Class 9 Maths - Linear Equations In Two Variable

$y - 2 = 0$ is parallel to $x$ -axis.
If true enter 1 else 0.

Since,

$y - 2 = 0 \implies y = 2$
In the linear equation

$y = 2$ , there is no term for

$x$ . This means, that

$x$ can take any value. Points lying on the line

$y = 2$ can be

$(4, 2), (0, 2)$ or

$(-5, 2)$ .
Plotting these points on the graph paper and joining them, we get the graph for

$y = 2$ , which is parallel to

$x-$ axis.

$2y - 5 = 0$ is parallel to $x$ -axis.
If true enter 1 else 0.

Since,

$2y - 5 = 0 => y =$

$\dfrac {5}{2}$

$\implies y = 2.5$
In the linear equation

$y = 2.5$ , there is no term for

$x$ . This means, that

$x$ can take any value.

Points lying on the line

$y = 2.5$ can be

$(0, 2.5), (2, 2.5)$ or

$(-5, 2.5)$ .
Note that

$y = 2.5$ lies in the midway between

$y = 2$ and

$y = 3$ on the graph shown.
Plotting these points

$(0, 2.5), (2, 2.5)$ or

$(-5, 2.5)$ on the graph paper and joining them, we get the graph for

$y = 2.5$ , which is parallel to

$x-$ axis.

$y + 6 = 0$ is parallel to $x$ -axis.
If true enter 1 else 0.

Since,

$y + 6 = 0 => y = -6$
In the linear equation

$y = -6$ , there is no term for

$x$ . This means that

$x$ can take any value.

Points lying on the line

$y = -6$ can be

$(3, -6), (-1, -6)$ or

$(0, -6)$ .
Plotting these points on the graph paper and joining them, we get the graph for

$y = -6$ , which is parallel to

$x-$ axis.

The cost of notebook is twice the cost of a pen. Write a linear equation in two variable to represent this statement.

Let the cost of a pen be Rs.

$x$ and that of a notebook be Rs.

$y$ .

We are given that

$y = 2x$

Hence, the required linear equation.

$x - 5 = 0$ is parallel to $y-$ axis. Find the value of $x$ co-ordinate where it crosses $x-axis$ .

Given,

$x - 5 = 0 => x = 5$
In the linear equation

$x = 5$ , there is no term for

$y$ . This means, that

$y$ can take any value. Points lying on the line

$x = 5$ can be

$(5,0), (5, 4)$ or

$(5, -7)$ . Plotting these points on the graph paper and joining them, we get the graph for

$x = 5$ .

Thus, given line cut the

$x-$ axis at

$5.$

Give the geometric representation of $y = 3$ as an equation in one variable.

it would be a point only.on y axis.

Give the geometric representation of $x = -\dfrac{9}{2}$ in one variable.

A greeting card costs Rs. $3$ and a ream of photocopy paper costs Rs. $8$ . Amir paid Rs. $32$ for $x$ cards and $y$ reams of paper. Write a linear equation based on the information above.

Overall, Amir purchased cards and reams of paper.

Now, 1 card cost Rs. 3 and he bought x cards so in total, he purchased cards of worth 3x

Similarly, for reams of paper, he paid 8y

Total money he paid=Rs.32

Hence our Equation becomes:

$3x+8y=32$

The perimeter of $\triangle PQR$ is $30 cm$ .
Write a linear equation in two variables based on the information given above.

The perimeter of a triangle is sum of all the sides

Hence, given on the values:

Perimeter

$= x+ (x+6) +(2y-4) =30cm$

Perimeter

$= 2x+6 +2y-4 = 30cm$

Perimeter

$x+y = 14cm$

The length and width of a rectangle is $4x cm$ and $y cm$ respectively. If its length is $3 cm$ longer than its width, write an equation involving $x$ and $y$ .

Given:

Length of the rectangle

$=4xcm$

Width of the rectangle

$= ycm$

Now according to the question, the Equation will be:

$4x=y+3$

Or,

$4x-y=3$

Express the following linear equation in the form of ax + by + c = 0 and indicate the value of a, b and c.
$y = \dfrac{-3}{2} x$

$y=\cfrac{-3}{2}x$

$\Rightarrow \cfrac{3}{2}x+y+0=0$

$ax+by+c=0$

Compair both we get

$a=\cfrac{3}{2},b=1$ and

$c=0$

Given the function $h$ such that $h(x)=x^2+kx+5$ and $h(3)=2$ . Find the value of $k$ .

Given,

$h(x)={ x }^{ 2 }+kx+5,h(3)=2$

Thus

$h(3)={ 3 }^{ 2 }+k(3)+5$

$\Rightarrow 2=14+3k$

$\Rightarrow 3k=-12\\ \Rightarrow k=-4$

Write each of the following in the form of ax + by + c = 0 and find the value of a, b and c
(i) x = - 5
(ii) y = 2
(iii) 2x = 3
(iv) 5y = -3

(i)

$x=-5$

$\Rightarrow x+0y+5=0$

$a=1,b0$ and

$c=5$

(ii)

$y=2$

$\Rightarrow 0x+y-2=0$

$a=0 ,b=1$ and

$c=-2$

(iii)

$2x=3$

$\Rightarrow 2x+0y-3=0$

$a=2,b=0$ and

$c=-3$

(iv)

$5y=-3$

$\Rightarrow 0x+5y+3=0$

$a=0,b=5$ and

$c=3$

Express the following linear equation in the form of ax + by + c = 0 and indicate the value of a, b and c.
$2x = - 5y$

$2x=-5y$

$\Rightarrow 2x+5y+0=0$

$ax+by+c=0$

Compair both we get

$a=2,b=5$ and

$c=0$

Express each of the following equations in the form of $ax + by + c= 0$ and write the values of a, b and c.
$\dfrac{x}{2} + \dfrac{y}{2} = \dfrac{1}{6}$

$\dfrac{x}{2}+\dfrac{y}{2}=\dfrac{1}{6}$

$\Rightarrow \dfrac{1}{2}x+\dfrac{1}{2}y-\dfrac{1}{6}=0$

Compair with

$ax+by+c=0$ we get

$a=\dfrac{1}{2}$ ,

$b=\dfrac{1}{2}$ and

$c=-\dfrac{1}{6}$

Hema's age is 4 times the age of Mary. Write a linear equation in two variables to represent this information.

Let the age of Hema Mery is

$x$ year and Mary's age is

$y$ years

Then as per given Hema's age is 4 times the age of Mary

Hence,

$x=4y$

$x-4y=0$

Express each of the following equations in the form of $ax + by + c= 0$ and write the values of a, b and c.
$3x = y$

$3x=y$

$\Rightarrow 3x-y+0=0$

$ax+by+c=0$

Compair both we get

$a=3,b=-1\ and\ c=0$

Express the following linear equation in the form of ax + by + c = 0 and indicate the value of a, b and c.
$\dfrac{x}{3} + \dfrac{y}{4} = 7$

$\frac{x}{3}+\frac{y}{4}=7$

$\Rightarrow \frac{1}{3}x+\frac{1}{4}y-7=0$

$ax+by+c=0$

$a=\frac{1}{3},b=\frac{1}{4}$ and c=-7

Express the following linear equation in the form of ax + by + c = 0 and indicate the value of a, b and c.
$8x + 5y - 3 = 0$

Given

$8x+5y-3=0$

$ax+by+c=0$

Compair both we get

$a=8,b=5$ and

$c=-3$

Sachin and Sehwag scored $137$ runs together. Express the information in the form of an equation.

Let Sachin scored

$x$ runs and Sehwag scored

$y$ runs.

Then Sachin's and Sehwag's total scored runs

$=x+y$

But given that Sachin and Sehwag scored

$137$ runs together.

So,

$x+y=137$

Express the following linear equation in the form of ax + by + c = 0 and indicate the value of a, b and c.
$\dfrac{y}{7} = 3$

$\frac{y}{7}=3$

$\Rightarrow 0x+\frac{1}{7}y-3-=0$

$ax+by+c=0$

Compair both we get

a=0,

$b=\frac{1}{7}$ and c=-3

Express the following statements as a linear equation in two variable.
The cost of a pencil is Rs. 2 and one ball point pen costs Rs.Sheela pays Rs. 100 for the pencils and pens she purchased.

Let Sheela purchase

$x$ pencils and

$y$ ball point pens

Given the cost of pencil is

$\text{Rs.}2$ and one ball point pen is

$\text{Rs.}15$

Then, cost of

$x$ pencils =

$2\times x=2x$

And cost

$y$ ball point pens=

$15\times y=15y$

Total cost

$=2x+15y$

Sheela paid a total of

$\text{Rs.}100$ for the purchase.

$\therefore 2x+15y=100$

$2x+15-100=0$ is the linear equation obtained.

Express the following statements as a linear equation in two variable.
The cost of a ball pen is Rs. 5 less than half the cost of a fountain pen.

Let, the cost of a fountain pen be

$Rs.x$

Half the cost of fountain pen

$=Rs.\dfrac{1}{2}x$

Let, the cost of a ball pen be

$Rs.y$

According to the question

$y=\dfrac{1}{2}x-5$

Express the following statements as a linear equation in two variable
The sum of two numbers is 34.

Let the number one is

$x$ and number two is

$y$

Then sum of both numbers

$=x+y$

But the sum of number given is

$34$

$x+y=34$ ( the required linear equation)

Express the following linear equation in the form of $ax + by + c = 0$ and indicate the values of $a, b$ and $c$ .
$y - 2 = 0$

The given equation is

$y-2=0$

It can be written as,

$\Rightarrow 0\times x+y-2=0$

Now, comparing with

$ax+by+c=0$ , we get,

$a=0,b=1$ and

$c=-2$

Express the following statements as a linear equation in two variable.
Bhargavi got 10 more marks than double of the marks of Sindhu.

Let Bhargavi got marks x and Sindhu got marks y

Given Bhargavi got 10 more marks then double of the marks Sindhu

$\therefore x=10+2y$

$\Rightarrow x-2y-10=0$

Express the following statements as a linear equation in two variable.
Yamini and Fatima of class IX together contributed Rs. 200/- towards the Prime Minister's Relief Fund.

Let Yamini contribute Rs

$x$ and Fatima Contribute Rs

$y$ for Prime Minister's Relief Fund

Then total amount contibuted by Yamini and Fatima for Prime Minister's Relief Fund

$=x+y$ Rs

But given Both contributed Rs 200 towards Prime Minister's Relief Fund

$x+y=200$

$x+y-200=0$

Express the following linear equation in the form of ax + by + c = 0 and indicate the value of a, b and c.
2x = 5

$2x=5$

$\Rightarrow 2x+0y-5=0$

$ax+by+c=0$

Compair both we get

a=2,b=0 and c=-5

Express the following linear equation in the form of ax + by + c = 0 and indicate the value of a, b and c.
$3x + 5y = 12$

$3x+5y=12$

$\Rightarrow 3x+5y-12=0$

$ax+by+c=0$

Compair both we get

$a=3,b=5$ and

$c=-12$

Write the linear equations in two variables using ' $x$ ' and ' $y$ ' to represent the following information :
"The sum of the two numbers is $30$ ."

Let the numbers be

$x$ and

$y$ .

Given that the sum of the two numbers is

$30$ ''
Therefore,

$x + y = 30$ .

Determine whether $(4, 6)$ is the solution of linear equation $x + y = 10$ .

Linear equation

$x + y = 10$
Given,

$x = 4, y = 6$

$4 + 6 = 10$

$10 = 10$

$L.H.S. = R.H.S$
Hence,

$(4, 6)$ is the solution of linear equation.

By using two variables, write the following statement in mathematical form:
"The cost of two tables and five chairs is Rs. $2,200.$ "

Let the cost of a table be Rs.

$x$ and that of a chair be Rs.

$y$ .

So, the given statement can be written in the form of an expression as

$2x + 5y = 2200$

The sum of two numbers is $28$ . Write this information in linear equation of two variable. (Use variable $x$ and $y$ )

Let two numbers be

$x$ and

$y$ respectively.
According to question,

$x + y = 28$

Draw the graph of the line $y=-\dfrac{5}{3}x+2$ .

Substituting

$x=-3, 0, 3$ in the equation of the line, we find the values of

$y$ as follows

Plot the points

$\left( -3,7 \right) ,\left( 0,2 \right)$ and

$\left( 3,-3 \right)$ and draw a line passing through the plotted points. This is the required graph of the equation

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

1028019_618902_ans_a07b02cbded343ed9ba8987f066c9437.jpg

Write any one linear equation using the variables ' $x$ ' and ' $y$ '.

The equation:

$2x+3y=15$

Write any one linear equation using the variables $x$ and $y$ .

A linear equation in

$x$ and

$y$ will be of the form

$ax+by=c$

For example:

$2x+3y=15$ .

In linear equation x + y = 5, if x = 3, then find the value of y.

x + y = 5
Put x = 3 in equation (i), we get

$3 + y = 5$

$\therefore y = 5 - 3$

$\therefore y = 2$

Write the following equation in the general form of a linear equation in two variables:
$7x = 3y + 23$

Given equation is

$7x=3y+23$

The given equation in the general form of a linear equation in two variables is,

$7x - 3y = 23$ .

Write L.H.S and R.H.S of the following simple equations:
$4z+1=8$

$LHS=4z+1$

$RHS=8$

Find the limits within which $c$ must lie in order that the equation $19x+14y=c$ may have six solutions, zero solutions being excluded.

The greatest value of

$c=(n+1)ab-a-b=7\times19\times14-19-14=1829$

The least value of

$c=(n-1)ab+a+b=5\times19\times14+19+14=1363$

Here,

$n=$ no of solutions

$a$ and

$b$ are coefficents of

$x$ and

$y$ respectively.

Hence, the value of

$c$ lies between

$1829$ and

$1363$ .

Prove that the coefficient of $x^r$ in the expansion of $(1-4x)^{- \frac{1}{2} }$ in $\dfrac {|\underline{2r}|}{(|\underline{r}|)^2}$

$(1+x)^{n}=1+nx+(n)(n-1)\dfrac{x^{2}}{2!}+ . . . . . . (n!)\dfrac{x^{n}}{n!}$

Where n is negative or fraction

given that

$(1-4x)^{\dfrac{-1}{2}}=1+(\dfrac{-1}{2})x+(\dfrac{-1}{2})(\dfrac{-1}{2}-1)\dfrac{x^{2}}{2!}+ . . . . . .(\dfrac{-1}{2})(\dfrac{-1}{2}-1)...(\dfrac{-1}{2}-r+1) \dfrac{x^{r}}{r!}$

coefficient of $x^{r}$ is

$\dfrac{(-1).(-3).(-5)....(1-2r)}{2^{r}.r!}$

Write L.H.S and R.H.S of the following simple equations:
$7m=14$

$LHS=7m$

$RHS=14$

Draw the graph of $y=4x-1$ .

Substituting the values

$x=-1, 0, 1$ in the given equation of line, we find the values of

$y$ as follows

$y=4x-1$

$x$	$-1$	$0$	$1$
$4x$	$-4$	$0$	$4$
$y=4x-1$	$-5$	$-1$	$3$

Plot the points

$\left( -1,-5 \right) ,\left( 0,-1 \right)$ and

$\left( 1,3 \right)$ in a graph sheet and draw a line passing through the plotted points. We now get the required linear graph.
1028026_618903_ans_ed27ce17c8db44b9b0503f9cbe2bb2ce.jpg

1028026_618903_ans_ed27ce17c8db44b9b0503f9cbe2bb2ce.jpg

Solve the following equation without transposing:
$x+5=9$

$x+5=9$

Subtract

$5$ from both sides,

$\begin{aligned}{}x + 5 - 5& = 9 - 5\\x &= 4\end{aligned}$

Write L.H.S and R.H.S of the following simple equations:
$5p+3=2p+9$

$LHS= 5p+3$

$RHS= 2p+9$

Write L.H.S and R.H.S of the following simple equations:
$8=q+5$

$LHS=8$

$RHS=q+5$

Write L.H.S and R.H.S of the following simple equations:
$2x=10$

$LHS=2x$

$RHS=10$

Write L.H.S and R.H.S of the following simple equations:
$2x-3=9$

$LHS=2x-3$

$RHs=9$

Solve the following equation.
$\sqrt[4]{1+5x}+\sqrt[4]{5-y}=3, 5x-y=11$ .

$x=0, y=-11$ or

$x=3, y=4$
Hint: Put

$\sqrt[4]{1+5x}=u$ and

$\sqrt[4]{5-y}=v$ . Then

$u+v=3, u^4+v^4=17$
Solving these symmetric equations, we shall get

$u=1, v=2$ or

$u=2, v=1$ etc

Romila went to a stationary stall and purchased $2$ pencils and $3$ erasers for Rs. $9$ . Her friend Sonali saw the new variety of pencils and erasers with Romila, and she also bought $4$ pencils and $6$ erasers of the same kind for Rs. $18$ . Represent this situation algebraically.

Let the cost of

$1$ pencil be Rs. x and that of one eraser be Rs. y.

It is given that Romila purchased

$2$ pencils and

$3$ erasers for Rs.

$9$ .

$\therefore 2x+3y=9$

It is also given that Sonali purchased

$4$ pencils and

$6$ erasers for Rs.

$18$ .

$\therefore 4x+6y=18$

The algebraic representation of the given situation is
(1)

$2x+3y=9$
(2)

$4x+6y=18$

The diameter of front and rear wheels of a tractor are $80cm$ and $2m$ respectively. Find the number of revolutions that rear wheel will make in covering distance in which the front wheel makes $1400$ revolutions.

Given diameter of front wheels

$(2r)=80cm\Rightarrow r=40cm$

diameter of rear wheels

$(2r)=2m\Rightarrow r=1m=100cm$

In one revolution area covered is

$=2\pi r$

$\therefore$ Distance covered by front wheel in

$1400$ revolutions

$=1400\times2\pi(40)$

Let us assume that to cover same distance rear wheel will take

$x$ revolutions

$=x\times 2\pi(100)$

As both wheel will cover same distance

$\therefore 1400\times2\pi(40)=x\times 2\pi(100)$

$x=\dfrac{1400\times 40}{100}=560$

That means rear wheel will make

$560$ revolutions to cover the required distance.

Solve the following linear equations for $x$ and $y$
$3x+2y=5$
$2x-3y=7$

$3x+2y=5.........(i)$

$2x-3y=7.........(ii)$

Solving

$(i)$ and

$(ii)$

$3\times(i)+2\times (ii)$

$9x+6y+4x-6y=15+14$

$13x=29$

$x= \dfrac {29}{13}$ and

$y= \dfrac {-11}{13}$

The temperature of certain place fell by $2^0$ C ( a charge of - $2^0$ C ) each day for a week. What is the change of temperature over the whole week?

Given for each day there is a temperature fall by

$2^{°}C$

In a whole week (7 days), temperature fall

$=7×2^{°}C=14^{°}C$

Solve the following equations :
$\cfrac{x-ab}{a+b} + \cfrac{x-ac}{a+c} + \cfrac{x-bc}{b+c}$ = a+ b + c

The equation can be written as
$(\cfrac{x-ab}{a+b}-c) + (\cfrac{x-ac}{a+c}- b) + (\cfrac{x-bc}{b+c}-a)$ = 0
or $\cfrac{x-ab-bc -ca}{a+b} + \cfrac{x-ab-bc -ca}{a+c} + \cfrac{x-ab-bc -ca}{b+c}$ =0
$(x -ab -bc- ca)(\cfrac{1}{a+b} +\cfrac{1}{a+c} + \cfrac{1}{b+c})$ =0
Assuming $(\cfrac{1}{a+b} +\cfrac{1}{a+c} + \cfrac{1}{b+c})$ $\neq$ 0,
$x= ab + bc +ca$ as the root of the equation. If, however,
$(\cfrac{1}{a+b} +\cfrac{1}{a+c} + \cfrac{1}{b+c})$ = 0, the given equation turns into an identity which holds for every value of x.

There are 'a' trees in the village Lat. If the number of trees increases every year by 'b', then how many trees will there be after 'x' years?

Every year

$b$ tree increase in

$x$ years it becomes

$bx$

after

$x$ year

$a+bx$

After $12$ years, Kanwar shall be $3$ times as old as I was $4$ years ago. Find my present age.

Let the present age of the person be

$x\ years$ .

Therefore,

$4\ years$ ago his age will be

$(x-4)\ years$ .

$12\ years$ after his age will be

$(x+12)\ years$ .

According to the question,

$3(x-4)=x+12$

$3x-12=x+12$

$2x=24$

$x=12$

Therefore, the present age of the person is

$12\ years$ .

A man is twice as old as his son. Twelve years ago, the man was thrice as old as his son. Find their present ages.

