the expression given.\dfrac{3}{4}x+\dfrac{1}{4}\dfrac{3}{4}x+\dfrac{1}{4}.

Add 14x+10y-12xy-13, 18-7x-10y+8xy, 4xy14x+10y-12xy-13, 18-7x-10y+8xy, 4xy.

Add -7mn+5, 12mn+2, 9mn-8, -2mn-3-7mn+5, 12mn+2, 9mn-8, -2mn-3.

Add 4x^2y, -3xy^2, -5xy^2, 5x^2y4x^2y, -3xy^2, -5xy^2, 5x^2y.

Simplify combining like terms.5x^2y-5x^2+3yx^2-3y^2+x^2-y^2+8xy^2-3y^25x^2y-5x^2+3yx^2-3y^2+x^2-y^2+8xy^2-3y^2.

Add 5m-7n, 3n-4m+2, 2m-3mn-55m-7n, 3n-4m+2, 2m-3mn-5.

Add x^2-y^2-1, y^2-1-x^2, 1-x^2-y^2x^2-y^2-1, y^2-1-x^2, 1-x^2-y^2.

Simplify combining like terms.(3y^2+5y-4)-(8y-y^2-4)(3y^2+5y-4)-(8y-y^2-4).

Identify the terms and factors in the expression given.0.1p^2+0.2q^20.1p^2+0.2q^2.

Add a+b-3, b-a+3, a-b+3a+b-3, b-a+3, a-b+3.

Add:
$6ax-2by+3cz, 6by-11ax-cz$ and $10cz-2ax-3by$

Add:
$6p+4q-r+3, 2r-5p-6, 11q-7p+2r-1$ and $2q-3r+4$

Subtract :
$-2a + b + 6d$ from $5a -2b -3c$

Subtract
$5x^2 -3xy + y^2$ from $7x^2 - 2xy -4y^2$

Subtract :
$a - 2b -3c$ from $-2a + 5b -4c$

Subtract
$-11 x^2y^2 + 7xy - 6$ from $9x^2y^2 -6xy +9$

Simplify :
$x^4 - 6x^3 + 2x - 7 + 7x^3 - x + 5x^2 + 2 - x^4$

Add :
$2x^3–3x^2+7x–8,\ \ –5x^3+2x^2–4x+1,\ \ 3–6x+5x2–x^3$

Add
$2 + x -x^2+6x^3,\ -6-2x+4x^2-3x^3,\ \ 2 +x^2,\ \ 3-x^3+4x-2x^2$

Subtract :
$6x^3 - 7x^2 + 5x - 3$ from $4 - 5x + 6x^2 -8x^3$

Subtract:
$-8xy$ from $7xy$

Subtract $x - 2y - z$ from the sum of $3x - y + z$ and $x + y - 3 z$ .

Enclose the given terms in brackets as required:
$x^{2}- y^{2} + z^{2} + 3x - 2y = x^{2} - (..............)$

Enclose the given terms in brackets as required:
$x^{2} - xy^{2} - 2xy - y^{2} = x^{2} - (..........)$

Simplify :
$k / 2 + k / 3 + 2k / 5$

Simplify :
$2p / 3 - 3p / 5$

Simplify :
$y / 5 + y - 19y / 15$

Simplify :
$x + x / 2 + x / 3$

Simplify :
$3y / 4 - y / 5$

Simplify :
$y / 4 + 3y / 5$

Simplify:

$\dfrac{2x}{5} +\dfrac{3x}{4} - \dfrac{3x}{5}$

Simplify :
$6k / 7 - (8k / 9 - k / 3)$

Simplify :
$4a / 7 - 2a / 3 + a / 7$

Simplify :
$x / 2 - x / 8$

Simplify :
$x / 5 + (x + 1)/ 2$

Write the number of terms in following polynomial: $ax-by+y \times z$

If $a = 3x - 5y, b = 6x + 3y$ and $c = 2y - 4x$ , then $2a - 3b + 4c.$ $=$

Subtract: $8x - 7y - 8p + 10q$ from $10x + 10y - 7p + 9q$ .

Subtract: $a- b - 2c$ from $4a + 6b - 2c$ .

If $a = 3x - 5y, b = 6x + 3y$ and $c = 2y - 4x$ , then $a + b - c$ $=$

Subtract the sum of $2x-{x}^{2}+5$ and $-4x-3+7{x}^{2}$ from $5$ is

How much does $-5a^{2}+3a-5b^{2}$ exceed $7a^{2}+4a-9b^{2}$ ?