Let, the present age of a man

$= x$ years

and the present age of his son

$= y$ years

According to the question,

$x = 2y$ and

$x - 12 = 3(y - 12)$

$\because x - 12 = 3(y - 12)$

$\implies 2y - 12 = 3y - 36$

$\implies 3y - 2y = 36 - 12$

$\implies y = 24$

Hence,

$x = 2y = 2(24) = 48$

Let, the present age of a man

$= 48$ years

and the present age of his son

$= 24$ years

Given $A= 2(a +b)h$ find $a$ if $A=144, b=1, h=12$

Given

$A=2(a+b)h$

$A=144$ ,

$b=1$ ,

$h=12$

Put values in above equation

$144=2(a+1)12$

$144=24(a+1)$

$144=24a+24$

$144-24=24a$

$120=24a$

$a=\dfrac{120}{24}$

$a=5$

Find the value of $k$ for which the system of equations $kx+y=2, x+ky=1$ is undefined.

The system of equation

$kx+y=2, x+ky=1$ will be undefined if

$\dfrac{k}{1}=\dfrac{1}{k}\ne\dfrac{2}{1}$

or,

$\dfrac{k}{1}=\dfrac{1}{k}$

or,

$k^2=1$

or,

$k=\pm 1$ .

So for

$k=\pm 1$ the given system of equations will be undefined.

State which of the following are equation (with variable). Given reasons for your an mention which variable is used from the equation with a variable.
$9x+6=12$

$9x+6=12$

Yes, It is a equation with variable and variable is

$x$

Express the given statements in the form of linear equations in two variables.

$1$ is added to the numerator and $4$ is subtracted from the denominator, the fraction becomes 1.

Let Numerator be

$x$ and denominator be

$y$

According to the question,

$\dfrac{x+1}{y-4}=1$

$\Rightarrow x+1=y-4$

$\Rightarrow x-y+5=0$

State which of the following are equation (with variable). Given reasons for your an mention which variable is used from the equation with a variable.
$5x=0$

$\dfrac{7}{4}+3>5$

No it is not an equation with variable this is an inequality

A train $100 m$ long is moving with a velocity of $60 km /hr$ . Find the time it takes to
cross the bridge one km long

Train 100m long and moving with velocity

$60km/hr=60\times\dfrac{5}{18}m/sec=\dfrac{50}{3}m/sec$ .

We have to find time train will take to cross a brige

$1km$ i.e

$1000m$ long.

Length of train be

$lt$ , length of bridge be

$lb$ and velocity of train be

$v$

So time taken =

$\dfrac{lt+lb}{v}$

$=\dfrac{100+1000}{\dfrac{50}{3}}$

$=66sec$

Verify $a-(-b)=a+b$ for the following value of $a$ and $b$ .
(i). $a=28,b=1$
(ii). $a=118,b=125$
(iii). $a=75,b=84$
(iv). $a=28,b=1$

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

(i)

$a = 21, b = 18$ then

$LHS = 21 - (- 18) = 21 + 18 = 39$

$RHS = 21 + 18 = 39$

(ii)

$a = 118, b = 125$ then

$LHS = 118 - (- 125) = 118 + 125 = 243$

$RHS = 118 + 125 = 243$

(iii)

$a = 75, b = 84$ then

$LHS = 75 - (- 84) = 75 + 84 = 159$

$RHS = 75 + 84 = 159$

(iv)

$a = 28, b = 1$ then

$LHS = 28 - (- 1) = 28 + 1 = 29$

$RHS = 28 + 1 = 29$

Hence verified.

The sum of the two digits of a $2-$ digit number is $9$ . When the digits are reversed, the number deceases by $9$ . Find the original number.

Let the two digit number be

$10x+y$

$x+y=9$

$x=9-y$ (1)

Also

$10x+y=10y+x+9$ (2)

Substituting (1) in (2)

$10(9-y) +y=10y+9-y+9$

$90-10y+y=9y+18$

$72=18y$

$y=4$ (3)

Substituting (3) in (1) gives

$x=5$

Hence the number is

$54$

What will be the ratio of the ages after 5 years?
Rina lost her weights inn the ratio 5:Her original wights was 80 kg. What is her new weight.

Reena's weight

$=80 kg$

Let her reduced weight

$= x$

Ratio of weight

$=5:3$ (given)

$\therefore 80:x=5:3$

product of means

$=$ Product of extremes

$x\times 5=80\times 3$

$x=\dfrac{80\times 3}{5}$

$x=48$

Here, new reduced weight

$=x = 48 kg$ .

A works twice as fast as B. They work together and complete a job in 8 days. How long would A take to complete the job?

${{\text{Ratio of Rate of working of A & B = 2:1}}}$

${{\text{Ratio of time taken is = 1:2}}}$

${{\text{Let x day taken by A}}{\text{.}}}$

${\text{So,;2 day taken to work }}{\left( {\dfrac{1}{x}} \right)^{th}}$

${\text{and 2x day by B}}$

${\text{1 day }} \to {\left( {\dfrac{1}{{2x}}} \right)^{th}}$

${\text{1 day }} \to \dfrac{1}{x} + \dfrac{1}{{2x}} = \dfrac{3}{{2x}}$

${\text{8 day }} \to \dfrac{{8 \times 3}}{{2x}}$

$\dfrac{{8 \times 3}}{{2x}} = 1$

${x = 12{\text{ day}}}$

State which of the following are equation (with variable). Given reasons for your an mention which variable is used from the equation with a variable.
$17q+3=54$

$17q+3=54$

Yes It is an equation with variable and variable is

$q$

Find the values of $x$ and $y$ if:
$(x - 1, y + 3) = (4, 4)$ .

$x-1=4\Rightarrow x=5\\y+3=4\Rightarrow y=1$

Solve $\dfrac{{2x - 3}}{5} + \dfrac{{x + 3}}{4} = \dfrac{{4x + 1}}{7}$

$\dfrac{{2x - 3}}{5} + \dfrac{{x + 3}}{4} = \dfrac{{4x + 1}}{7}$

$\dfrac{{4(2x - 3) + 5(x + 3)}}{{20}} = \dfrac{{4x + 1}}{7}$

$7(8x - 12 + 5x + 15) = 80x + 20$

$56x - 84 + 35x + 105 = 80x + 20$

$91x - 80x = 20 - 105 + 84$

$11x = - 1$

$x = \dfrac{{ - 1}}{{11}}$

Solve $10p\, - \,\left( {3p - 4} \right)\, = \,4\left( {p + 1} \right)\, + \,9$

Consider the given equation.

$10p-(3p-4)=4(p+1)+9$

$10p-3p-4=4p+4+9$

$10p-3p-4p=4+4+9$

$(10-3-4)p=4+4+9$

$3p=17$

$p=\dfrac{17}{3}$

Hence, the value is

$p=\dfrac{17}{3}$ .

Solve:
$8x+4=3(3c-1)+7$

$\begin{array}{l} 8x+4=3\left( { 3c-1 } \right) +7 \\ 8x=3\left( { 3c-1 } \right) +7-4 \\ x=\dfrac { { 3\left( { 3c-1 } \right) +3 } }{ 8 } \\ x=\dfrac { { 9c-3+3 } }{ 8 } \\ x=\dfrac { { 9c } }{ 8 } \end{array}$

Hence,

$x=\dfrac { { 9c } }{ 8 }$

Solve $\dfrac{p}{4} + \dfrac{p}{3} = 55 - \dfrac{{p + 40}}{{50}}$

$\dfrac{7p}{12}+\dfrac{p}{50}=55-\dfrac{4}{5}$

$\dfrac{350p+12p}{600}=\dfrac{271}{5}$

$362p=271\times120$

$p=89.834$

Express the following statement as a linear equation in two variables.
"Seetha got $8$ mark more than double of the marks of Ramya"

Let the marks obtained by seetha and ramya be

$x$ and

$y$ respectively then accordingly to question

$x = 2y + 8$

$= x - 2y = 8$

In the given expression $-2y+3x=14$ , express $y$ is terms of $x$ .

Given expression:

$-2y+3x=14$

To express

$y$ in terms of

$x$ ,

$3x=2y+14$

$2y=3x-14$

$y=\cfrac{3x}{2}-\cfrac{14}{2}$

$\displaystyle y=\cfrac{3x}{2}-7$

If four is subtracted from $3$ times the square of a number to get $6$ . Write this in the form of a linear equation.

Let the number be

$x$

$\therefore 3{ x }^{ 2 }-4=6$

$3x^2=10$

is the required linear equation

Solve for $x$ and $y$ :

$\dfrac{6}{x -1} - \dfrac{3}{y - 2} = 1$

$\dfrac{5}{x-1} + \dfrac{1}{y-2} =2$ , where $x \neq 1, y \neq 2$ .

$\dfrac{6}{x-1}-\dfrac{3}{y-1}=1$ …………

$(1)$

$\dfrac{5}{x-1}+\dfrac{1}{y-2}=2$ ……..

$(2)$

$[$ Here

$x\neq 1, y\neq 2]$ .

Multiply

$(2)\times 3$ .

$(2)\times 3\rightarrow \dfrac{15}{x-1}+\dfrac{3}{y-2}=6$

$(1)=\dfrac{6}{x-1}-\dfrac{3}{y-2}=1$

Add

____________________________

$\dfrac{21}{x-1}=7$

$21=7x-7$

$7x=28$

$x=4$

Put x in

$(1)$

$\dfrac{6}{4-1}-\dfrac{3}{y-2}=1$

$2-\dfrac{3}{y-2}=1$

$1=\dfrac{3}{y-2}$

$y-2=3$

$y=5$ .

Solve the equation $\dfrac {x}{3}+2=2x-3y$

$\dfrac{x}{3}+2=2x-3y$

$x+6=6x-9y$

$5x=9y+6$

$\displaystyle x=\frac{9y+6}{5}$

Factorize the following using appropriate identities:
(i) $9x^2+6xy+y^2$
(ii) $4y^2-4y+1$
(iii) $x^2-\dfrac{y^2}{100}$

$\left(i\right)9{x}^{2}+6xy+{y}^{2}={\left(3x\right)}^{2}+2\times 3x\times y+{\left(y\right)}^{2}$ using

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

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

$\left(ii\right)4{y}^{2}-4y+1={\left(2y\right)}^{2}-2\times 2y\times 1+{1}^{2}$ using

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

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

$\left(iii\right){x}^{2}-\dfrac{{y}^{2}}{100}={x}^{2}-\dfrac{{y}^{2}}{{10}^{2}}$ is of the form

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

$=\left(x+\dfrac{y}{10}\right)\left(x-\dfrac{y}{10}\right)$

Solve $2x+3=5x+6$ & represent it as
i) One variable
ii) Two variable

Given,

$2x+3=5x+6$

$3-6=5x-2x$

$3x=-3$

$x=-1$

1) In one variable

$2x+3=5x+6\rightarrow 1$

$5x-2x+6-3=0$

$3x+3=0$

$x+1=0$

$x+1=0$ is the equation in one variable

2)In two variable

$2x+3=5x+6\rightarrow 1$

Let, consider the whole equation is equal to y

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

$y=3x+3$

$3x-y+3=0$

$3x-y+3=0$ is the equation in two variables.

The marks of Urmila of $4$ subjects are $52,70,66,60$ respectively. If the total average marks is equal to $60$ . How many marks should urmila obtain in fifth subject.

Let the fifth subject marks be

$x$

$\dfrac{52+70+66+60+x}{5}=60$

${52+70+66+60+x}=60\times 5$

${52+70+66+60+x}=300$

$x=54$

Find $k$ , if $kx+3y=5$ passes through $(1,5)$ .

put

$x=1,y=5$

$k+15=5$

$k = 5-15$

$k=-10$

Solve: $\dfrac{2x}{7} + \dfrac{x - 1}{5} = \dfrac{x}{3} + \dfrac{x}{7}$ .

$\dfrac{2x}{7}+\dfrac{x-1}{5}=\dfrac{x}{3}+\dfrac{x}{7}$

$\dfrac{2x}{7}-\dfrac{x}{7}=\dfrac{x}{3}-\left(\dfrac{x-1}{5}\right)$

$\dfrac{x}{7}=\dfrac{5x-3x+3}{15}$

$\dfrac{x}{7}=\dfrac{2x+3}{15}$

$\Rightarrow 15x=14x+21$

$x=21$ .

Solve following equation:
$x^2 + \dfrac{x}{\sqrt{2}} + 1 = 0$

The given equation is

$x^{2}+\dfrac{x}{\sqrt{2}}+1=0$

$\Rightarrow \sqrt{2}x^{2}+x+\sqrt{2}=0$

The roots of the quadratic equation

$Ax^{2}+Bx+C=0$ are

$x=\dfrac{-B\pm \sqrt{B^{2}-4AC}}{2A}$

for the given equation,

$B^{2}-4AC$ is

$=(1)^{2}-4(\sqrt{2})(\sqrt{2})$

$=1-8=-7$

So the roots are

$x=\dfrac{-1\pm \sqrt{-7}}{2\sqrt{2}}$

$x=\dfrac{-1\pm \sqrt{7}i}{2\sqrt{2}}$

Find the value of $x$ in terms of $y$ $4x-4y-20=0$

$4x-4y-20=0$

$x-y-5=0$

$x=y+5$

Give two equations passing through $(3,10)$

a line through point

$(x_{1}y_{1})$ and having slope

$m$ is -

$\dfrac{y-y_{1}}{x-x_{1}-}=m$

$\Rightarrow x=4$

(ii)

$2x+7=13$

Add

$-7$ to both sides,

$2x+7-7=13-7$

$2x=6$

Divide both sides by

$2$ , we get

$\dfrac{2x}{2}=\dfrac{6}{2} \therefore x=3$

(iii)

$-y+36=32$

Add

$-36$ both sides, we get

$-y+36-36=32-36$

$\Rightarrow -y=-4$

Multiply

$\left(-1\right)$ to both sides we get

$\Rightarrow \left(-1\right)\left(-y\right)=\left(-1\right)\left(-4\right)$

$\Rightarrow y=4$

(iv)

$2y-6=8$

Add

$6$ to both sides we get

$2y-6+6=8+6$

$\Rightarrow 2y=14$

Divide both sides by

$2$ , we get

$\dfrac{2y}{2}=\dfrac{14}{2}$

$\therefore y=7$

Father is thrice as old as his son. The sum of their ages is $88$ years. Find their ages.

Let the ages of father and son be

$F,S$ respectively.

Given that

$\Rightarrow F=3S$ and

$\Rightarrow F+S=88$

$\Rightarrow 3S+S=88$

$\Rightarrow 4S=88$

$\Rightarrow S=22$

$\Rightarrow F=66$

Therefore, ages o father and son are

$66, 22 years$

Write a linear equation in two variables x and y such that at $x=4, y=4$ .

$\begin{array}{l} x=4,\, \, y=4 \\ x-4+A\left( { y-4 } \right) =0 \\ x+Ay+4A-4=0 \end{array}$

The cost of $2kg$ of apples and $3kg$ of grapes is Rs. $190$ . Represent this information as linear equation in two variables.

Let cost of apples be

$x$ and grapes be

$y$

$\Rightarrow$ Total cost

$=2x+3y$

Given that cost is

$190$

Therefore,

$2x+3y=190$

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

Consider the given expression ${{\left( 2x+y \right)}^{3}}+{{\left( 2x-y \right)}^{3}}$

We know that ${{A}^{3}}+{{B}^{3}}=\left( A+B \right)\left( {{A}^{2}}+{{B}^{2}}-AB \right)$

$\therefore \,\,A=2x+y,B=2x-y$

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

$=4x\left[ 4{{x}^{2}}+{{y}^{2}}+4xy+4{{x}^{2}}+{{y}^{2}}-4xy-\left( 4{{x}^{2}}-2xy+2xy-{{y}^{2}} \right) \right]$

$=4x\left[ 8{{x}^{2}}+2{{y}^{2}}-4{{x}^{2}}-{{y}^{2}} \right]=4x\left( 4{{x}^{2}}-{{y}^{2}} \right)$

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

The present age of $A$ and $B$ are in the ratio $4:5$ . Ten years ago their ages were in the ratio $3:4$ . What are their present ages?

Let the age of A and B are

$4x$ and

$5x$

According to question,

ten years ago ,

$\dfrac { 4x-10 }{ 5x-10 }$

$=$

$\dfrac { 3 }{ 4 }$

$(4x-10)4=(5x-10)3$

$16x-40=15x-30$

$x=10$

So their present ages are

$4(10)=40$ ,

$5(10)=50$ .

Write the standard or general form of linear equation with two variables.

The general form of linear equation with two variables is

$Ax+By=C$

Find the value of $20 + {\left( {27} \right)^{\frac{{ - 1}}{3}}}$

$20+(27)^{-1/3}$

$=20+\dfrac{1}{(27)^{1/3}} 20+\dfrac{1}{(3)^{3\times 1/3}}$

$=20+\dfrac{1}{3}$

$=\dfrac{60+1}{3}$

$=\dfrac{61}{3}$

If $x:y=1:7$ and $y:z=4:5$ , find
$x:z$

Given

$\dfrac{x}{y}=\dfrac{1}{7}...........(1)$

$\dfrac{y}{z}=\dfrac{4}{5}...........(2)$

Multiply (1) and (2)

$\dfrac{x}{y}\times\dfrac{y}{z}=\dfrac{1}{7}\times\dfrac{4}{5}$

$\dfrac{x}{z}=\dfrac{4}{35}$

Therefore,

$x:z=4:35$

Given Munnus's age to be $x$ years, ca you guess what $(x-2)$ may show .can you guess what $(x+4)$ may show? what $(3x+7)$ may show?

1348530_1134477_ans_7b2b1ecee5b0400c8ad55d606a6e2c6d.jpg

Solve $\dfrac{{5x}}{3} + \dfrac{2}{5} = 1$

$\dfrac{5x}{3} +\dfrac{2}{5} = 1$

$\dfrac{5x}{3} = 1 - \dfrac{2}{5}=\dfrac{3}{5}$

$x=\dfrac{3}{5}*\dfrac{3}{5} =\dfrac{9}{25}$

Meera's mother, in four times as old as Meera. After $5$ years, her mother will be three times as old as she will be then. What are their present ages.

Let present age of Meera

$=x$ years

Let present age of Meera's mother

$=y$ years

After five years age of Meera

$=(x+5)$ years

After five years age of Meera's mother

$=(y+5)$ years

According to problem,

$y=4x$ …………..

$(1)$

$y+5=3(x+5)$ ………

$(2)$

Put

$(1)$ in

$(2)$ , we get

$4x+5=3x+15$

$4x-3x=15-5$

$x=10$

$y=40$

$\therefore$ Meera's present age

$=10$ years

Meera's mother's present age

$=40$ years.

The points $A(a,b)$ and $B(b,0)$ lie on the line $y=8x+3$ then
(i)Find the value if $a\ and\ b$
(ii)Determine if $(2,0)$ is a solution of $y=8x+3$ .

(i)Since

$(a,b)$ and

$(b,0)$ lie on the line

$y=8x+3$ , they must satisfy this equtaion .

$b=8a+3$

$0=8b+3$

Solving we get

$b=\dfrac{-3}{8}$ and

$a=\dfrac{-27}{64}$

(ii) Since

$(2,0)$ doesn't satisfy

$y=8x+3$ ,therefore it is not its solution .

simplify
$3x + \dfrac{1}{2} = \dfrac{3}{8} + \dfrac{x}{8}$

$3x+\dfrac{1}{2}=\dfrac{3}{8}+\dfrac{x}{8}$

$\Rightarrow\,3x-\dfrac{x}{8}=\dfrac{3}{8}-\dfrac{1}{2}$

$\Rightarrow\,\dfrac{24x-x}{8}=\dfrac{3}{8}-\dfrac{4}{8}$

$\Rightarrow\,\dfrac{23x}{8}=\dfrac{3-4}{8}$

$\Rightarrow\,23x=-1$

$\therefore\,x=-\dfrac{1}{23}$

(i) $3(Y+3)+4=28$ . Find Y.

$3(Y+3)+4=28$

$3(Y+3)=28-4$

$3(Y+3)=24$

$(Y+3)=\dfrac{24}{3}$

$(Y+3)=8$

$Y=8-3$

$Y=5$

If $2x+8=-2$ , find the value of $4x+5$ .

$2x+8=-2$

$2x=-10$

$x=-5$

Therefore,

$4x+5=4(-5)+5=-20+5=-15$

Solve: $4(2 - x) = 8$

$4(2-x)=8$

$8-4x=8$

$8-8=4x$

$0=4x$

$\therefore x=0$ .

x-5=6

$x-5=6\\x=11$

$\frac{1}{10}-\frac{7}{x}=35$

$\dfrac{1}{10} - \dfrac{7}{x} = \dfrac{35}{1}$

$\Rightarrow \dfrac{x - 70}{10x} = \dfrac{35}{1}$

$\Rightarrow x - 70 = 350 x$

$\Rightarrow 350x - x = -70$

$\Rightarrow 349 x = -70$

$\Rightarrow x = \dfrac{-70}{349}$

Identify like terms in the following:
$10pq,7p,8q,-{ p }^{ 2 }{ q }^{ 2 },-7qp,100q,-23,12{ p }^{ 2 }{ q }^{ 2 },-5{ p }^{ 2 },41,2405p,78qp,13{ p }^{ 2 }{ q }^{ 2 },{ q }{ p }^{ 2 },701{ p }^{ 2 }$

$10pq,7p,8q,-p^2q^2,-7q,100q,-23,12p^2q^2,-5p^2,41,2405p,78qp,13p^2q^2,qp^2,701p^2$

Like terms are

$-p^2q^2,12p^2q^2,13p^2q^2\\-5p^2,701p^2\\10pq,-7pq,78pq\\7p,2405p,8q,100q,-23,41$

$......$ contains only one variable with highest power $1$ .
A) Simple equation
B) linear equation
C) Linear polynomial
D) multi equation

$1)$ Simple equation:Linear equation which involves one variable is called a linear equation in one variable or Simple equation.

Example:

$3x+5=x+2$

$x+2=10$

$2)$ Linear Equation:An equation in which the highest index of the variables present is

$1$ is called a linear equation

Example:

$x+y+z=1$

$3)$ Linear Polynomial:If the degree of a polynomial is

$1$ then the polynomial is said to be a linear polynomial.Its general form in one variable

$x$ is

$ax+b$ where

$a,b\in R$ and

$a\neq 0$

For example:

$2x+5$

$4)$ Multi-equation:If the degree of a polynomial is negative then the polynomial is said to be a multinomial.Its general form in one variable

$x$ is

$\dfrac{a}{x}+b$ where

$a,b\in R$ and

$a\neq 0$

For example:

${x}^{3}-\dfrac{4}{x}$ ,

$\sqrt{x}-5$ etc

Thus, from the above definitions, we conclude that a simple equation contains only one variable with highest power

$1$

Identify like terms in the following:
$-{ xy }^{ 2 },-4{ yx }^{ 2 },-{ 8x }^{ 2 },2{ xy }^{ 2 },{ 7y,-{ 11x }^{ 2 },11yx,-{ 20x }^{ 2 }y,-6{ x }^{ 2 },{ y,-{ 2xy,3x } } }$

$-xy^2,-4yx^2,-8x^2,2xy^2,7y,-11x^2,11yx,-20x^2y,-6x^2,y,-2xy,3x$

Like terms

$-xy^2,2xy^2\\-4yx^2,-20x^2y\\-8x^2y,-11x^2,-6x^2\\11yx,-2xy$

Find the value of y in terms of x:
$2x+3y+5=0$

Solution:-

$2x + 3y + 5 = 0$

$3y = -5 - 2x$

$\Rightarrow y = \cfrac{-\left( 5 + 2x \right)}{3}$

$\Rightarrow y = \cfrac{-2x}{3} + \cfrac{-5}{3}$

Find $x$ :
$\dfrac { 5 - x } { 5 + x } = \dfrac { 2 } { 5 }$

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

$5(5-x)=2(5+x)$

$25-5x=10+2x$

$15=7x$

$x=\dfrac{15}{7}$

Solve
$5x = 7x - \frac{{10}}{7}$

$5x=7x-\dfrac{10}{7}$

$7x-5x-\dfrac{10}{7}=0$

$2x-\dfrac{10}{7}=0$

$2x=\dfrac{10}{7}$

$x=\dfrac{5}{7}$

Obtain pair of linear equations in two variables: In a cricket match, rohit makes his score twice Dhawan's scored. Both of them together make a total score 150 runs.

Let Rohit's score be x

while Dhawan's score be y.

then,

At.Q : x = 2y

Total runs made by them,

x + y = 150.

1144447_1250482_ans_fc6e11a1fff24965b975a5a8ec4338cb.jpg

The cost of a toy elephant is the same as cost of $3$ balls. Express the statement as a linear equation in two variables.

Let y be the cost of a toy elephant

and x be the cost of a ball.

Given that

cost of a toy elephant is same as cost of

3 balls.

$\Rightarrow y = 3x.$

Therefore the linear equation is

$y - 3x = 0$

1211160_1250308_ans_86faae20462e470983745c5c92b6019a.jpg

Find the value of the unknown variable in the following equation.
$\dfrac{-7p+13}{6}=2p-1$

$\frac { -7p+13 }{ 6 } =2p-1\\ \Rightarrow -7p+13=6\times \left( 2p-1 \right) \\ \Rightarrow -7p+13=12p-6\\ \Rightarrow -19p=-19\\ \Rightarrow p=1$

Solve the following equation without transporting
$3x + 8 = 5x + 2$ .

$3x+8=5x+2.$

$2x=6$

$x= 3$

1117425_1278888_ans_6eab9dfb75f04dafa4aa67d223af11d7.JPG

Write constant term
$2x^2-x$

This equation doesnot contain any constant term so constant term is

$0$

If $3x-2y=24$ and $2x-3y=18$ then find $x+y$ and $x-y$

3x-2y=24 _____ (1)

2x-3y=18 _____ (2)

multiply (1) by 2

6x - 4y = 48

multiply (2) by 3

6x-9y=54

subtracting

6x - 4y = 48

6x - 9y = 54

- + -

___________

5y = -6

$y=-\frac{-6}{5}$

$3x-2(-\frac{6}{5})=24$

$=3x+\frac{12}{5}=24$

$=3x=24-\frac{12}{5}=\frac{120-12}{5}=\frac{108}{5}$

$\therefore x=\frac{36}{5}$

$x+y=\frac{36-6}{5}=\frac{30}{5}=6$

$x-y=\frac{36+6}{5}=\frac{42}{5}$

1182857_1293344_ans_94809cb1acdf4707a01f053925de058c.jpg

Solve $3(y+8)=10(y-4)+8$ .

Given,

$3(y+8)=10(y-4)$

$3y+24=10y-40+8$

$7y=24+40-8$

$7y=56$

$y=8$

Find x:
x+5=12

$x+5=12$

$\Rightarrow x=12-5$

$\Rightarrow x=7$ .

1134031_1245717_ans_747370bff1194f67a2122116b3c8942f.jpg

Write constant term
$x^2y+xy^2+5xy-4$

Constant termof

$x^2y+xy^2+5xy-4$ is

$-4$

$4(x-\frac{1}{3})=28$

1134005_1248728_ans_9f45b375b40040dabc016b878b74a613.jpg

State which of the following are equation (with a variable) given reason and tell your answer. identify the variable from the equation with a variable.
$\dfrac{4}{2} = 2$

An equation is the one in which variable, constants & a sign of equality/inequality should be present.

$\dfrac{4}{2}=2$ {no variable presents)

$2=2$

$\therefore$ Its an identity not equation.

1213260_1300319_ans_5d52a6dcd1824ab0b5f48d5abc3af110.JPG

State which of the following are equation (with a variable) given reason and tell your answer. identify the variable from the equation with a variable.
$5\times 4-8=2x$

$5\times 4-8=2x$

$x\rightarrow$ variable

$20-3=2x$

$2x=12$

$5, 4, 8, 2\rightarrow$ constant

$x=\dfrac{12}{2}=6$

Its an equation, as it contains constant variables and a sign of equality/inequality.

State which of the following are equation (with a variable) given reason and tell your answer. Identify the variable from the equation with a variable.
$2n+1=11$

$2n+1=11$

$n\rightarrow$ variable

$2n=10$

$2, 1, 11\rightarrow$ constant

$n=5$

It is an equation because it consists of a constant, a variable and a sign of equality.

Write one solution of the equation $2x+y=10$ .

Solution means the value of

$x$ &

$y$ which stasfies the given eqation

$x=1, y=8$ will be one of it's solution

State which of the following are equation (with a variable) given reason and tell your answer. identify the variable from the equation with a variable.
$2m<30$

$2m < 30$

$m\rightarrow$ variable

$m < 15$

$30\rightarrow$ constant

The above is defined as an equation as it contains all

$3$ parameters i.e., a constant, a variable & a sign of inequality/equality.

State which of the following are equation (with a variable) given reason and tell your answer. identify the variable from the equation with a variable.
$x-2=0$

$x-2=0$

$x\rightarrow$ variable

$\Rightarrow x=2$

$2\rightarrow$ constant

An equation is defined as the one in which constants, variables and a sign of equality/inequality must present.

If $p = \frac{{4xy}}{{x + y}}$
Prove that :
$\frac{{p + 2x}}{{p - 2x}} + \frac{{p + 2y}}{{p - 2y}} = 2$

$P=\frac{4xy}{x+y}$

$\frac{p+2x}{p-2x}+\frac{p+2y}{p-2y}$

$\Rightarrow \frac{\frac{4xy}{x+y}+2x}{\frac{4xy}{x+y}-2x}+\frac{\frac{4xy}{x+y}+2y}{\frac{4xy}{x+y}-2y}$

$\Rightarrow \frac{2x^{2}+6xy}{2xy-2x^{2}}+\frac{2y^{2}+6xy}{2xy-2y^{2}}$

$\Rightarrow \frac{x^{2}+3xy}{xy-x^{2}}+\frac{y^{2}+3xy}{xy-y^{2}}$

$\Rightarrow \frac{x+3y}{y-x}+\frac{y+3x}{x-y}$

$\Rightarrow \frac{x+3y}{y-x}-\frac{y+3x}{y-x}$

$\Rightarrow \frac{x+3y-y-3x}{y-x}=2$

$\Rightarrow \frac{2y - 2x}{y - x} = 2$

Hence proved

1193003_1294310_ans_e5f356af88ed4eb4b3f6571bb9a06b39.jpg

Simplify $27^{y}=9/3^{y}$ , find $y$ .

$27^{y}= 9/3^{y}$

$= 27^{y}\times 3^{y} = 9$

$= (3^{3})^{y} \times 3^{y} = 3^{2}$

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

$3y + y = 2$

$\therefore 4y = 2$

$\Rightarrow y = 1/2$

1193918_1295809_ans_6a35bbff5d534d65bd24b1c2b48e319e.jpg

State which of the following are equation (with a variable) given reason and tell your answer. identify the variable from the equation with a variable.
$(7*3)-19=8$

$(7\times 3)-19=8$ (variable absent)

$\Rightarrow 21-19=8$

$2\neq 8$ absurdity

It is an absurdity not equation as it doesn't have a variable.

State which of the following are equation (with a variable) given reason and tell your answer. identify the variable from the equation with a variable.
$20=5y$

$20=5y$

$y\rightarrow$ variable

$y=\dfrac{20}{5}$

$\Rightarrow y=4$

$4\rightarrow$ constant

above can be said a equation, as it contains a constant, a variable and a sign of equality/inequality.

1213272_1300397_ans_9ea478a3819544e49be7512d9cc74609.JPG

Verify that $x = 2 , y = - 1 ,$ is a solution of the linear equation $7 x + 3 y = 11$

$\begin{array}{l} We\, have \\ 7x+3y=11 \\ Now\, put\, x=2\, and\, y=-1 \\ 7\left( 2 \right) +3\left( { -1 } \right) \\ =14-3 \\ =11 \\ Hence, \\ L.H.S=R.H.S \\ So,\, x=2,\, and\, \, y=-1\, \, is\, a\, solultion\, of\, the\, linear\, equation\, 7x+3y=11. \end{array}$

$X$ is $5\$ of $y$ , $y$ is $24%$ of $z$ . If $x=480$ , find the values of $y$ and $x$

$x=5y$

$y=24z$

$x=480$

$y=\dfrac{x}{5}=\dfrac{480}{5}=96$

$z=\dfrac{y}{24}=\dfrac{96}{24}=4$ .

1206389_1301500_ans_7b9e2daf5d9c4ce6807dab302a89e2c3.jpg

The cost of $2$ pencils is same as the cost of $5$ erasers. Express the statement as a linear equation in two variables.

Cost of

$2$ pencils=Cost of

$5$ erasers

Let the cost of each pencil be

$x$

and the cost of each eraser be

$y$

$\therefore 2\times x=5\times y$

$\Rightarrow 2x=5y$

$\Rightarrow 2x-5y=0$

Hence, the required equation is

$2x-5y=0$

Find the values of $x$ in the following diagram.

By the helps of given figure

$2x-55+3x+10=180$ [angle at the points is

$180^o$ ]

$5x-45=180$

$5x=180+45$

$5x=225$

$x=45$

Hence, the value of

$x=45$ .

Give the geometrical representation of the equation $y = 3$ as an equation.
(i) In one variable

In one variable- there is only

$Y-axis$ (No x-axis)

Here

$y=3$ is a point

Hence, we can say that

in one variable

$y=3$ is a point.

1228582_1322292_ans_5276bb1d572a448cb036240b954f52da.JPG

Find the solution of the equation $\cfrac{{p - 3}}{{1.5p + 9}} = \cfrac{{ - 7}}{5}$ .

We have,

$\cfrac{{p - 3}}{{1.5p + 9}} = \cfrac{{ - 7}}{5}$

$5p - 15=-10.5p-63$

$5p+10.5p=15-63$

$15.5p=-48$

$p=-3.1$

Hence, this is the answer.

Solve the following linear equation :
$\dfrac{2(x+1)}{3}=\dfrac{3(x-2)}{5}$

$\dfrac{2x+2}{3}=\dfrac{3x-6}{5}$

$\Rightarrow 10x+10=9x-18$ (cross multiplication)

$\Rightarrow x=-28$ .

1194215_1316749_ans_272aff404ccf4c939c96f3e717bbb43d.jpg

Solve for $x$ and $y$
$\dfrac { x + 5 } { 2 } = 3.5$ and $\dfrac { y + 2 } { 2 } = 4$

$\dfrac{x+5}{2}=3.5$

$\dfrac{y+2}{2}=4$

$\Rightarrow x+5=3.5\times 2$

$\Rightarrow y+2=8$

$\Rightarrow x+5=7$

$\Rightarrow y=8-2=6$ .

$\Rightarrow x=7-5=2$

$\therefore x=2, y=6$ .

1222994_1332562_ans_350c2e1160eb415ea7c1610d02c1de22.PNG

Find the product
(ii) $\left( 2x-3y \right) \left( 2x+3y \right) \left( { 4x }^{ 2 }+{ 9y }^{ 2 } \right)$

$(2x-3y)(2x+3y)(4x^{2}+9y^{2})$

$= (4x^{2}-9y^{2})(4x^{2}+9y^{2})$

$;(\because a^2-b^2=(a-b)(a+b)$

$= 16x^{4}-81y^{4}$

Solve:
$4p - 3 = 13\left( {p - 4} \right)$

We have,

$4p - 3 = 13\left( {p - 4} \right)$

$4p - 3 = 13p - 52$

$13p-4p = 52-3$

$9p = 49$

$p = \dfrac{49}{9}$

Hence, this is the answer.

Solve the following
$3x+y=-2$ & $x+y=2$

2x+3y=2 ...(1)

x+y=2 ...(2)

_________

2x= -4

[x=-2] and from equation (1) 3x+y=-2

y=-2-3x=-2+6

[y=4]

$\therefore [x=-0]$ & [y=4]

1212519_1363737_ans_2b737da2712f47e7ac4f0d1206f554f9.JPG

Convert $x+6y=6$ in the form of $y=mx+c$

$x+6y=6$

$\Rightarrow -x+x+6y=-x+6$

$\Rightarrow 6y=-x+6$

$\Rightarrow y=\dfrac{-1}{6}x+\dfrac{6}{6}$

$\Rightarrow y=\dfrac{-1}{6}x+1$ is of the form

$y=mx+c$ where

$m=\dfrac{-1}{6},\,\,c=1$

Find x, y in:
$4x+3y-4=0$
$6x+5y+8=0$

1338753_1344754_ans_771d7ad55fc342b188a6dcda1fb547eb.jpg

If x = 4, find the value of 5x.

For

$x=4$ ,

$5x =5\times4= 20$ .

$2+\sqrt { x } =9$ the value of x ?

$2+\sqrt{x} =9$

$\sqrt{x} =7$

$x=49$

By using variables $x$ and $y$ form any two linear equation in two variable.

We are to construct two linear equation with two variables

$x$ and

$y$ .

They are,

$x+y=3$ and

$2x-3y=5$ .

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

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

Combine like terms

$\Rightarrow 5xy^{2}+3x^{2}y-5x^{2}+x^{2}-3y^{2}-y^{2}+3y^{2}+8xy^{2}$

$\Rightarrow 8x^{2}y-4x^{2}-10y^{2}+8xy^{2}$

$\Rightarrow -[4x^{2}+10y^{2}-8x^{2}y-8xy^{2}]$

$\Rightarrow -[4x^{2}+10y^{2}-8xy[x+y]]$

1211791_1363023_ans_fbed55f5acba4cc4ba80a8e659a289f3.JPG

Solve :
${ (0.7p-0.6q) }^{ 2 }$

${ (0.7p-0.6q) }^{ 2 }$ =

$0.49p^{2} +0.36q^{2}-0.84pq$

Solve $\dfrac { x+5 }{ 2 } +\dfrac { x }{ 3 } =20$

1308510_1368079_ans_a7a3b4a14fd046d3a8eede9bee851050.jpeg

Solve:
$\dfrac{5}{x}+\dfrac{1}{x}=2$

$\dfrac{5}{x}+\dfrac{1}{x}=2$

$\Rightarrow \dfrac{6}{x}=2$

$\Rightarrow 2x=6$

$\therefore x=\dfrac{6}{2}=3$

Find p if $\dfrac23p=\dfrac32$

$\dfrac23p=\dfrac32\\p=\dfrac32\times\dfrac32\\p=\dfrac94$

Express 15 = 2x in the form of ax + by + c = 0 and find the values of a , b, c.

Given

$15=2 x$

$\implies 2 x-15=0$

$\implies 2 x+0 y-15=0$

This is in the form of

$a x+b y+c=0$

$\implies a=2,b=0,c=-15$

Solve the following equation for x :
$\sqrt{(\dfrac{3}{5})^{1-2x}}=4\dfrac{17}{27}$

$\sqrt{(\dfrac{3}{5})^{1-2x}}=4\dfrac{17}{27}$

$[(\dfrac{3}{5})^{1-2x}]^{1/2}=\dfrac{125}{27}$

$(\dfrac{3}{5})^{\dfrac{1-2x}{2}} = (\dfrac{5}{3})^{3}$

$\dfrac{1-2x}{2}=-3$

$1-2x=-6$

$2x=7$

$x=\dfrac{7}{2}$

Solve: $x+7=10$

$x+7=10$

$\Rightarrow x+7-7=10-7$

$\Rightarrow x=3$

$\therefore x=3$

Write four solution for each of the following equation:
$2x+y=7$

$2x+y = 7 \Rightarrow y = -2x+7$

Let

$x = 1\Rightarrow y=5; x=2$

$\Rightarrow y = 3; x = 3$ ;

$\Rightarrow y = 1 x = 4$

$\Rightarrow y = -1 \Rightarrow (1,5) , (2,3), (3,1), (4, -1)$

1226978_1398577_ans_3a07d72c70eb43e09e580df838b5f922.jpg

Write four solution for each of the following equation:
$\pi x+y=9$

$y = 9 - \pi x$

for

$x = i, \ y = 9-\pi; \quad x=2, \ y= 9-2\pi;\quad x = 3, \ y =9-3\pi; \quad x=4, y = 9-4\pi$

Hence

$(1, 9, -\pi), (2, 9-2\pi), (3, 9-3\pi)(4, 9-4\pi)$ would be four solutions possible.

1226979_1398578_ans_7e2d9ebd176c49d898db2e0ef0271941.jpg

Write four solutions for each of the following equations:
$x=4y$

For

$x = 1, y = \dfrac{1}{4}; \quad x=2, y = \dfrac{1}{2}; \quad x = 3, y = \dfrac{3}{4}; \quad x = 4, y = 1$

$\Rightarrow \left(1, \dfrac{1}{4}\right), \left(2, \dfrac{1}{2}\right), \left(3, \dfrac{3}{4}\right)$ and

$(4,1)$ are four possible solutions.