Simplify: $4x+3-(2x-5y)$

$(n-10)^2 + (10-n)$

Factorize $\left( 1+\dfrac { 1 }{ x } +\dfrac { 1 }{ { x }^{ 2 } } +\dfrac { 1 }{ { x }^{ 3 } } \right)$

$-3{x}^{2}y+4{x}^{2}y-5{x}^{2}{y}^{2}$ What is a coficient of $x$

Subtract the sum of $a+2b-3c$ and $2c-3b-4a$ from the sum of $5b-4c+a$ and $2c-3b-4a$ .

Simplify
$\left( a+b \right) \left( c-d \right) +\left( a-b \right) \left( c+d \right)$

Simplify : ${ (4m+5n) }^{ 2 }-{ (4m-5n) }^{ 2 }$

What is the result when $2x^2+5x-6$ is subtracted from $4x^2-9x+6$

Enclose the given terms in brackets as required:
$4a - 9 + 2b - 6 = 4a - (..........)$

Simplify :
$a/10 + 2a/5$

Enclose the given terms in brackets as required:
$-2a^{2} + 4ab - 6a^{2}b^{2} + 8ab^{2} = -2a (...........)$

Simplify :
$2x - (x + 2y - z)$

Simplify :
$x / 2 + x / 4$

Simplify :
$8 (2a + 3b - c) - 10 (a + 2b + 3c)$

Add the following:
$2x^3-9x^2+8,3x^2-6x-5,7x^3-10x+1$ and $3+2x-5x^2-4x^3$

Add:
$8ab, -5ab, 3ab, -ab$

Simplify: $p+2(q- \overline { r+p } )$

Add: $p(p-q),q(q-r)$ and $r(r-p)$

Combine the following : $( x - 3 ) ( x + 3 ) \left( x ^ { 2 } + 9 \right)$

Simplify: $(3.5e-4.5f)(1.5e+4f+ef)-4.5e+10f$ .

Identify terms and factors in the expression given.
$-4x+5y$ .

Identify the numerical coefficients of terms (other than constants) in the following expression
$(5-3t)$

Simplify :
$a^{2}+4a^{2}-3a+6-10+5a$

Solve:
$z+z=?$

Simplify the following expression.
$6xy+13x-2yx-5x$ .

Subtract: { $4(2p - 3q)$ } - { $4(2p - 3q)$ }

Simplify: $9{\left( {x - 2y} \right)^2} - 13 + 4\left( {x - 2} \right)$

Find and correct errors of the following mathematical expressions:
$2x+3y =5xy$

Find and correct errors of the following mathematical expressions:
$x+2x+3x = 5x$

Find and correct errors of the following mathematical expressions:
$5y+2y+y-7y =0$

Simplify the polynomial and write it in standard form:
$2(x+3)-(-3x+7)$

The length and breadth a rectangular field are $( 2 x + 3 y )$ units and $( 3 x - y )$ respectively. Find the perimeter of the field in the form of an algebraic expression .

Solve :
$2a - 3b - [4a - 3b - \{ a - 2c - (a - 2b - c)\} ]]$

Add:-
$(4x^2-12xy+9y^2)+(25x^2+4xy-32y^2)$

Evaluate: $(l+m) -4m\quad$

From the sum of $3x-y+11$ and $-y-11$ subtract $3x-y-11$

Find the difference between $P\left( x \right) ={ x }^{ 4 }-{ 3x }^{ 2 }+4x+5,g\left( x \right) ={ x }^{ 2 }+1-x$

Identify the terms and factors in the expression given.
$\dfrac{3}{4}x+\dfrac{1}{4}$ .

Add $14x+10y-12xy-13, 18-7x-10y+8xy, 4xy$ .

Add $-7mn+5, 12mn+2, 9mn-8, -2mn-3$ .

Add $4x^2y, -3xy^2, -5xy^2, 5x^2y$ .

Simplify combining like terms.
$5x^2y-5x^2+3yx^2-3y^2+x^2-y^2+8xy^2-3y^2$ .

Add $5m-7n, 3n-4m+2, 2m-3mn-5$ .

Add $x^2-y^2-1, y^2-1-x^2, 1-x^2-y^2$ .

Simplify combining like terms.
$(3y^2+5y-4)-(8y-y^2-4)$ .

Identify the terms and factors in the expression given.
$0.1p^2+0.2q^2$ .

Add $a+b-3, b-a+3, a-b+3$ .

Subtract $6xy$ from $-12xy$ .