1226982_1398581_ans_780740cf5ee84248b93038686c5f1ab4.jpg

If $\left( {x + \dfrac{1}{x}} \right):\left( {x - \dfrac{1}{x}} \right) = 5:4\$ then the value of $x$ is

Given,

$\left( {x + \frac{1}{x}} \right):\left( {x - \frac{1}{x}} \right) = 5:4\$

or,

$\dfrac{4}{x}+4x=-\dfrac{5}{x}+5x$

or,

$x=\dfrac{9}{x}$

or,

$x^2=9$

or,

$x=\pm 3$ .

if $\dfrac{4}{a}+\dfrac23=\dfrac{22}{15}$ find $a$

$\dfrac4a+\dfrac23=\dfrac{22}{15}\\\dfrac{12+2a}{3a}=\dfrac{22}{15}\\15(12+2a)=22(3a)\\30a+180=66a\\36a=180\\a=5$

Find x if $\dfrac 1x+\dfrac 3x=1$

$\dfrac 1x+\dfrac 3x=1\\1+3=x\\x=4$

solve $4x=28$

$4x=28$

$\Rightarrow x=\dfrac{28}{4}=7$

$\therefore x=7$

Ram covered a distance of 200km in 10 hrs . The first part of his journey is covered by auto ,then he hired a car .The speed of the auto and car is 15 km/hr and 30 km /hr resp. Find the ratio of distance covered by auto and car

Let the distance be covered by auto

$X$ in

$Y$ time

Then distance covered by car

$200-X$ in

$10-Y$ time

According to question,

$15\times Y=X$ and

$30\times (10-Y)=(200-X)$

By solving them we get

$Y=6$

$hours$

$40$

$minutes$ ,

$10-Y=3$

$hours$

$20$

$minutes$ ,

$X=100km$ and

$200-X=100km$

Ratio of distance covered by auto and car

$=100:100=1:1$

Rs. 69 were divided among 115 students so that each girl gets 50 paise less than a boy. Thus each boy received twice the paise as each girl received. The number of girls in the class is:

Given that

$Rs. \ 69$ are divided among a total of

$115$ students.

Each girl gets

$50\ paise$ less than a boy.

To find out: Number of girls in the class.

Let the number of boys be

$x$ and the number of girls be

$y$ .

Hence,

$x+y=115\quad .....(i)$

Let the amount received by each boy be

$b \ paise$ .

Hence, amount received by each girl will be

$(b-50) \ paise$ .

It is also given that a boy receives twice the paise as each girl.

Hence,

$b=2(b-50)$

$\Rightarrow b=2b-100$

$\therefore \ b=100 \ paise$ and

$b-50=100-50=50\ paise$ .

Hence, each boy receives

$100\ paise$ , while each girl receives

$50\ paise$ .

Total amount

$=Rs.\ 69=6900 \ paise$

$\therefore \ 100\times x+50\times y=6900$

$\Rightarrow 100x+50y=6900$

$\Rightarrow 2x+y=138\quad .....(ii)$

Subtracting

$(ii)$ from

$2\times (i)$ , we get:

$2[x+y=115]-[2x+y=138]$

$\Rightarrow (2x+2y=230)-(2x+y=138)$

$\Rightarrow 2x+2y-2x-y=230-138$

$\Rightarrow y=92$

Hence, the number of girls in the class is

$92$ .

Check whether the following is solution of the equation $x-2y=4$ or not
$\left(2, 0\right)$

$\left( 2, 0 \right)$ is solution of equation

$x - 2y = 4$ , then it must satisfy the equation.

Therefore,

$2 - 2 \times 0 = 2 \ne 4$

Hence

$\left( 2, 0 \right)$ is not a solution of equation

$x - 2y = 4$ .

Find b
$\dfrac { 2b }{ 3 } -5=3$

$\dfrac{2b}{3}-5=3$

$\Rightarrow \dfrac{2b}{3}=3+5=8$

$\Rightarrow 2b=8\times 3$

$\Rightarrow 2b=24$

$\Rightarrow b=\dfrac{24}{2}=12$

$\therefore b=12$

Solve $\dfrac { { x }^{ 2 }-(x+1)(x+2) }{ 5x+1 } =6$

We have

$\dfrac{x^2-(x+1)(x+2)}{5x+1}=6$

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

$x^2-x^2-3x-2=30x+6$

$-3x-2=30x+6$

$-3x-30x=6+2$

$-33x=8$

$x=\dfrac{-8}{33}$

Rearrange the formula to make $w$ the subject
$5w-3y+7=0$

$5w-3y+7=0$

$5w=3y-7$

$w=\dfrac{3y-7}{5}$

Check whether the following is solution of the equation $x-2y=4$ or not:
$\left(\sqrt{2}, 4\sqrt{2}\right)$

$\left( \sqrt{2}, 4 \sqrt{2} \right)$ is solution of equation

$x - 2y = 4$ , then it must satisfy the equation.

Therefore,

$\sqrt{2} - 2 \times 4 \sqrt{2} = - 7 \sqrt{2} \ne 4$

Hence

$\left( \sqrt{2}, 4 \sqrt{2} \right)$ is not a solution of equation

$x - 2y = 4$ .

Solve :- $\dfrac {y-4} {3} + 3y = 4$

We have

$\dfrac{y-4}{3} + 3y = 4$ ...(1)

Multiplying 3 in (1) we have

$y - 4 + 9y = 12$

$10y - 4 = 12$

$10y = 16$

$y = \dfrac{16}{10}$

Hence

$y = \dfrac{8}{5}$ or

$1.6$

Solve : $\dfrac { 2 x - 1 } { 3 } - \dfrac { x + 2 } { 2 }=0$

$\dfrac { 2x-1 }{ 3 } -\dfrac { x+2 }{ 2 } =0$

$\Rightarrow \dfrac { 2x-1 }{ 3 } =\dfrac { x+2 }{ 2 }$

$\Rightarrow 4x-2=3x+6$

$\Rightarrow x=8$

Check whether the following equation is with a variable or not. In case of an equation with a variable, identity the variable.
$17+x=5$

$17+x=5$ is an equation

$\to L.H.S =R.H.S \to$ Related variable

$x$ .

State which of the following are equations with a variable. In case of an equation with a variable, identity the variable.
$2b-3=7$

$2b-3=7$ is an equation

$\to L.H.S =R.H.S \to$ Related variable

$b$ .

Show that the points $A (1,2), B (-1,-16)$ and $C (0, -7)$ lies on the graph of the linear equation $y= 9x - 7.$

For

$A (1,2),$ we have

$2 = 9(1) - 7 = 9 - 7 = 2$
For

$B ((-1, -16),$ we have -

$16 = 9(-1) - 7 = -9 - 7 = - 16$
For

$C (0,-7),$ we have -

$7 = 9(0) - 7 = 0 - 7 = - 7$
We see that the line

$y = 9x - 7$ is satisfied by the points

$A (1,2), B (-1, -16)$ and

$C (0, -7).$
Therefore,

$A (1,2), B (-1, -16)$ and

$C (0, -7)$ are solutions of the linear equation

$y = 9x - 7$ and therefore, lie on the graph of the linear equation

$y = 9x - 7.$

Draw the graph for each equation, given below: $5x + y + 5 = 0$ .

$5x + y + 5 = 0$

$y = -5x - 5$
When

$x = 0; y = -5x \times - 5 = -0 - 5 = -5$
When

$x = -1; y = -5\times (-1) - 5 = 5 - 5 = 0$
When

$x = -2; y = -5\times (-2) - 5 = 10 - 5 = 5$

$x$	$0$	$-1$	$-2$
$y$	$-5$	$0$	$5$

1800691_1843764_ans_e00e728fc64a4d5292cfeba392b5dd35.png

State which of the following are equations with a variable. In case of an equation with a variable, identity the variable.
$\dfrac 93 =3$

$\dfrac 93 =3$
Is not an equation

$\to L.H.S\neq R.H.S$ .
It has no variable.

Write any three linear equations in two variables using the variables $x$ and $y?$

We can write many equations which are linear and have the variables x and y.

For example,

$x + y = 3$

$x -2y = 6$

$2x + 3y = 18$

For the graph given below, write down their x-y equation

We know that general equation of straight line ,

$y=mx+c$

From the graph, intercept

$c=2$ and slope,

$m=\tan\theta=\tan135=\tan(90+45)=-\cot45=-1$

The required equation will be

$y=-x+2$

The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement.

$\textbf{Hint: Using the linear equation}$

$\textbf{Step1: Find the linear equation}$

$\text{Let the cost of a notebook be the Rs.x}$

$\text{and the cost of pen be Rs.y}$

$\text{It is given that}$

$\text{Cost of notebooks}$

$=2 \times$

$\text{Cost of pen}$

$x=2y$

$x-2y=0$

$\textbf{ x-2y=0 is the linear equation of two variables}$

If the point $\left(3, 4\right)$ lies on the graph of the equation $3y=ax+7$ , find the value of $a$ .

$3y=ax+7$

As given

$x=3$ and

$y=4$

Put the value in the equation

$\Longrightarrow 3\times 4=a\times 3+7$

$\Longrightarrow 12=3a+7$

$\Longrightarrow 3a=12-7$

$\Longrightarrow 3a=5$

$\Longrightarrow a=\dfrac{5}{3}$

From the choices given below, choose the equation whose graphs are given in Fig. (i) and Fig. (ii).

For fig. (i) For fig. (ii)
(i) $y=x$ (i) $y=x+2$
(ii) $x+y=0$ (ii) $y=x-2$
(iii) $y=2x$ (iii) $y=-x+2$
(iv) $2+3y=7x$ (iv) $x+2y=6$

In the given fig. (i), the solutions of the equation are

$(-1,1),(0,0)$ and

$(1,-1)$ .

$\therefore$ the equation which satisfies these solutions is the correct equation.

Equation (ii)

$x + y = 0$ , satisfies these solutions.

Proof:

If we put the value of

$x = -1$ and

$y = 1$ in the equation

$x + y = 0$

$= x + y = -1 + 1 = 0$

$\therefore L.H.S= R.H.S$

if we put the the value of x = 0 and y = 0

$= x + y = 0 + 0 = 0$

$\therefore L.H.S= R.H.S$

If we put the value of

$x = 1$ and

$y = -1$

$= x + y = 1 + (-1) = 1 – 1 = 0$

$L.H.S= R.H.S$

Hence, option (ii)

$x + y = 0$ is correct.

In the given Fig.(ii) the solutions of the equation are

$(-1,3),(0,2)$ and

$(2,0)$ .

$\therefore$ The equation which satisfies these solutions is the correct equation.

Equation (iii)

$y=-x+2$ , satisfies these solutions.

Proof:

If we put the value of

$x = -1$ and

$y = 3$ in the equation

$y = -x+2$

$y= -x +2$

$3=-(-1)+2$

$3=3$

$\therefore L.H.S= R.H.S$

if we put the the value of

$x = 0$ and

$y = 2$

$y=-x+2$

$2=-0+2$

$2=2$

$\therefore L.H.S= R.H.S$

If we put the value of

$x = 2$ and

$y = 0$

$y= -x+2$

$0=-2+2$

$0=0$

$L.H.S= R.H.S$

Hence, option (iii)

$y=-x+2$ is correct.

The taxi fare in a city is as follows: For the first kilometre, the fare is $8$ and for the subsequent distance it is $5$ per km. Taking the distance covered as $x$ km and total fare as Rs. $y$ , write a linear equation for this information and draw its graph.

$x$	$0$	$1$	$2$
$y$	$3$	$8$	$13$

Taxi fare for first kilometer

$=$ Rs.

$8$

Taxi fare for subsequent distance

$=$ Rs.

$5$

Total distance covered

$= x$

Total fare

$= y$

Since the fare for first kilometer

$=$ Rs.

$8$

According to problem,

Fare for

$(x – 1)$ kilometer

$=$

$5(x-1)$

So, the total fare

$y = 5(x-1) + 8$

$\Rightarrow y = 5(x-1) + 8$

$\Rightarrow y = 5x – 5 + 8$

$\Rightarrow y = 5x + 3$

$Hence$ ,

$y = 5x + 3$

$\text{is the required linear equation.}$

Now the equation is

$y = 5x + 3$ ...(1)

Now, putting the value

$x = 0$ in (1)

$y = 5 \times 0 + 3$

$y = 0 + 3 = 3$ So the solution is

$(0, 3)$

Putting the value

$x = 1$ in (1)

$y = 5 \times 1 + 3$

$y = 5 + 3 = 8$ . So the solution is

$(1, 8)$

Putting the value

$x = 2$ in (1)

$y = 5 \times 2 + 3$

$y = 10 + 3 = 13$ . So the solution is

$(2, 13)$

Express the following linear equations in the form $ax+by+c=0$ and indicate the values of $a, b$ and $c$ in each case:

(i) $2x+3y=9.3\overline { 5 }$
(ii) $x-\dfrac{y}{5}-10=0$
(iii) $-2x+3y=6$
(iv) $x=3y$
(v) $2x=-5y$
(vi) $3x+2=0$
(vii) $y-2=0$
(viii) $5=2x$

(i)

$2x+3y=9.3\bar{5}$

$\Rightarrow 2x+3y-9.3\bar{5}$

Comparing the equation with

$ax+by+c=0$

$a=2,b=3 ,c=-9.3\bar{5}$

(ii)

$x-\dfrac{y}{5}-10=0$

Comparing the equation with

$ax+by+c=0$

$a=1,b=-\dfrac{1}{5} ,c=-10$

(iii)

$-2x+3y=6$

$\Rightarrow -2x+3y-6=0$

Comparing the equation with

$ax+by+c=0$

$a=-2,b=3 ,c=-6$

(iv)

$x=3y$

$\Rightarrow x-3y=0$

Comparing the equation with

$ax+by+c=0$

$a=1,b=-3 ,c=0$

(v)

$2x=-5y$

$\Rightarrow 2x+5y=0$

Comparing the equation with

$ax+by+c=0$

$a=2,b=5 ,c=0$

(vi)

$3x+2=0$

$\Rightarrow 3x+0y+2=0$

Comparing the equation with

$ax+by+c=0$

$a=3,b=0,c=2$

(vii)

$y-2=0$

$\Rightarrow 0x+y-2=0$

Comparing the equation with

$ax+by+c=0$

$a=0,b=1 ,c=-2$

(viii)

$5=2x$

$\Rightarrow -2x+0y+5$

Comparing the equation with

$ax+by+c=0$

$a=-2,b=0 ,c=5$

Give the geometric representations of $y=3$ as an equation
(i) in one variable
(ii) in two variable

(i)

$y = 3$

(ii)

$y = 3$

$\Rightarrow 0x+y=3$

$\Rightarrow 0x + y – 3 = 0$

which is in fact

$y = 3$

It is a line parallel to

$x$ -axis at a positive distance of

$3$ from it. We have two of its solution as

$(0, 3), (1, 3).$

Yamini and Fatima, two students of Class $IX$ of a school, together contributed $Rs.100$ towards the Prime Minister's Relief Fund to help the earthquake victims. Write a linear equation which satisfies this data. (You many take their contributions as $Rs.x$ and $Rs.y$ .) Draw the graph of the same.

$x$	$0$	$50$	$100$
$y$	$100$	$50$	$0$

Let the contribution of Yamini be

$x$ and that of Fatima be

$y$ .

According to problem,

$x + y = 100$ ...(1)

Now , putting the value

$x = 0$ in (1)

$0 + y = 100$

$y = 100$

So, the solution is

$(0, 100)$

Putting the value

$x = 50$ in (1)

$50 + y = 100$

$y = 100 – 50$

$y = 50$

So, the solution is

$(50, 50)$

Putting the value

$x = 100$ in (1)

$100 + y = 100$

$y = 100 – 100$

$y = 0$

So, the solution is

$(100, 0)$

Express each of the following equations in the form of $ax + by + c= 0$ and write the values of a, b and c.
$x - 5 = \sqrt 3 y$

$x-5=\sqrt{3}y$

$\Rightarrow x-5-\sqrt{3}y=0$

$\Rightarrow x-3\sqrt{3}y-5=0$

$ax+by+c=0$

Compair both we get

$a=1,b=-\sqrt{3}\ and\ c=-5$

A number is 27 more than the number obtained by reversing its digits. If its unit's and ten's digits are x and y respectively, write the linear equation representing the above statement.

Let the number's unit's digit is

$x$ and ten's digit is

$y$

Then the number

$=x+10y$

If we reverse the number's digit then the number

$=y+10x$

Given that the number is

$27$ more than the reversed number

So,

$(x+10y)-(y+10x)=27$

$x+10y-y-10x=27$

$9y-9x-27=0$

$y-x-3=0$

$A, B$ and $C$ lie on a line, as shown. The length of $\overline{AB}$ is $(x-4)$ and the length of $\overline{AC}$ is $(x+6)$ . What is the length of $\overline{BC}$ ?

From the figure,

$AC=AB+BC$

$AC=x+6$ and

$AB=x-4$

$BC=AC-AB=(x+6)-(x-4)$

$BC=10$

Prove that $(1+x)^n = 2^n \left\{ 1-n \dfrac{1-x}{1+x} +\dfrac{n(n+1)}{1 \cdot 2 } \left( \dfrac {1-x}{1+x} \right)^2 \cdot \cdot \cdot \right\}$

given

$LHS=(1+x)^{n}$

multiply and divide by 2 inside the bracket

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

denomenator can be manipulated as

$=2^{n}.(\dfrac{1+x}{1+x+1-x})^{n}$

$=2^{n}.(\dfrac{1+x+1-x}{1+x})^{-n}$

$=2^{n}.(1+\dfrac{1-x}{1+x})^{-n}$

now expand it

$=2^{n}.[1-n.\dfrac{1-x}{1+x}+n(n+1).(\dfrac{1-x}{1+x})^{2} - . . . . . . .]$

$=RHS$

Hence proved

Draw the graph of y = 7x.

1714536_619149_ans_482af5663f684b8ca25ad6b8744bfb6f.jpg

Draw a linear graph for the following data.
x 1 2 3 4 5
y 1 2 3 4 5

1718877_619144_ans_838ee59e20d140f89f38621af8428323.jpg

Express the following linear equation in the form of ax + by + c = 0 and indicate the value of a, b, c.
$28 x - 35 y = -7$

The given equation is

$28x-35y=-7$

We can write it as

$28x-35y+7=0$

Comparing it with this equation

$ax+by+c=0$ , we get

$a=28,b=-35$ and

$c=7$

Represent $3x - 7 = 0$ in the form of $ax+by+c=0$ and find the values of $a$ , $b$ and $c$

Given

$3x-7=0$

$3x+0y-7=0$

$ax+y+c=0$

Compare to get

$a=3, b=0 and c=-7$

If x be so small that its square and higher powers may be neglected, find the value of :
$\dfrac { \left( 1 + \dfrac{2}{3} x \right)^{-5} \times (4+3x)^{ \frac{1}{2} } } { (4+x)^{\frac{3}{2} } }$

$y=(1+\dfrac{2x}{3})^{-5} (4+3x)^{\dfrac{1}{2}}(4+x)^{-\dfrac{3}{2}}$

since it is given that

x be so small that its square and higher powers may be neglected

so we expand each bracket upto single power only

$=[1+\dfrac{2x}{3}(-5)] \times 2(1+\dfrac{3x}{4}\times \dfrac{1}{2})\times \dfrac{1}{8}[1+\dfrac{x}{4}({-\dfrac{3}{2}})]$

again neglecting square terms

$=\dfrac{1}{4}[1-\dfrac{10x}{3}] [1]$

$=\dfrac{1}{4}-\dfrac{5x}{6}$

The sum of a two digit number and the number obtained by reversing the order of its digits isIf the digits in unit's and ten's place are 'x' and 'y' respectively.

Given units digit is x and tens digit is y

Hence the two digit number = 10y + x

Number obtained by reversing the digits = 10x + y

Given that sum of a two digit number and the number obtained by reversing the order of its digits is 121.

Then

$(10y+x)+(10x+y)=121$

$\Rightarrow 10y+x+10x+y=121$

$\Rightarrow 11x+11y=121$

$\Rightarrow x+y=11$

Thus the required linear equation is x + y = 11.

If $x$ be so small that its square and higher powers may be neglected, find the value of :
$\sqrt{4-x} \cdot \left( 3 - \dfrac {x}{2} \right)^{-1}$

$y=(4-x)^{0.5}.(3-\dfrac{x}{2})^{-1}=\dfrac{2}{3}(1-\dfrac{x}{4})^{\dfrac{1}{2}}.(1-\dfrac{x}{6})^{-1}$

$=\dfrac{2}{3}[1-\dfrac{1}{2}.\dfrac{x}{4}-\dfrac{1}{4}.\dfrac{x^{2}}{16.2!} . . . . . . . ][1+\dfrac{x}{6}+2\times \dfrac{x^{2}}{36.2!} . . . . . . .]$
since higher powers are neglected , we can neglect power 2 terms as well

$=\dfrac{2}{3}[1+\dfrac{x}{6}-\dfrac{x}{8}]$

$=\dfrac{2}{3}[1+\dfrac{x}{24}]$

If $x$ be so small that its square and higher powers may be neglected, find the value of :
$(1-7x)^{\frac{1}{3}} (1+2x)^{-\frac{3}{4}}$

let

$y=(1-7x)^{\dfrac{1}{3}}(1+2x)^{\dfrac{-3}{4}}$

$= (1-\dfrac{1}{3}.7x-\dfrac{2}{9}\dfrac{49x^{2}}{2!}.....)(1-\dfrac{3}{4}.2x+\dfrac{21}{16}\dfrac{4x^{2}}{2!}...)$

x be so small that its square and higher powers may be neglected

$= (1-\dfrac{1}{3}.7x)(1-\dfrac{3}{2}.x)$

again neglect square terms

$= 1-\dfrac{23x}{6}$

In a class test $(+3)$ marks are given for every correct answer and $(-2)$ marks are given for every incorrect answer and no marks for not attempting any question. (i) Radhika scored $20$ marks. If she has got $12$ correct answers, how many questions has she attempted incorrectly? (ii) Mohini scores $(-5)$ marks in this test, though she has got $7$ correct answers How many questions has she attempted incorrectly.

1) Radhika scored

$12\times3=36$ marks. as

$3$ marks are for correct answer

final score of radhika is

$20$ .

she got

$16$ marks negative. as

$-2$ marks are for incorrect answer

therefore she made

$\dfrac{16}{2}=8$ incorrect answer.

2) Mohini scored

$7\times3=21$ marks. as

$3$ marks are for correct answer

final score of radhika is

$-5.$

she got

$26$ marks negative. as

$-2$ marks are for incorrect answer

therefore she made

$\dfrac{26}{2}=13$ incorrect answer.

How many two- digits positive integers N have the property that the sum of N and the number obtained by reversing the order of the digits of is a perfect square

Let units digit

$x$ and tens digit be

$y$

Number

$=10y+x$

Reverse number

$=10x+y$

Sum

$=11x+11y$

$=11(x+y)$

For sum to be square

$x+y=11$

$\therefore$ Number of ways

$={ ^{ 7+2-1 }{ C }_{ 2-1 } }=^{ 8 }{ C }_{ 1 }=8$

Form a pair of linear equations in two variables for the following information.
"There are some $50$ paisa and $25$ paisa coins in a bag. The total number of coins are $140$ and the value of all coins is $Rs. 50$ ."

Let the

$50$ paisa coins be

$x$ and

$25$ paisa coins be

$y$ , then

$x + y = 140$

and

$\dfrac {50x}{100} + \dfrac {25y}{100} = 50$

$\Rightarrow \dfrac x2 + \dfrac y4 = 50$

$\Rightarrow 2x+ y = 200$

Ten students of class X took part in Mathematics quiz. If the number of girls is $4$ more than the number of boys. Represent this situation algebraically.

Formulation:

Let the number of girls be x and the number of boys be y. It is given that a total of ten students took part in the quiz.

$\therefore$ Number of girls

$+$ Number of boys

$=10$
i.e.,

$x+y=10$

It is also given that the number of girls is

$4$ more than the number of boys.

$\therefore$ Number of girls

$=$ Number of boys

$+4$
i.e.,

$x=y+4$
or,

$x-y=4$

Algebraic Representation:

Thus, the algebraic representation of the given situation is

$x+y=10$ (i)

$x-y=4$ (ii)

Find all numbers x such that
$\left | x \, + \, 1 \right | \, - \,|x|\, + \, 3|x \, - \, 1||x \, - \, 2| \, = \, x \, + \, 2$

The given equation can be re - written as
|x - 1| - |x| + 3 | (x - 1) (x + 1)| + x + 2
the change points as marked on real line are - 1 , 0 , 1, 2 .
cansider five cases
(a) x < - 1 (b) -1 < x < 0
(c) 0 < x < 1
(d) 1< x< 2
(e) x > 2
In case of (a) i . e . x < -1 the equation reduce to
- (x + 1) - (-x ) + 3 ( x - 1) ( x - 2) = x + 2
or

$3x^2$ - 10x + 3 = 0 or (x - 3) ( 3x - 1) = 0

$\therefore x = 3 \, or \, \dfrac {1}{3}$
Since these value of x do not satisfy condition (a) the original equation has not solution in this case
In case (b) , i . e . -1 < x < 0 the equation reduce to
- (x + 1) - (-x ) + 3 ( x - 1) ( x - 2) = x + 2
or

$3x^2$ - 10x + 3 = 0 or (x - 3) ( 3x - 1) = 0

$\therefore x = 3 \, or \, \dfrac {5}{3}$
Since these value of x do not satisfy condition (b) the original equation has not solution in this case
In case (c) , i. e. 0 < x< 1 that equation is
x + 1 - x + 3 (x - 1)(x - 2) = (x + 2)

$3x^2$ - 10x + 3 = 0

$\therefore x = \dfrac {10 \, \pm \, \sqrt{100 \, - \, 60 }}{6} \, = \dfrac {10 \, \pm \,\sqrt[2]{10}}{6} \, = \dfrac {10 \, \pm \, \sqrt{10 }}{6} \,$
the root of

$x = \dfrac{5 \, - \, \sqrt 10}{3}$ satisfied the equation (c) whereas
the root

$x = \dfrac{5 \, + \, \sqrt 10}{3}$ does not satisfy the condition
Hence

$x = \dfrac{5 \, - \, \sqrt 10}{3}$ is the only root of the real equation in this case
In case (d) , i. e. 1 < x< 2 , the equation becomes
x + 1 - x + 3 (x - 1)(x - 2) = (x + 2)

$3x^2$ - 8x + 7 = 0
Which gives the imaginary values of x .
Hence there is no solution in this case
In case , i. e. x > 2 . the equation is reduce to
x + 1 - x + 3 (x - 1)(x - 2) = (x + 2)

$3x^2$ - 10x + 5 = 0
as in case (c) the root of

$x = \dfrac{5 \, + \, \sqrt 10}{3}$
Since the root of

$x = \dfrac{5 \, + \, \sqrt 10}{3}$ is does not satisfy the condition (e) where the root

$x = \dfrac{5 \, + \, \sqrt 10}{3}$ satisfy it the root of origin equation is at

$x = \dfrac{5 \, + \, \sqrt 10}{3}$
Thus from the above discussion , we see that roots of the given equation are at

$x = \dfrac{5 \, + \, \sqrt 10}{3}$

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

$\cfrac { 5x+4 }{ 4 } -\cfrac { 3x-2 }{ 2 } =5\\ \Rightarrow \cfrac { 5x+4-2(3x-2) }{ 4 } =5\\ \Rightarrow 5x+4-6x+4=20\\ \Rightarrow \quad -x+8=20\\ \Rightarrow \quad x=-12$

Solve the following linear equation :
$\dfrac{x}{2} + 7x - 6 = 7x + {1 \over 4}$

$\cfrac { x }{ 2 } +7x-6=7x+\cfrac { 1 }{ 4 }$

$\Rightarrow \cfrac { x }{ 2 } =6+\cfrac { 1 }{ 4 }$

$\Rightarrow x=\cfrac { 25 }{ 2 }$

If each one of the following graphs are the graph of a polynomial, then identify which one corresponds to a linear polynomial and which one corresponds to a quadratic polynomial?

(i) We observe that the graph y = f(x) is a parabola opening upwards. Therefore,f(x) is a quadratic polynomial in which coefficient of

$x^{2}$ is positive.

(ii) We find that the graph y = g (x) is a straight line. So, g (x) is a linear polynomial.

(iii) Here, p (x) is neither linear nor quadratic.

(iv) Here, q(x) is a quadratic polynomial in which coefficient of

$x^{2}$ is negative because the graph is a parabola opening downwards.

A bag contains $25$ paise and $50$ paise coins whose total value is $Rs.\ 30$ . If the number of $25\ paise$ coins is four times that of $50\ paise$ coins, find the number of each type of coins.

Total amount in the bag

$=$ Rs.

$30 = 3000$ paise

Let the number of

$50$ paise coins be

$x$

Then the number of

$25$ paise coins

$= 4x$

Now, according to the question

$25 (4x) + 50x = 3000$

$100x + 50x = 3000$

$150x = 3000$

$x = \dfrac{3000}{150}$

$x = 20$

$4x = 80$

Therefore, number of

$50$ paise coins

$= 20$

Number of

$25$ paise coins

$= 80$

Solve the following linear equation :
$2x - \dfrac{1 }{3} = \dfrac{1 }{5} - x$

$2x-\cfrac { 1 }{ 3 } =\cfrac { 1 }{ 5 } -x$

$\Rightarrow 3x=\cfrac { 1 }{ 3 } +\cfrac { 1 }{ 5 }$

$\Rightarrow x=\cfrac { 8 }{ 45 }$

Solve for x: $\cfrac{p+q-x}{r} +\cfrac{q+r-x}{p} + \cfrac{r+p-x}{q} +\cfrac{4x}{p+q+r}$ =1

The given equation can be re-written as
$\cfrac{p+q-x}{r}+1 +\cfrac{q+r-x}{p} +1 + \cfrac{r+p-x}{q}+ 1$ = $4 - \cfrac{4x}{p+q+r}$
or $(p + q + r-x)$ $(\cfrac{1}{p} +\cfrac{1}{q} + \cfrac{1}{r})$ = $\cfrac{4(p+q+r-x)}{p+q+r}$
or $(p + q + r-x)$ $(\cfrac{1}{p} +\cfrac{1}{q} + \cfrac{1}{r})$ - $\cfrac{4}{p+q+r}$ =0
Hence $p + q + r-x = 0$ or $x=p + q + r.$

Aftab tells his daughter, "Seven years ago, I was seven times as old as you were then. Also, Three years from now, I shall be there times as old as you will be." Represent this situation algebraically.

Let Aftab's present age is

$x$ years and his daughter's is

$y$ years.

Seven years ago Aftab's age was

$x-7$ years and that of his daughter was

$y-7$ .years.

According to the problem,

$(x-7)=7(y-7)$

or,

$x-7y+42=0$ ......(1).

Three years from now Aftab's age will be

$x+3$ years and that of his daughter will be

$y+3$ .

According to the problem,

$x+3=3(y+3)$

or,

$x-3y=6$ ........(2).

(1) and (2) represents the given problem algebraically.

The cost of a notebook is twice the cost of a pen. Write a linear equation in two variable to represent this statement.
( Take the cost of a notebook to be $Rs.x$ and that of a pen to be $Rs. y$ ).

Let the cost of a notebook be Rs.

$x$ and that of a pen be Rs.

$y$ .

Given the cost of a notebook is twice the cost of a pen.

Then

$x=2y$

or,

$x-2y=0$ .

This is the required linear equation in two variable.

The sum of a two digit number and the number obtained by reversing the digits is $66$ . If the digit of the number differ by $2$ , find the number. How many such numbers are there?

Let the two digit be

$x$ and

$y$

$\therefore$ Number (

$2-$ digit)

$=10\times x+y$

Sum of

$2-$ digit and reverse of it

$\implies 10x+y+10y+x=66$ (given)

$\therefore x+y=6\quad$ (equation

$1$ )

Digits differ by

$2$

$\therefore x-y=2\quad$ (equation

$2$ )

On adding,

$2x=8$

$\implies x=4$

$\therefore y=2$

or,

$x+y=6$ From

$1$ equation

and

$y-x=2$ From

$2$ equation

On adding:

$2y=8$

$y=4$

$x=2$

$\therefore$ Two digit numbers are:

$42$ and

$24$
There are two such numbers.

Solve $\dfrac{{x - b - c}}{a} + \dfrac{{x - c - a}}{b} + \dfrac{{x - a - b}}{c} = 3$

$\dfrac{x-b-c}{a}+\dfrac{x-c-a}{b}+\dfrac{x-a-b}{c}=3$

put

$x=a+b+c$

$\Rightarrow \dfrac{a+b+c-b-c}{a}+\dfrac{a+b+c-c-a}{b}+\dfrac{a+b+c-a-b}{c}=3$

$\Rightarrow \dfrac{a}{a}+\dfrac{b}{b}+\dfrac{c}{c}=3$ .

${\dfrac{3t - 2} 4} - {\dfrac{2t + 3} 3} = {\dfrac2 3} - t$

$\cfrac { 3t-2 }{ 4 } -\cfrac { 2t+3 }{ 3 } =\cfrac { 2 }{ 3 } -t\\ \Rightarrow \cfrac { 3\left( 3t-2 \right) -4\left( 2t+3 \right) }{ 12 } =\cfrac { 2-3t }{ 3 } \\ \Rightarrow 9t-6-8t-12=8-12t\\ \Rightarrow 8t-18=8-12t\\ \Rightarrow 20t=26\\ \Rightarrow t=\cfrac { 13 }{ 10 }$

Reducing equation to simple from :- $\dfrac{6x+1}{3}=\dfrac{x-3}{6}$

given

$\dfrac { 6x+1 }{ 3 } =\dfrac { x-3 }{ 6 }$

cross multiplying we get

$\Rightarrow 6(6x+1)-3(x-3)=36x+6-3x+9=0$

$\Rightarrow 33x+15=0$

$\Rightarrow x=-\dfrac{15}{33}=\dfrac{-5}{11}$ or

$11x+5=0$

Find the value of $y$ :
$15(y-4) - 2(y-9) + 5(y+6)=0$

$15(y-4)-2(y-9)+5(y+6)=0$

$\Rightarrow 15y-2y+5y=60-18-30$

$\Rightarrow 18y=12$

$\Rightarrow y=\cfrac { 2 }{ 3 }$

Represent the solution of each of the following inequalities on the real number line:
$4x - 1 > x + 11$

$4x-1>x+11$

$4x>x+12$

$3x>12$

$x>4$

$\begin{bmatrix}\mathrm{Solution:}\:&\:x>4\:\\ \:\mathrm{Interval\:Notation:}&\:\left(4,\:\infty \:\right)\end{bmatrix}$

1140272_1025840_ans_5fc109abac2c4be5a469641a2e82de18.png

If $\dfrac{1}{5} - x = -\dfrac{4}{5}$ , then $x$ is ...............

$\cfrac { 1 }{ 5 } -x=-\cfrac { 4 }{ 5 } \\ \Rightarrow -x=-\cfrac { 4 }{ 5 } -\cfrac { 1 }{ 5 } \\ \Rightarrow -x=-\cfrac { 5 }{ 5 } \\ \Rightarrow -x=-1\\ \therefore \quad x=1$

A ball falls from a height of 100 mts. on a floor. If in each rebound, it describe (4/5)th height of the previous falling height, then the total distance travelled by the ball before it come to rest is

The total height it travel before it comes to rest is

$S=100+2(100)\dfrac{4}{5}+2(100)\left(\dfrac{4}{5}\right)+.....................\infty$

$S=100+2(100)\dfrac{4}{5}\left(1+\dfrac{4}{5}+\dfrac{4^2}{5^2}+.............+\infty \right)$

$S=100+160(\dfrac{1}{\tfrac{1}{5}})=900m$

At a certain time in a deer park, the number of heads and the number of legs of deers and human visitors were counted and it was found there were $39$ heads & $132$ legs. If the number of deers and human visitors in the park are $'d'$ and $'h'$ respectively, Then find $\left|\dfrac{d-h}{3}\right|$ .

Solution:

No. of legs of a deer

$=4$

No. of legs of a human

$=2$

No. of head of a deer

$=1$

No. of head of a human

$=1$

$d$ no. of deers and

$h$ no. of humans are there

So, total number of heads,

$d+h=39$

$\Rightarrow h=39-d$ ………

$(1)$

and, total number of legs,

$4d+2h=132$

$\Rightarrow 4d+2(39-d)=132$ [from

$(1)$ ]

$\Rightarrow4d-2d+78=132$

$\Rightarrow2d=54$

$\Rightarrow d=27$

So, from

$(1)$ ,

$h=39-27=12$

Now,

$\left|\dfrac{d-h}{3}\right|=\left|\dfrac{27-12}{3}\right|=\left|\dfrac{15}{3}\right|=5$ .

Hence, the required value is

$5$

Solve following equation:
$x^2 - x + 2 = 0$

The roots of the quadratic equation

$Ax^{2}+Bx+C=0$ are

$x=\dfrac{-B\pm \sqrt{B^{2}-4AC}}{2A}$

The given equation is

$x^{2}-x+2=0$

for the given equation,

$B^{2}-4AC=(-1)^{2}-4(1)(2)=-7$

So the roots are

$x=\dfrac{-(-1)\pm \sqrt{-7}}{2}$

$x=\dfrac{1\pm \sqrt{7}i}{2}$

Raju's salary was first increased by $10$ %, then decreased by $20$ %. If the latest salary is Rs. $17600$ , then find his original salary.

Let the original salary of Raju be

$x$

Salary after

$10\%$ increment

$=x+x\times 10\%$

$=x+\dfrac{10x}{100}$

$=\dfrac{11x}{10}$

Now salary is decreased by

$20\%$ of new salary

$\implies$ Latest salary

$=\dfrac{11x}{10}-\dfrac{11x}{10}\times 20\%$

$17600=\dfrac{11x}{10}-\dfrac{11x \times 20}{10 \times 100}$

$17600=\dfrac{11x}{10}-\dfrac{11x}{50}$

$17600=\dfrac{55x-11x}{50}$

$17600=\dfrac{44x}{50}$

$\implies x=\dfrac{17600\times 50}{44}$

$\implies x=\text{Rs }20,000$

$\therefore$ The original salary of Raju was

$\text{Rs }20,000$

what is linear equation?with example.

Linear equation is an equation which only contain 1st degree terms
like x + y = 5

x = 3 not like

$xy = 3$

$4xy+ y = 5$

${x^2}\, + \,3x\, = \,6$ or

${x^3}\, + \,{y^3} = 7$

Solve the equation:
$\dfrac{{\sqrt {4x + 1} + \sqrt {x + 3} }}{{\sqrt {4x + 1} - \sqrt {x + 3} }} = \dfrac{4}{1}$

Consider the given equation.

$\dfrac{\sqrt{4x+1}+\sqrt{x+3}}{\sqrt{4x+1}-\sqrt{x+3}}=\dfrac{4}{1}$

$\dfrac{\sqrt{4x+1}+\sqrt{x+3}}{\sqrt{4x+1}-\sqrt{x+3}}\times \dfrac{\sqrt{4x+1}+\sqrt{x+3}}{\sqrt{4x+1}+\sqrt{x+3}}=\dfrac{4}{1}$

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

$\dfrac{4x+1+x+3+2\sqrt{\left( 4x+1 \right)\left( x+3 \right)}}{\left( 4x+1 \right)-\left( x+3 \right)}=\dfrac{4}{1}$

$\dfrac{5x+4+2\sqrt{4{{x}^{2}}+12x+x+3}}{4x+1-x-3}=\dfrac{4}{1}$

$\dfrac{5x+4+2\sqrt{4{{x}^{2}}+13x+3}}{3x-2}=\dfrac{4}{1}$

$5x+4+2\sqrt{4{{x}^{2}}+13x+3}=12x-8$

$2\sqrt{4{{x}^{2}}+13x+3}=7x-12$

$4\left( 4{{x}^{2}}+13x+3 \right)=49{{x}^{2}}+144-168x$

$16{{x}^{2}}+52x+12=49{{x}^{2}}-168x+144$

$33{{x}^{2}}-220x+132=0$

$x=\dfrac{220\pm \sqrt{{{\left( 220 \right)}^{2}}-4\times 33\times 132}}{66}$

$x=\dfrac{220\pm \sqrt{48400-17424}}{66}$

$x=\dfrac{220\pm \sqrt{30976}}{66}$

$x=\dfrac{220\pm 176}{66}$

$x=\dfrac{220+176}{66},\dfrac{220-176}{66}$

$x=6,\dfrac{2}{3}$

Hence, the value of x is $6,\dfrac{2}{3}$ .

If $4a - 3 = 13,$ then find the value of $10{a^2} - 5a + 6.$

$\begin{aligned}{}4a - 3 &= 13\\ 4a&= 13 + 3\\ &= 16\\a& = 4\end{aligned}$

Let,

$p(a)=10a^2-5a+6$

Substitute

$4$ for

$a$ in

$p(a),$

$\begin{aligned}{}p\left( 4 \right) &= 10{\left( 4 \right)^2} - 5\left( 4 \right) + 6\\& = 10\left( {16} \right) - 20 + 6\\ &= 160 - 20 + 6\\ &= 146\end{aligned}$

Solve the given equation $\dfrac{9x+7}{2}-\left(x-\dfrac{x-2}{7}\right)=36$

$\cfrac { 9x+7 }{ 2 } -\left( x-\cfrac { x-2 }{ 7 } \right) =36\\ =>\cfrac { 9x+7 }{ 2 } -x+\cfrac { x-2 }{ 7 } =36\\ =>7(9x+7)-14x+2(x-2)=36\times 14\\ =>63x+49-14x+2x-4=504\\ =>51x+45=504\\ =>x=\cfrac { 504-45 }{ 51 } =9$

if $p(x)=2x+3$ , then find $P(1),P(3),P(-3)$ and $P(-1)$

The given function is $p\left( x \right) = 2x + 3$ .

$p\left( 1 \right) = 2\left( 1 \right) + 3$

$= 5$

$p\left( 3 \right) = 2\left( 3 \right) + 3$

$= 9$

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

$= - 3$

$p\left( { - 1} \right) = 2\left( { - 1} \right) + 3$

$= 1$

$\dfrac{8x-3}{3x}=2$ .

We are given,

$\dfrac{8x-3}{3x} =2$

$\therefore 8x-3 = 6x$

$\therefore 8x-6 x = 3$

$\therefore 2x =3$

$\therefore x =\dfrac{3}{2}$

The sum of the ages of Anup and his father isWhen he will be as old as his father now, he will be five times as old as his son Anuj is now. Also, Anuj will be eight years older than Anup is now at that time. What are their ages right now?

Let the present ages of Anup, his father and his son Anuj be x, y and z, respectively.

Given that, the sum of ages of Anup and his father is

$100$

$\therefore \ x + y = 100$ ..........(1)

Let their new ages after few years be x', y' and z', respectively.

Given that, after those few years, new age of Anup = present age of father = 5

$\times$ present age of Anuj

$\therefore\ x' = y = 5z$ ............(2)

Also, given that new age of Anuj= 8+present age of Anup

$\therefore\ z'=8+x$ .....(3)

$\because\$ Number of years = New age - present age for each person

$\therefore\$ Number of years

$= x'-x = y'-y = z'-z$

Taking first and last term we have,

$x'-x = z'-z$ ..........(4)

Using equations (1), (2), and (3), we get,

$x'=y,\ x=100-y,\ z'=8+x=8+100-y=108-y,\ and\ z=\dfrac{y}{5}$

Substituting these values in equation (4), we get:

$\Rightarrow y - ( 100 - y ) = ( 108-y ) - \dfrac{y}{5}$

$\Rightarrow 2y - 100 = 108-y - \dfrac{y}{5} \Rightarrow 2y + \dfrac{6y}{5} = 108 + 100$

$\Rightarrow \dfrac{16y}{5} = 208$

$\Rightarrow 16y=208\times 5$

$\Rightarrow y=65$

Now, from equation (1) ,

$x = 100 - y = 100 -65 =35$

And from equation (2),

$z= \dfrac{y}{5} = \dfrac{65}{5} = 13$

Hence, the present age of Anup is

$35$ years, that of his father is

$65$ years and that of his son, Anuj is

$13$ years.

If $\dfrac{\sqrt{11}-\sqrt{7}}{\sqrt{11}+\sqrt{7}}=x-y\sqrt{77}$ , find the values of $x$ and $y$ ?

$\dfrac{\sqrt {11}-\sqrt 7}{\sqrt {11}+\sqrt 7}$

$=\dfrac{\sqrt {11}-\sqrt 7}{\sqrt {11}+\sqrt 7}\times$

$\dfrac{\sqrt {11}-\sqrt 7}{\sqrt {11}-\sqrt 7}$ (on rationalizing the denominator)

$=\dfrac {11+7-2\sqrt {77}}{11-7}$

$=\dfrac{18-2\sqrt{77}}{4}$

$=\dfrac{18}{4}-\dfrac{\sqrt{77}}{2}$

$=\dfrac{9}{2}-\dfrac{\sqrt{77}}{2}$ -------(1)

Now, on comparing (1) with

$x-y\sqrt{77}$

we get,

$x=\dfrac{9}{2}$ and

$y=\dfrac{1}{2}$

Solve
$x+\dfrac{x}{2}+\dfrac{x}{3} <11$

$x+\dfrac{x}{2}+\dfrac{x}{3} <11$

$\dfrac{11x}{6} <11$

$x<6$

$x \in (-\infty ,6)$

Simplify $3x(4x-5)+3$ and find its value for $x=3$ .

$3x(4x-5)+3=12x^2-15x+3$

$x=3$ gives the value

$12(9)-15\times 3 +3$

$=108-45+3$

$=66$

Give the geometrical representation of $2x+5=0$ as an equation in two variables.

$\begin{matrix} 2x+5=0 \\ let \\ y=2x+5 \\ Put\, \, x=0 \\ y=2\times 0+5 \\ y=5 \\ x=1 \\ y=2\times 1+5\, \, \, =7 \\ and\, \, \, x=2 \\ y=2\times 2+5\, \, =9 \\ It\, \, is\, \, straight\, \, line\, \, \\ \end{matrix}$

Find the coefficient of the term independent of x in $\left(x-\dfrac{1}{x^2}\right)^{3n}$ .

We can write the expression as

$x^{3n}\left(1-\dfrac{1}{x^3}\right)^{3n}$ which can be written as

$x^{3n}\displaystyle\sum^{3n}_{p=0}$

$^{3n}C_r\left(\dfrac{-1}{x^3}\right)^{3n-p}$
Coefficient of

$x^{3n+3p-9n}$ is

$(-1)^{(n-p)3n}C_p$
To find coefficient of

$x^{\theta}$ , put

$0=3n+3p-9n\Rightarrow p=2n$

$\therefore$ The coefficient of term independent of x is

$(-1)^n$

$^{3n}C_{2n}$ .

$x = \dfrac{a^n - a^{-n}}{a^n + a^{-n}}$ then n =

$x=\cfrac { { a }^{ n }-{ a }^{ -n } }{ { a }^{ n }+{ a }^{ -n } }$

$x=\cfrac { { a }^{ n }-\cfrac { 1 }{ { a }^{ n } } }{ { a }^{ n }+\cfrac { 1 }{ { a }^{ n } } }$

$x=\cfrac { { a }^{ 2n }-1 }{ { a }^{ 2n }+1 }$

${ a }^{ 2n }x+x={ a }^{ 2n }-1$

${ a }^{ 2n }\left( 1-x \right) =x+1$

${ a }^{ 2n }=\cfrac { x+1 }{ 1-x }$

$2n=\log { \cfrac { x+1 }{ 1-x } }$

$n=\cfrac { 1 }{ 2 } \log { \cfrac { 1+x }{ 1-x } }$

Pooja and Ritu can do a piece of work in $17\cfrac { 1 }{ 7 }$ days. If one day work of Pooja be three fourth of one day work of Ritu; find in how many days each will do the work alone.

Amount of time to do the work by both

$=17\cfrac { 1 }{ 7 } =\cfrac { 120 }{ 7 }$

Work done in one day by both

$=\cfrac { 7 }{ 120 }$

Let Pooja do

$x$ amount of work in one day

Work done by Ritu

$=\cfrac { 4 }{ 3 } x$

$\therefore x+\cfrac { 4 }{ 3 } x=\cfrac { 7 }{ 120 }$

$\therefore \cfrac { 7 }{ 3 } x=\cfrac { 7 }{ 120 }$

$x=\cfrac { 1 }{ 40 }$

$\therefore$ Pooja will take

$40$ days to complete the work while Ritu will take

$30$ days to complete the work alone.

People of sundargram planted tree in the village garden. Some of the trees were fruit trees. The number of non-fruit trees were two more than three times the number of fruits trees. What was the number of fruit trees planted if the number of non-fruit trees planted was $77$ ?

Let the number of fruit trees is

$x$ and the number of nonfruit trees is

$y.$

According to the given condition,

$y = 2+3x$

If,

$y = 77$ then,

$77 = 2+3x$

$\Rightarrow 3x = 75$

$\Rightarrow x = 25$

The number of fruit trees planted is

$25.$

Twice of the numerator of a fraction is $1$ more than the denominator. If $5$ is added to both numerator and denominator, the fraction is $\dfrac{7}{60}$ more than the original one. Find the original fraction.

Let the fraction $=\dfrac{x}{y}$

Then, According to the given question,

$2x=y+1$

$x=\dfrac{y+1}{2}\,\,\,........\,\,\,\,\,\left( 1 \right)$

Now, again according to the given question,

$\dfrac{x+5}{y+5}=\dfrac{7}{60}+\dfrac{x}{y}$

$\Rightarrow \dfrac{x+5}{y+5}=\dfrac{7y+60x}{60y}$

$\Rightarrow 60xy+300y=7{{y}^{2}}+60xy+35y+300x$

$\Rightarrow 7{{y}^{2}}+35y+300x=300y$

$\Rightarrow 7{{y}^{2}}+35y+300x-300y=0$

Put the value of $x$ by equation (1) to (2) and we get,

$7{{y}^{2}}+35y+300\left( \dfrac{y+1}{2} \right)-300y=0$

$\Rightarrow 7{{y}^{2}}+35y+150y+150-300y=0$

$\Rightarrow 7{{y}^{2}}-115y+150=0$

$\Rightarrow 7{{y}^{2}}-\left( 105+10 \right)y+150=0$

$\Rightarrow 7{{y}^{2}}-105y-10y+150=0$

$\Rightarrow 7y\left( y-15 \right)-10\left( y-15 \right)=0$

$\Rightarrow \left( y-15 \right)\left( 7y-10 \right)=0$

$\Rightarrow y-15=0\,\,\,\,\,\,\,\,\,\,\,\,7y-10=0$

$\Rightarrow y=15,\,\,\,\,\,\,\,\,\,\,\,\,y=\dfrac{10}{7}$

Put the value of $y=15$ in equation (1) and we get,

$x=\dfrac{y+1}{2}$

$x=\dfrac{15+1}{2}$

$x=8$

Hence, the fraction is $=\dfrac{x}{y}=\dfrac{8}{15}$ ans.

Give the geometrical representation of $2x+5=0$ as an equation in one variable.

1372835_1126670_ans_1e641cae901a4d4aa0221195f3ed7128.jpg

Give the equations of two lines passing through $(2, 14)$ . How many more such lines are there, and why?

From point (2,14), infinitely many lines can pass because to define a unique line require at least two parameters like 2 point from which line will pass (or) one point and slope of line (or) slop of line and anyone intercept.

Since anyone parameter is provided in the question it is not possible to define a line passing through it.

A lending library has a fixed per day charge for the first three days and an additional charge for each day thereafter. Arushi paid Rs. 27 for a book kept for seven days. The fixed charges are Rs. x per day and additional charges are Rs. y per day. Write the linear equation representing the above information.

Given that, a library has a fixed per day charge for the first three days and an additional charge for each day thereafter.

The fixed charges are

$Rs.\ x$ per day and the additional charges are

$Rs.\ y$ per day.

Also, Arushi paid

$Rs.\ 27$ for a book she kept for

$7$ days.

Since the book was kept for

$7$ days, fixed charges will be applicable for each of the first three days and an additional charge of

$Rs. \ y$ per day for the remaining days.

Hence, per day charges after three days

$=Rs.\ (x+y)$ .

$\therefore \ 3x+(7-3)(x+y)=27$

$\Rightarrow 3x+4x+4y=27$

$\Rightarrow 7x+4y=27$

Hence, the linear equation representing the given information is

$7x+4y=27$ .

Find the point where the graph of the linear equation $2x + 3y = 6$ cuts the y-axis.

1369150_1159843_ans_242f36f6780c45a9ae632aec2cf74dee.jpg

Find $x$
$2x+3 = 3x+5$

We have,

$2x+3 = 3x+5$

$2x-3x=5-3$

$-x=2$

$x=-2$

Hence, this is the answer.

Add: $3mn, - 5mn,8mn, - 4mn$

Here we need to add

$3mn,-5mn,8mn,-4mn$

$3mn-5mn+8mn-4mn=(3-5+8-4)mn\\=2mn$

$\dfrac{1}{2}$ of a boat is under water. When $\dfrac{1}{5}$ of it is pulled out, 2 ft is still under water. What is the height of the boat?

$\frac{1}{2}$ of the boat is under water

$\frac{1}{5}$ of it is pulled out

i.e

$\frac{1}{5}\left ( \frac{x}{2} \right )$ is pulled out

$\frac{x}{2}+\frac{x}{10}$ is out of water

$x-\left ( \frac{x}{2}+\frac{x}{10} \right )$ is under water

$\therefore x-\frac{x}{2}-\frac{x}{10}=2$

$\frac{10x-5x-x}{10}=2$

$\frac{4x}{10}$ =2

2x=10

x=5

Simplify the equation: $\dfrac { 1 } { a + b + x } = \dfrac { 1 } { a } + \dfrac { 1 } { b } + \dfrac { 1 } { x } ; a + b \neq 0$

$\dfrac{1}{a+b+x} = \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{x}$

$a+b\neq 0$

$= \dfrac{1}{a+b+x} = \dfrac{bx+ax+ab}{abx}$

$= abx = (a+b+x)(bx+ax+ab)$

$\Rightarrow abx = (a+b)^{2}x+(a+b)ab+x^{2}(a+b)+abx$

$\Rightarrow (a+b)^{2}x+(a+b)x^{2}+a^{2}b+ab^{2} = 0$

$\Rightarrow (a+b)x^{2}+(a+b)^{2}+(a+b)ab = 0$

$\Rightarrow x^{2}+(a+b)x+ab = 0$

$x = \dfrac{-(a+b)\pm \sqrt{(a+b)^{2}-4ab}}{2} = \dfrac{-(a+b)\pm (a-b)}{2}$

$x = \dfrac{-a-b+a-b}{2}, x = \dfrac{-a-b-a+b}{2}$

$x = -b$

$x = -a$

$x = -a,-b$

1086895_1166955_ans_4e9a9260142b476c84cf3cc086aa6270.png

Solve for $x$ and $y.$

$\begin{array}{l}\dfrac{x}{b} + \dfrac{y}{a} = 2 , \, ax - by = {a^2} - {b^2}\end{array}$

Given,

$\dfrac{x}{a}+\dfrac{y}{b}=2,ax-by=a^2-b^2$

$ax+by=2ab,ax-by=a^2-b^2$

adding both, we get

$2ax=2ab+a^2-b^2$

$x=b+\dfrac{a^2-b^2}{2a}$

subtracting both, we get

$2by=2ab+b^2-a^2$

$y=a+\dfrac{b^2-a^2}{2b}$

Find the equations of the two lines through the origin which
intersect the line $\dfrac{{x - 3}}{2} = \dfrac{{y - 3}}{1} = \dfrac{z}{1}$ at angle of $\dfrac{\pi }{3}$ each.

1222679_1143189_ans_958f840bfa8840f486c7a7de0957bc66.jpg

If $51x+49y=150$ and $49x+51y=50$ then obtain the value of $x-y:x+y$

1208746_1227771_ans_67300e72035b4b74a993a6567a8beb0f.jpg

Solve to find $x$ :

$\displaystyle {{3x} \over 5} - 4 = 8$

$\dfrac{3x}{5}-4=8$

$\Rightarrow \dfrac{3x}{5}=12$

$\Rightarrow x=20$

'When $5$ is subtracted from the length and breadth of the rectangle, the perimeter becomes $26$ .' What is the mathematical form of the statement

${\textbf{Step 1: Find the equation for the variables}}{\textbf{.}}$

${\text{We know perimeter of rectangle = 2}}\left( {l + b} \right)$

${\text{Let original length and breadth be }}l{\text{ and }}b{\text{.}}$

$\Rightarrow 2\left( {l - 5 + b - 5} \right) = 26$

${\textbf{Step 2: Solve the equation to find the mathematical statement}}{\text{.}}$

$\Rightarrow 2\left( {l + b - 10} \right) = 26$

$\Rightarrow 2\left( {l + b} \right) = 46$

$\Rightarrow l + b = 23$

${\textbf{Hence, the mathematical statement is}}$

$\mathbf{ l + b = 23.}$

Solve for x: $x-12=9$

Given

$x-12 = 9$

$x = 9+12$

$x = 21$

Solve the equation $2x + 6 < 12$

$2x+6<12\Rightarrow x<\dfrac{12-6}{2}\Rightarrow x<3\Rightarrow x\epsilon(-\infty,3)$

Solve:
$5 = 6 + 11x$

Solution:-

$5 = 6+11x$

$\Rightarrow 5-6 = 11x$

$\Rightarrow -1 = 11x$

$\Rightarrow x = \dfrac{-1}{11}$

Simplify
$\dfrac { 25 }{ 4 } { x }^{ 2 }-\dfrac { { y }^{ 2 } }{ 9 }$

$\cfrac{25}{4}x^2-\cfrac{y^2}{9}={(\cfrac{5}{2}x)}^2-{(\cfrac{y}{3})}^2=(\cfrac{5}{2}x+\cfrac{y}{3})(\cfrac{5}{2}x-\cfrac{y}{3})$

The ratio of girls and boys are in a class is $1:3$ .Set up an equation between the students of class and boys .

1345785_1236626_ans_4eb415d6994346fca620ab0cafe85b9c.PNG

Solve it :-
$\frac{7}{{x - 5}} = \frac{5}{{x - 7}}$

$\dfrac{7}{{x - 5}} = \dfrac{5}{{x - 7}}\Rightarrow 7x-49=5x-25\Rightarrow 2x=24\Rightarrow x=12$

Solve:
$0.16\left( {5x - 2} \right) = 0.4x + 7$

Solution:-

Given

$0.16(5x-2) = 0.4x+7$

$\Rightarrow 0.8x-0.32 = 0.4x+7$

$\Rightarrow 0.8x-0.4x = 7+0.32$

$\Rightarrow 0.4x = 7.32$

$\Rightarrow x = \dfrac{732\times 10}{4\times 100} = 1.83$

Solve the following simultaneously equation using elimination method
$4m+6n=54; 3m+2n=28$

$Mulitply$ second equation by

$3$ and

$subract$ first equation from it,

$3(3m+2n)-(4m+6n)=3(28)-54$

$9m+6n-4m-6n=30$

$5m=30$

$m=6$

Substitute in first equation,

$4(6)+6n=54$

$6n=54-24$

$6n=30$

$n=5$

Thus

$m=6$ and

$n=5$

A number consist of two digits, the difference of whose digits is $5$ . If $8$ times the number is equal to $3$ times the number obtained by reversing the digits, Find the number.

let number is

$xy$ , where

$x$ is ten's place and

$y$ is unit place

so it can be written as

$(10x + y)$

given that

$|x - y| = 5$

that is

$x - y = +5 ........eq(1)$

$x-y = -5 .........eq(2)$

Now,

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

$80x +8y = 30y +3x$

$77x -22y = 0$

$7x-2y = 0 ........eq (3)$

solving

$eq(1)$ and

$eq(3)$

$7x -2y = 0$

$-(2x-2y = 10)$

it gives

$5x = -10$

hence

$x = -2$

put in eq(1)

$-2-y = 5$

$y =-7$

so number

$, 10x + y = 10(-2) + (-7)$

$N = -27$

solving eq(2) and eq(3)

$7x-2y = 0$

$-(2x-2y = -10)$

it gives

$, 5x = 10$

$x =2$

put in eq(1)

$2 - y = -5$

$y = 7$

$,N = 10x + y = 10*2 + 7$

$N = 27$

Solve
$5x-4y=12,for(y)$

We have,

$5x-4y=12$

$-4y=-5x+12$

divide

$(-4)$ to isolate

$\therefore \ y=\dfrac {5}{4}x-3$

Hence this is the answer.

Find values of a,b and c:
$a-b+2c=0$
$a+2b-c=0$

$a-b+2c=0$ ………….

$(1)$

$a+2b-c=0$ ………….

$(2)$

$\dfrac{a}{-3}=\dfrac{b}{3}=\dfrac{c}{3}=1$

$\dfrac{a}{\begin{matrix} -1 & 2\\ 2 & -1\end{matrix}}=\dfrac{b}{\begin{matrix} 2 & 1\\ -1 & 1\end{matrix}}=\dfrac{c}{\begin{matrix} 1 & -1\\ 1 & 2\end{matrix}}$ [From equation

$(1)$ and

$(2)$ , by cross-multiplication method]

$\dfrac{a}{1-4}=\dfrac{b}{2+1}=\dfrac{c}{2+1}=1$

$\dfrac{a}{-3}=1$

$\therefore a=-3, b=3, c=3$ .

1162344_1281080_ans_ac19bb7c6b8c48e9b271f2908f02f96d.jpg

Solve for $\displaystyle x , 2 \left( \frac { 2 x + 3 } { x - 3 } \right) - 25 \left( \frac { x - 3 } { 2 x + 3 } \right) = 5$

We have,

$2\left(\dfrac{2x+3}{x-3}\right)-25\left(\dfrac{x-3}{2x+3}\right)=5$

$\Rightarrow$

$\dfrac{2(2x+3)^2-25(x-3)^2}{(x-3)(2x+3)}=5$

$\Rightarrow$

$2(2x+3)^2-25(x-3)^2=5(x-3)(2x+3)$

$\Rightarrow$

$2(4x^2+9+12x)-25(x^2+9-6x)=5(2x^2+3x-6x-9)$

$\Rightarrow$

$8x^2+18+24x-25x^2-225+150x=10x^2+15x-30x-45$

$\Rightarrow$

$-17x^2+174x-207=10x^2-15x-45$

$\Rightarrow$

$27x^2-189x+162=0$

$\Rightarrow$

$3x^2-21x+36=0$

$\Rightarrow$

$3x^2-21x+36=0$

$\Rightarrow$

$3x^2-9x-12x+36=0$

$\Rightarrow$

$3x(x-3)-12(x-3)=0$

$\Rightarrow$

$(x-3)(3x-12)=0$

$\Rightarrow x=3; \ x=4$

What must be subtracted from $\frac{{ - 2}}{3}{x^2} + \frac{3}{2}xy + 2{y^2}$ to get $3{x^2} - 2xy - \frac{7}{2}{y^2}$

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

$\rightarrow$ we have to find

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

$=-\dfrac{11}{3}x^2+\dfrac{7}{2}xy+\dfrac{11}{2}y^2$ .

and ${ x }^{ 2 }-bx+a=0$ , then find $a+b$ .

We have,

${{x}^{2}}+ax+b=0\,\,......\,\,\left( 1 \right)$

${{x}^{2}}+bx+a=0\,\,......\,\,\left( 2 \right)$

Subtract $\left( 1 \right)$ from $\left( 2 \right)$ and we get,

$x\left( a-b \right)+\left( b-a \right)=0$

$x\left( a-b \right)=-\left( b-a \right)$

$x\left( a-b \right)=\left( a-b \right)$

$x=1$

Put the value of $x$ in equation (1) and we get,

$1+a+b=0$

$a+b=-1$

Hence, this is the answer.

If $12x+13y=29$ and $13x+12y=21$ find $x+y$

$12x+13y=29$ .....(1)

$13x+12y=21$ ....(2)

Adding (1) and (2)

$25x+25y=50$

$x+y=2$

The sum of two numbers is $4000$ . If $15\%$ of one is equal to $25\%$ of the other, find the numbers.

Let one number be

$x$ ,

then, other number

$= (4000 - x)$

$ATQ,$

$15\%$ of

$x = 25\%$ of

$(4000 - x)$

$\Rightarrow \dfrac{15}{100} \times x = \dfrac{25}{100} (4000 - x)$

$\Rightarrow15 x = 100000 - 25 x$

$\Rightarrow15x + 25 x = 100000$

$\Rightarrow40 x = 100000$

$\Rightarrow x = \dfrac{100000}{40}$

$\Rightarrow x = 2500$

Other number

$= 4000 - x$

$= 4000 - 2500$

$= 1500$

Hence, the numbers are

$2500$ and

$1500$

The linear equations that converts Fahenheit $(^{o}F)$ to Celsius $(^{o}C)$ is given by the relation:
$^{o}C=\dfrac {5^{o}F-160}{9}$
What is the numerical value of the temperature is same in both the scale?

$^{o}C=\dfrac {5^{o}F-160}{9}$

Let

$C^{o}=F^{o}$

$^{o}C=\dfrac {5^{o}C-160}{9}$

$9^{o}C= 5^{o}C-160$

$9^{o}C- 5^{o}C=160$

$4^{o}C=160$

$^{o}C=40$

Solve $\cfrac { 1 }{ x } -\cfrac { 4 }{ y } =2,\cfrac { 1 }{ x } -\cfrac { 3 }{ y } =9$ by reducing it into the pair of linear equations

$\cfrac { 1 }{ x } -\cfrac { 4 }{ y } =2\quad ,\quad \cfrac { 1 }{ x } -\cfrac { 3 }{ y } =9$

Let

$\cfrac { 1 }{ x } =u$ and

$\cfrac { 1 }{ y } =v$

then we have

$u-4v=2 \quad -(i) \\ u-3v=9 \quad -(ii)$

Subtracting eq.

$(ii)$ from

$(i)$

$v=7$

Also,

$u- 3 \times 7 =9 \Rightarrow u=21+9=30$

Now,

$v=\cfrac { 1 }{ y } \Rightarrow y=\cfrac { 1 }{ v } =\cfrac { 1 }{ 7 }$

and

$u=\cfrac { 1 }{ x } \Rightarrow x=\cfrac { 1 }{ u } =\cfrac { 1 }{ 21 }$

Draw the graph of the equation $y=3x$

Given equation is

$y=3x$

When

$x=1,\,\Rightarrow y=3\times 1=3$

When

$x=2,\,\Rightarrow y=3\times 2=6$

When

$x=3,\,\Rightarrow y=3\times 3=9$

$x$	$1$	$2$	$3$
$y$	$3$	$6$	$9$

Plot the points

$A\left(1,\,3\right),\,B\left(2,6\right),\,C\left(3,\,9\right)$ on the graph paper.Join these points by a straight line

$AC$

1342372_1363411_ans_fdf9c00bf35143169d4d24dcaa03cb80.PNG

If the number obtained by interchanging the digits of a two digit number is 18 more than the original number and the sum of the digits is 8 then what is the original number?

Let the given number be

$xy$

$xy$ can be written as

$10x+y$

According to the given data

$xy = yx + 18$

$\Rightarrow 10x+y=10y+x+18$

$\Rightarrow 9y-9x=18$

$\Rightarrow y-x=2$ ………..

$(1)$

and also its given that sum of digits = 8

$\Rightarrow x+y=8$ ………

$(2)$

Solving equation

$(1)$ &

$(2)$ we get

$x+y=8$

$x-y=-2$

__________________

$2x=6$

$x=3$ and

$y=5$

Hence the given number is

$35$ .

Draw the graph of the equation $2x+3y+6=0$

$2x+3y+6=0$

$\Rightarrow y=-\left(\dfrac{6+2x}{3}\right)$

When

$x=-3,\,y=-\dfrac{6+2\times -3}{3}=-\dfrac{6-6}{3}=0$

When

$x=0,\,y=-\dfrac{6+2\times 0}{3}=-\dfrac{6}{3}=-2$

When

$x=3,\,y=-\dfrac{6+2\times 3}{3}=-\dfrac{6+6}{3}=-4$

$x$	$-3$	$0$	$3$
$y$	$0$	$-2$	$-4$

Plot the points

$A\left(-3,\,0\right),\,B\left(0,-2\right),\,C\left(3,\,-4\right)$ on the graph paper. Join these points by a straight line

$AC$

1342369_1363397_ans_93297325e97d4a25bfd66593911be120.png

Solve and verify the result :
$3x-5=13$

Given,

$3x-5=13 \\ 3x=13+5 \\ x=\dfrac { { 18 } }{ 3 } =6$
Verification :

$3\times 6-5=13 \\ 18-5=13 \\ LHS=RHS$

Give the geometric representation of $y=3$ as an equation in two variables.

In two variables, we have

$x$ and

$y$ axis.

Hence, writing equation

$y=3$ in terms of

$x$ and

$y$

$y=3$

$\Rightarrow 0+y=3$

$\Rightarrow 0.x+y=3$ .....

$(1)$

Putting

$x=0$ in

$(1)$ we get

$0\left(0\right)+y=3$

$\Rightarrow 0+y=3$

$\Rightarrow y=3$

So, the point is

$\left(0,3\right)$

Putting

$x=1$ in

$(1)$ we get

$0\left(1\right)+y=3$

$\Rightarrow 0+y=3$

$\Rightarrow y=3$

So,the point is

$\left(1,3\right)$

Now if we plot these points on a graph paper we will obtain a line parallel to

$x-$ axis

Hence we can say that,in two variables,

$y=3$ is a line which is parallel to

$x-$ axis

The sum of the digits of a two digit number is $13$ . The number obtained by interchanging the digits of the given number exceeds the number by $27$ . Find the number.

1272362_1352812_ans_a2182d6c9931412ca5423254938e2c1c.jpg

For what value of $y,10 y=50$ satisfy the equation.

$\begin{array}{l} 10y=50 \\ y=\dfrac { { 50 } }{ { 10 } } =5 \\ y=5 \end{array}$

State which of the following are equations (with a variable). Give reasons for your answer. Identify the variable from the equations with a variable.
$17=x+7$

It is equation with one variable, i.e.,

$x$

Construct a word problem on simultaneous linear equation in two variables so that the value of one of the variables will be $10$ ( person, rupees, metres, years etc) and solve it.

Example:The sum of present ages of Madhu and Raju is

$11$ years.Madhu is elder than Raju by

$9$ years.Find their present ages.

Solution:Let the present age of Madhu be

$x$ years and the age of Raju be

$y$ years.

$\therefore x+y=11$ .....

$(1)$

$x-y=9$ .......

$(2)$

Adding

$(1)$ and

$(2)$ we get

$2x=20$

$\therefore x=\dfrac{20}{2}=10$

And

$x+y=11$

$\therefore 10+y=11$

$\therefore y=11-10=1$

$\therefore$ Present age of Madhu is

$10$ years and of Raju is

$1$ year.

Find the product: $\left(2x-\dfrac{3}{y}\right)\left(2x+\dfrac{3}{y}\right)$

Given:

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

Multiply term wise:

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

$=4{x}^{2}+\dfrac{6x}{y}-\dfrac{6x}{y}-\dfrac{9}{{y}^{2}}$

$=4{x}^{2}-\dfrac{9}{{y}^{2}}$

Solve $\cfrac { 4m+2 }{ 6m+10 } =\cfrac { 1 }{ 2 }$ .

Hint: Use cross multiplication to sove for required value of m

$$2(4m+2) = 1(6m+10)

$8m+4 = 6m+10$
$2m=6$

$m=3$

Solve: $2y+9\ge 4$

$2y+9\ge 4$

$\Rightarrow\,2y+9-9\ge 4-9$

$\Rightarrow\,2y\ge -5$

$\Rightarrow\,y\ge \dfrac{-5}{2}$

$\therefore\,y\in\left[\dfrac{-5}{2},\,\infty\right)$

Draw the graph of the equation $x=1,\,y=2,\,x+3=0,\,y+4=0$
$x=1$ is the line $AB$ parallel to $y-$ axis at a distance of $1$ units to the right of the origin $O$

$x=1$ is the line

$AB$ parallel to

$y-$ axis at a distance of

$1$ units to the right of the origin

$O$

$y=2$ is the line

$CD$ parallel to

$x-$ axis at a distance of

$2$ units to the above the origin

$O$

$x+3=0$ or

$x=-3$ is the line

$EF$ parallel to

$y-$ axis at a distance of

$3$ units to the left of the origin

$O$

$y+4=0$ or

$y=-4$ is the line

$GH$ parallel to

$x-$ axis at a distance of

$4$ units to the below the origin

$O$

1342377_1363424_ans_cf88a28f35eb4891b89ff7d9ac079f6f.PNG

Find the equation of the straight line which is parallel to $2x-9y=11$ and makes $x-$ intercept $4$ units.

The given line is

$2x-9y=11$ .......

$(1)$

The equation of the family of lines parallel to

$(1)$ is

$2x-9y+k=0$ .....

$(2)$ where

$k$ is a parameter.

To find

$x-$ intercept,put

$y=0$ in

$(2)$ we get

$2x+k=0$

$\Rightarrow\,x=-\dfrac{k}{2}$

For the required member of the family which makes

$x-$ intercept

$4$ units,

$-\dfrac{k}{2}=4\Rightarrow\,k=-8$

Substituting this value of

$k$ in

$(2)$ , the equation of the required line is

$2x-9y-8=0$

Solve: $15(y-4) -2(y-9)+5(y+6)=0$

$\textbf{Hint: Use the method of solution of linear equation in one variable}$

Solution:

$\textbf{Step 1: Simplify the given equation}$

$\quad\quad\quad$ We have given:

$15\left( {y - 4} \right) - 2\left( {y - 9} \right) + 5\left( {y + 6} \right) = 0$

$\Rightarrow 15y - 60 - 2y + 18 + 5y + 30 = 0$

$\textbf{ Step 2: Bring the like terms together}$

$\Rightarrow \left( {15y + 5y - 2y} \right) + \left( {18 + 30 - 60} \right) = 0$

$\Rightarrow \left( {20{\rm{y}} - 2y} \right) + \left( {48 - 60} \right) = 0$

$\quad\quad\quad$ $\Rightarrow 18y - 12 = 0$

$\Rightarrow 18y = 12$

$\quad\quad\quad$ $\Rightarrow y = \dfrac{{12}}{{18}}$

$\quad\quad\quad$ $\Rightarrow y = \dfrac{2}{3}$

$\textbf{ Hence, the solution of the given equation is }$ $y = \dfrac{2}{3}$ .

$2$ tables and $3$ chairs cost Rs. $2000$ .Where as $3$ tables and $2$ chairs cost Rs. $2500$ .Find the total cost of $1$ table and $5$ chairs.

Let the cost of a table be Rs.

$x$ and that of a chair be Rs.

$y$ .Then

$2x+3y=2000$ ........

$(1)$

$3x+2y=2500$ ........

$(2)$

Multiplying equation

$(1)$ by

$3$ and

$(2)$ by

$1$ , we get

$6x+9y=6000$ .........

$(3)$

$6x+4y=5000$ .........

$(4)$

Subtracting

$(4)$ from

$(3)$ we get

$6x+4y-6x-9y=5000-6000$

$\Rightarrow -5y=-1000$

$\Rightarrow y=\dfrac{-1000}{-5}=200$

$\therefore y=200$

Put

$y=200$ in

$(3)$ we get

$2x+3\times 200=2000$

$\Rightarrow 2x=2000-600=1400$

$\Rightarrow x=\dfrac{1400}{2}=700$

So,Cost of a table

$=$ Rs.

$700$ and cost of a chair

$=$ Rs.

$200$

Hence cost of

$1$ table and

$5$ chairs

$=1\times 700+5\times 200=700+1000=$ Rs.

$1700$

Solve:
$\dfrac{6}{x}-\dfrac{3}{x}=1; x\neq 0$

$\dfrac{6}{x}-\dfrac{3}{x}=1$

$\Rightarrow \dfrac{3}{x}=1$

$\therefore x=3$

Solve: $-3\le\,4-\dfrac{7x}{2}\le\,2$

$-3\le\,4-\dfrac{7x}{2}\le\,2$

$\Rightarrow \,-3\le\,\dfrac{8-7x}{2}\le\,2$

$\Rightarrow\,-6\le\,8-7x\le \,4$

$\Rightarrow \,6\ge 7x-8\ge\,-4$

$\Rightarrow\,6+8\ge 7x\ge\,-4+8$

$\Rightarrow\,14\ge 7x\ge \,4$

$\Rightarrow \,4\le\,7x\le\,14$

$\Rightarrow\,\dfrac{4}{7}\le\,x\le\,2$

$\therefore\,x\in\left[\dfrac{4}{7},\,2\right]$

Find the equation of the straight line which is parallel to $x-5y=9$ and makes $x-$ intercept $4$ units.

The given line is

$x-5y=9$ .......

$(1)$

The equation of the family of lines parallel to

$(1)$ is

$x-5y+k=0$ .....

$(2)$ where

$k$ is a parameter.

To find

$x-$ intercept,put

$y=0$ in

$(2)$ we get

$x+k=0$

$\Rightarrow\,x=-k$

For the required member of the family which makes

$x-$ intercept

$4$ units,

$-k=4\Rightarrow\,k=-4$

Substituting this value of

$k$ in

$(2)$ , the equation of the required line is

$x-5y-4=0$

Let $x+y=5$ and $x-y=3$ then find the value of $\dfrac{x}{2}.\dfrac{y}{2}$

Given,

$x+y=5$ .....(1) and

$x-y=3$ ....(2).

We've,

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

$4xy=5^2-3^2$

$4xy=16$

$\dfrac{xy}{4}=\dfrac{16}{16}$

$\dfrac{x}{2}.\dfrac{y}{2}=1$ .

$x+y=17\\2x-y=-2$

$x+y=17\ldots(1)\\2x-y=-2\ldots(2)\\(1)+(2)\\3x=15\\x=\dfrac{15}{3}=5\\5+y=17\\y=17-5=12$

Draw the graph of each of the following linear equations in two variables:
$y=3x$

x	0	1	2	-1
y	0	3	6	-3

$y=3x$

1830678_1400283_ans_2fe09d0cc75042aa819e36f73c7e1c7e.png

Find the value of $\dfrac{a-2}{3}+\dfrac{a-1}{2}+\dfrac{a+5}{6}=1$

$\dfrac{a-2}{3}+\dfrac{a-1}{2}+\dfrac{a+5}{6}=1$

$\Rightarrow \,\dfrac{2\left(a-2\right)}{3}+\dfrac{3\left(a-1\right)}{2}+\dfrac{a+5}{6}=1$

$\Rightarrow \,\dfrac{2a-4}{6}+\dfrac{3a-3}{6}+\dfrac{a+5}{6}=1$

$\Rightarrow \,\dfrac{2a-4+3a-3+a+5}{6}=1$

$\Rightarrow \,\dfrac{6a-2}{6}=1$

$\Rightarrow \,6a-2=6$

$\Rightarrow \,6a=6+2=8$

$\Rightarrow\,a=\dfrac{8}{6}=\dfrac{4}{3}$

Points are to be recognised on X-Y plane to draw graph of equation $2x + 3y = 12$ . So fill the boxes given below and identify points.

Given the equation is

$2x + 3y = 12$

or,

$x=\dfrac{12-3y}{2}$ ......(1) and

$y=\dfrac{12-2x}{3}$ ......(2).

Now, for

$x=3,6$ respectively we've from (2) we get,

$y=2,0$ respectively.

Now for

$y=4,6$ respectively we have from (1) we get,

$x=0,-3$ respectively.

So the points are

$(0,4),(3,2), (6,0), (-3,6)$ .

Draw the graph of each of the following linear equations in two variables:
$x-y=2$

x	0	1	2	4
y	-2	-1	0	2

$x-y=2$

$-y = 2 - x$

$y = -2 + x$

$y = x - 2$
1830663_1400281_ans_38370b7dac8141189f8269077b052105.png

Solve: $8x+5>12$

$8x+5>12$

$\Rightarrow\,8x+5-5>12-5$

$\Rightarrow\,8x>7$

$\Rightarrow\,x>\dfrac{7}{8}$

$\therefore\,x\in\left(\dfrac{7}{8},\infty\right)$

Sudhir's present age is $5$ more than three times of age of Viru. Amit's age is half the age of Sudhir. If the ratio o the sum of Sudhir's and Viru's age to three times Amit's age is $5:6$ , then find Viru's age.

Let Viru`s age be $x$

Sudhir~s age is $3x+5$

Amit`s age is $\dfrac{3x+5}2$

Sum of sudhir and Viru`s age $x+3x+5=4x+5$

Three times Amit`s Age is $\dfrac32(3x+5)$

Their ratio is $4x+5:\dfrac 32(3x+5)=5:6\\\dfrac{8x+10}{9x+15}=\dfrac 56\\48x+60=45x+75\\3x=15\\x=5$

The sum of a two consecutive numbers isthen find the numbers

Let the numbers be

$x,x+1$

Their sum is

$111$

$\implies x+x+1=111\\2x=110\\x=55$

So the numbers are

$55,56$

What must be added to $2x-4x+9$ to get $5x-13$

Let

$f$ should be added to

$2x - 4x + 9$ to get

$5x - 13$ .

Therefore,

$\left( 2x - 4x + 9 \right) + f = 5x - 13$

$\Rightarrow f = 5x - 13 - \left( -2x + 9 \right)$

$\Rightarrow f = 5x - 13 + 2x - 9$

$\Rightarrow f = 7x - 22$

Hence

$\left( 7x - 22 \right)$ should be added to

$\left( 2x - 4x + 9 \right)$ to get

$\left( 5x - 13 \right)$ .

If there are 50 Mangoes in a box, how will you write the total number of mangoes in terms of the number of boxes? ( Use b for the number of boxes).

Let

$b$ be the

$number\, of \,boxes$ and

$n$ be the

$total\, number \,of \,mangoes$

then equation can be written as:-

$n=50b$

Find the equation of the perpendicular bisector of the line joining $\left(-5,-6\right)$ and $\left(2,-2\right)$

Let the point be

$A\left(-5,-6\right)$ and

$B\left(2,-2\right)$

Midpoint of

$AB=\left(\dfrac{{x}_{1}+{x}_{2}}{2},\,\dfrac{{y}_{1}+{y}_{2}}{2}\right)$

$=\left(\dfrac{-5+2}{2},\,\dfrac{-6-2}{2}\right)$

$=\left(\dfrac{-3}{2},\,\dfrac{-8}{2}\right)$

$=\left(\dfrac{-3}{2},\,-4\right)$

Slope of

$AB=\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}=\dfrac{-2+6}{2+5}=\dfrac{4}{7}$

$\therefore\,$ Slope of the perpendicular

$=\dfrac{-7}{4}$

Now equation is

$y-{y}_{1}=m\left(x-{x}_{1}\right)$

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

$\Rightarrow\,4y+16=-7x-\dfrac{21}{2}$

$\therefore\, 7x+\dfrac{21}{2}-4y-16=0$

$\Rightarrow 7x-4y+\dfrac{21-32}{2}=0$

$\Rightarrow\,7x-4y-\dfrac{11}{2}=0$

$\Rightarrow\,14x-8y-11=0$

$\therefore\,4x+7y+34=0$ is the equation of the line.

Prove $4$ is the root of $2x-3-5=0$ .

Given equation is

$2x-3-5=0$

$\Rightarrow$

$2x-8=0$

$\Rightarrow$

$2x=8$

$\Rightarrow$

$x=\dfrac 82$

$\therefore x=4$

Hence proved.

Check whether the following is solution of the equation $x-2y=4$ or not: $\left(1, 1\right)$

$\left( 1, 1 \right)$ is solution of equation

$x - 2y = 4$ , then it must satisfy the equation.

Therefore,

$1 - 2 \times 1 = -1 \ne 4$

Hence

$\left( 1, 1 \right)$ is not a solution of equation

$x - 2y = 4$ .

Check whether the point $\left(4, 0\right)$ satisfies the equation $x-2y=4$ or not.

Substitute

$4$ for

$x$ and

$0$ for

$y$ in the given equation,

$4 - 2 \times 0 = 4$

$\Rightarrow 4 = 4$ which is true

Hence, point

$\left( 4, 0 \right)$ satisfies the equation

$x - 2y = 4$ .

One third of a number added to itself gives $8$ . Find the number.

Let the required number be

$x$

Then, one third of number

$=\dfrac x3$

$\dfrac x3+x=8$

$\Rightarrow \dfrac{x+3x}{3}=8$

$\Rightarrow x+3x=8\times 3$

$\Rightarrow 4x=24$

$\Rightarrow x=\dfrac {24}{4}$

$\Rightarrow x=6$

$\therefore$ Hence, the required number be

$6$ .

Solve for X :
$\dfrac 1{2x+5}=\dfrac 43$

$\dfrac 1{2x+5}=\dfrac 43\\2x+5=\dfrac 34\\2x=\dfrac 34-5\\2x=\dfrac{-17}4\\x=\dfrac{-17}8$

The value of $x$ if $\dfrac 12x+\dfrac 14x=\dfrac 16$

$\dfrac 12x+\dfrac 14x=\dfrac 16\\\dfrac 34 x=\dfrac 16\\x=\dfrac 16\times \dfrac 43\\x=\dfrac 29$

$\dfrac{2}{3}rd$ of a number is $20$ less than original number. Find the original number.

1219438_1441612_ans_f30f9521992c4b91bf90b854d191480d.jpg

One pen costs $Rs\ 15$ and one pencil costs $Rs\ 3$ . What is the total cost of $x$ pens and $y$ pencils

Cost of one pen

$= Rs. 15$

Cost of

$x$ pens

$= 15 \times x = Rs. 15x$

Cost of

$1$ pencil

$= Rs. 3$

Cost of

$y$ pencils

$= 3 \times y = 3y$

Therefore,

Total cost of

$x$ pens and

$y$ pencils

$= 15x + 3y = 3 \left( 5x + y \right)$

A rabbit covers a distance of $10$ metres, in which it covers $y$ metres distance by walking in slow motion and the remaining by $x$ jumps, each jump contains $2$ metres. Express this information in linear equation.

$\textbf{Step 1: Form the linear equation using the given informations.}$

$\text{It is given that,}$

$\text{Total distance covered = 10 m}$

$\text{Distance covered by walking = y m}$

$\text{Remaining distance covered by x jumps and each jump contains 2 m.}$

$\text{Remaining distance = 2x m}$

$\text{Total distance = (y+2x) m}$

$\text{ATQ}$

$\Rightarrow y+2x=10$

$\Rightarrow 2x+y=10$

$\textbf{Hence, this is the required equation .}$

Solve: $3x - 2 = 19$

Given

$3x-2=19$

$\Rightarrow 3x=19+2$

$\Rightarrow 3x=21$

$\Rightarrow x=\dfrac{21}{3}$

$\therefore x=7$

Find $x$ : $\dfrac { 7x+ 5 }{ 5 } =7x$

$\dfrac{7x+5}{5} = \dfrac{7x}{1}$

$\Rightarrow 7x+5=7x\times 5$

$\Rightarrow$

$5=35x-7x$

$\Rightarrow$

$28x=5 \Rightarrow x = \dfrac{5}{28}$

Solve:
$3t + 25.8 = -4.2$

To solve:

$3t+25.8=-4.2$

Now,

$3t+25.8=-4.2$

$\Rightarrow 3t=-4.2-25.8$

$\Rightarrow 3t=-30$

$\Rightarrow t=-\dfrac{30}{3}$

$\Rightarrow t=-10$

Hence, the solution is

$t=-10$ .

Find $m$ in $\dfrac{{5m}}{2} - 4=0.$

$\dfrac{5m}{2}=4$

$\Rightarrow \dfrac{5m}{2}=4$

$\Rightarrow 5m=4\times 2$

$\Rightarrow 5m=8$

$\Rightarrow m=\dfrac 85$

Verify by substitution that the root of $3x - 5 = 7$ is $x = 4$ .

$3x-5=7$

Given root

$x=4$

$LHS=3(4)-5=12-5=7$

$RHS=7$

$LHS=RHS$

$\therefore x=4$ is the root

Show that $3$ is the root of $2(x+4)=14$ .

$2(x+4)=14$

$\Rightarrow 2x+8=14$

$\Rightarrow 2x=14-8$

$\Rightarrow 2x=6$

$\Rightarrow$

$x=\dfrac 62$

$\therefore x=3$ .

$\text{Solve : }$ $5 y = 15$

${\textbf{Step - 1: Rearranging the terms}}{\text{.}}$

${\text{We have }}5y = 15,{\text{ so we can rearrange it as,}}$

$y = \dfrac{{15}}{5}$

${\textbf{Step - 2: Finding the value of y.}}$

${\text{We can write }}y = \dfrac{{15}}{5}$

$\Rightarrow y= \dfrac{{3 \times 5}}{5}$

${\text{Cancelling 5 from both the numerator and denominator we get,}}$

$\Rightarrow y = 3$

${\textbf{Therefore, the value of }}\mathbf{y = 3.}$

If the point $(2, -2)$ lies on the graph of the linear equation $5x + ky = 4$ , find the value of $k$ .

Given equation of line is

$5 x+ k y=4\cdots(1)$

Also given that

$(2,-2)$ lies on the line

$\implies (2,-2)$ must satisfy the equation

$\implies 5(2)+k(-2)=4$

$\implies 10-2 k=4$

$\implies 2 k=6$

$\implies k=3$

The value of

$k$ is

$3$

Natasha thinks of number and subtracts $\dfrac{2}{3}$ from it and multiplies the result by $9$ . The product now obtained is $7$ times the numbers she thought of. What is the number?

Let no. be

$x$

ATQ

$(x-\cfrac{2}{3})\times 9=7x\\ \Rightarrow 9x-7x=2x=\cfrac{2}{3}\times9\\ \Rightarrow 2x=6 \\\Rightarrow x=3$

Draw the graph of each of equations given below. Also, find the coordinates of the points where the graph cuts the coordinate axes:
$6x - 3y = 12$ .

Given

$6 x-3 y=12$

$\implies 3(2 x-y)=12$

$\implies 2 x-y=4$

$\implies y=2 x-4\cdots(1)$

To draw the graph , we need atleast two solutions

Also to find the coordinates of the points where the graph cuts the coordinate axes,put

$x=0$ and

$y=0$ respectively

Put

$x=0$ in

$(1)$

$\implies y=2(0)-4=0-4=-4$

$(0,-4)$ is the solution of given line

Put

$y=0$

$\implies 0=2 x-4$

$\implies 2 x=4$

$\implies x=2$

$(2,0)$ is the solution of given line

With these two solutions we can draw graph of

$6x-3y=12$

$(2,0)\ \&\ (0,-4)$ are the points where the graph cuts the coordinate axes

1656176_1669850_ans_46ae33f625bb4a3a980579e5f094fd44.png

Draw the graph of each of the following linear equations in two variables:
$2y = -x + 1$ .

$2 y=-x+1\cdots(1)$

To draw the graph ,we need at least two solutions of the equation

Put

$x=0$ in

$(1)$

$2 y=-0+1$

$\implies 2 y=1$

$\implies y=\dfrac{1}{2}$

$\left(0,\dfrac{1}{2}\right)$ or

$(0,0.5)$ is a solution of the equation

Put

$y=0$ in

$(1)$

$2(0)=-x+1$

$\implies -x+1=0$

$\implies x=1$

$(1,0)$ is a solution of the equation

With these two solutions we can draw graph of

$2y=-x+1$

1655182_1669826_ans_538ab6b8a4434b92be445571e4b0e23f.png

Solve for $x$ : $7x-3=0$

$7x-3=0$

$7x=3$

$\therefore x=\cfrac{3}{7}$

Classify the following polynomials as polynomials in one-variable, two variables etc.:
$x^{2} - 2tx + 7t^{2} - x + t$ .

A variable is a quantity that may change within the context of a mathematical problem or experiment. Typically, we use a single letter to represent a variable.

$x^2-2tx+7t^2-x+t$

Here,

$x$ and

$t$ are two variables.

So, it is a polynomial in two variable.

Verify by substitution that the root of $3 + 2x = 9$ is $x = 3$ .

$3+2x=9$

Given

$x=3$

$LHS=$

$3+2(3)=9$

$RHS$

$=9$

$LHS=RHS$

$\therefore x=3$ is root

Draw the graph of each of the following linear equations in two variables:
$y = 2x$ .

$y=2 x\cdots(1)$

To draw the graph ,we need at least two solutions of the equation

Put

$x=0$ in

$(1)$

$y=2(0)=0$

$(0,0)$ is a solution of the equation

Put

$x=5$ in

$(1)$

$y=2(5)=10$

$(5,10)$ is a solution of the equation

With these two solutions we can draw graph of

$y=2x$

1655140_1669823_ans_556e0651c9184ae6b88e50fc7733065a.png

Verity by substitution that the root of $8 - 7y = 1$ is $y = 1$ .

$8-7y=1, y=1$

$LHS$

$=8-7(1)=1$

$RHS$

$=1$

$LHS=RHS$

$\therefore y=1$ is a root

Draw the graph of each of equations given below. Also, find the coordinates of the points where the graph cuts the coordinate axes:
$3x + 2y + 6 = 0$ .

Given

$3 x+ 2 y+6=0$

$\implies 2 y=-6 -3 x$

$\implies y=-3-\dfrac{3}{2} x$

To draw the graph , we need atleast two solutions

Also to find the coordinates of the points where the graph cuts the coordinate axes,put

$x=0$ and

$y=0$ respectively

Put

$x=0$ in

$(1)$

$\implies y=-3-\dfrac{3}{2}(0)=-3-0=-3$

$(0,-3)$ is the solution of given line

Put

$y=0$

$\implies 0=-3-\dfrac{3}{2} x$

$\implies \dfrac{3}{2} x=-3$

$\implies x=-2$

$(-2,0)$ is the solution of given line

With these two solutions we can draw linear graph of

$3x+2y+6=0$

$(-2,0)\ \&\ (0,-3)$ are the points where the graph cuts the coordinate axes

1656188_1669856_ans_65d99da6f634497d842dfa5e91c28678.png

Draw the graph of the equation $2x + y = 6$ . Shade the region bounded by the graph and the coordinate axes. Also, find the area of the shaded region.

$2 x+y=6\cdots(1)$

To draw the graph ,we need at least two solutions of the equation

Put

$x=0$ in

$(1)$

$2 (0)+y=6$

$\implies y=6$

$\left(0,6\right)$ is a solution of the equation

Put

$y=0$ in

$(1)$

$2 x+0=6$

$\implies 2 x=6$

$\implies x=3$

$(3,0)$ is a solution of the equation

The area enclosed by the line with coordinate axes is shaded in the graph

The shaded region forms the triangle with base

$3$ units and height

$6$ units

Area of the shaded region is

$\dfrac{1}{2} \text{base}\times \text{height}=\dfrac{1}{2}\times 3\times 6=9\ \text{sq.units}$

1656297_1669862_ans_657c1b7918d740808c78421a632e1489.png

Draw the graph of each of equations given below. Also, find the coordinates of the points where the graph cuts the coordinate axes:
$-x + 4y = 8$ .

Given

$-x+4 y=8$

$\implies 4 y=x+8$

$\implies y=\dfrac{ x}{4}+2\cdots(1)$

To draw the graph , we need atleast two solutions

Also to find the coordinates of the points where the graph cuts the coordinate axes,put

$x=0$ and

$y=0$ respectively

Put

$x=0$ in

$(1)$

$\implies y=\dfrac{0}{4}+2=0+2=2$

$(0,2)$ is the solution of given line

Put

$y=0$

$\implies 0=\dfrac{ x}{4}+2$

$\implies x=-2\times 4=-8$

$(-8,0)$ is the solution of given line

With these two solutions we can draw linear graph of

$-x+4y=8$

$(-8,0)\ \&\ (0,2)$ are the points where the graph cuts the coordinate axes

1656179_1669853_ans_b4fb0d504736446b8434cd615a0002a8.png

Solve the equation $2y - 1 = y + 1$ and represent it graphically on the coordinate plane.

Given equation is:

$2 y-1=y+1$

$2 y-y=1+1$

$\implies y=2$

On cartesian plane, equation represents all points for which

$y=2$

1655331_1669908_ans_4f19bd8a7ff742b99422224109bbcf10.png

Draw the graph of $y = |x| + 2$ .

Given equation is

$y=|x|+2$

$\implies y=\left\{\begin{matrix}x+2,&x\ge 0\\-x+2,&x<0\end{matrix}\right.$

For

$x\ge 0,$ plot the graph of

$y=x+2$

we need atleast two points on

$y=x+2$ for which

$x\ge 0$

For

$x=0$ ,

$y=2$

for

$x=1,y=3$

$(0,2)$ and

$(1,3)$ are the solutions of

$y=x+2,x\ge 0$

For

$x< 0,$ plot the graph of

$y=-x+2$

we need atleast two points on

$y=-x+2$ for which

$x< 0$

For

$x=-1$ ,

$y=-(-1)+2=3$

for

$x=-2,y=-(-2)+2=4$

$(-1,3)$ and

$(-2,4)$ are the solutions of

$y=-x+2,x< 0$

1656235_1669866_ans_2e1e6e447bfb4afeb7a5390af5d40226.png

If the point $(a, 2)$ lies on the graph of the linear equation $2x - 3y + 8 = 0$ , find the value of $a$ .

Given equation of line is

$2 x-3 y+8=0\cdots(1)$

Also given that

$(a,2)$ lies on the line

$(a,2)$ must satisfy the equation

$(1)$

$\implies 2(a)-3(2)+8=0$

$\implies 2 a-6+8=0$

$\implies 2 a=-2$

$\implies a=-1$

The value of

$a$ is

$-1$

Find the value of $k$ for which the point $(1, -2)$ lies on the graph of the linear equation $x - 2y + k = 0$ .

Given equation of line is

$x-2 y+k=0\cdots(1)$

Also given that

$(1,-2)$ lies on the line

$\implies (1,-2)$ must satisfy the equation

$(1)$

$\implies 1-2(-2)+k=0$

$\implies 1+4+k=0$

$\implies k=-5$

The value of

$k$ is

$-5$

Give the geometrical representation of $2x + 13 = 0$ as an equation in two variables.

Two variable graphical representation of

$2 x+13=0$

$\implies 2 x+0 y+13=0$

$\implies 2 x=-13$

$\implies x=-\dfrac{13}{2}$

$\implies x=-6.5$

On the cartesian plane, the equation represents all points for which

$x=-6.5$

1655203_1669879_ans_1477abcdbf804d58b04b6f94f124e68d.png

Solve the equation $3x - 2 = 2x + 3$ and represent the solution on the number line.

Given

$3 x-2=2 x+3$

$\implies 3 x-2 x=2+3$

$\implies x=5$

The representation of

$x=5$ is shown in the above figure

1657647_1669906_ans_7dfe78f8842042758481eb36ce4583d8.png

Draw the graph of the equation $x+2y-3=0$
From your graph,find the value of $y$ when
(i) $x = 5$
(ii) $x = -5$

1961112_1715234_ans_444ed7fea46d4d5090bbcf7900dc58ed.jpg

State which of the following are equations with a variable. In case of an equation with a variable, identity the variable.
$5\times 4-8=31$

$5\times 4-8=31$
Is an equation

$=L.H.S =R.H.S \to$ . Related variable

$t$ .

Draw the graph of the equation, $2x + y = 6$
Find the coordinates of the graph cut the x-axis.

1708511_1715295_ans_9cb390bea93043e78233054a7edf9689.jpg

State which of the following are equations with a variable. In case of an equation with a variable, identity the variable.
$(y-7)> 5$

$(y-7) > 5$
Is not an equation

$\to L.H.S\neq R.H.S$ .
It has no sign of equality

$(=)$

State which of the following are equations with a variable. In case of an equation with a variable, identity the variable.
$7\times 3x-19=2$

1975286_1787133_ans_52f7a46195fa45a8980f38ed179c0520.jpg

Draw the graph of the equation $y = 3x$
From your graph, find the value of y when
$(i)x=2$
$(ii) x =-2$

(i) It is given

$y = 3x$
Substituting

$x = 1$ in the given equation

$y = 3(1)$
so we get

$y = 3$
Substituting

$x = -1$ in the given equation

$y = 3 (-1)$
So we get

$y = -3$

Now draw a graph using the points

$A(1,3)$ and

$B(-1,-3)$
Join the points

$AB$ through a line and find the value of

$y$
Based on the graph.
i)Consider

$x = 2$ on

$X-axis$ as point

$M$
Now the line joining

$AB$ and point on the

$Y-axis$ gives the value of

$y$
Therefore, when

$x = 2$ the value of

$y = 6$

Substituting

$x = 2$ in the given equation

$y = 3(2)$
So we get

$y = 6$

Now draw a graph using the points

$A (1,3)$ and

$(2.6)$ Join the points

$PQ$ through a line and find the value of

$y$ .
Based on the graph
Consider

$x = 2$ on

$X-axis$ as point

$P$
Now the joining

$AB$ and a point on the

$Y-axis$ gives the value of

$y$
Therefore, when

$x = -2$ the value of

$y = -6$

1538715_1715195_ans_e04a26fa31cf4e34bfa7077f1bc49137.PNG

Draw the graph of the equation, $3x+2y = 6$
Find the coordinates of the point where the graph cuts the y-axis

It is given

$3x+2y = 6$

We can also write it as

$2y = 6 - 3x$

$y = \dfrac{6-3x}{2}$

Substituting

$x = 2$ in the given equation

$y = \dfrac{6-3(2)}{2}$

So we get

$y = \dfrac{6-6}{2}$

$y = 0$

Substituting

$X = 4$ in the given equation

$y = \dfrac{6-3(4)}{2}$

So we get

$Y = \dfrac{6-3(4)}{2}$

So we get

$y = \dfrac{6-12}{2}$

$y = \dfrac{-6}{2}$

By division

$y = -3$

x:	2	4
y:	0	-3

Now draw a grapg using the points

$(2,0) (4,3)$
Join the points through a line and find the coordinates which cuts y-axis.
From the graph we know that the line joining the two points cut the y-axis at a point P which is above x-axis at 3 units distance
Therefore, the coordinate of the point is

$P(0,3)$

1538774_1715324_ans_4b510b4af15445399998f4d80f5b98b6.PNG

Draw the graph of the equation, $2x - 3y = 5$
From your graph find
(i) The value of y when $x = 4$
(ii) The value of x when $y = 3$

It is given

$2x - 3y = 5$
we can write it as

$3y = 2x - 5$
So we get

$y = \dfrac{2x-5}{3}$

Substituting

$x = 4$ in the given equation

$y = \dfrac{2(4)-5}{3}$
So we get

$Y = \dfrac{8-5}{3}$

$y = \dfrac{3}{3}$
By division we get

$y = 1$

Substituting

$x =-2$ in the given equation

$y = \dfrac{2(-2)-5}{3}$
So we get

$y = \dfrac{-4-5}{3}$

$y = \dfrac{-9}{3}$
By division we get

$y = -3$

x	4	-2
y	1	-3

Now draw a graph using the points

$(4,1)$ and

$(-2,3)$
Join the point through a line and find the value of y
(i) Based on the graph
consider

$x = 4$ on X-axis
Now the line joining the two points and a point on the Y-axis gives the value of y
Therefore,when

$x = 4$ the value of

$y = 1$
(ii) Based on the graph,
Consider

$y = 3$ on Y-axis
Now the line joining the two points and a point on the X-axis the value of x
Therefore, when

$y = 3$ the value of

$x = 7$

1538743_1715260_ans_efc3f4fa48a94a84a332927ac2cbf97d.PNG

State which of the following are equations with a variable. In case of an equation with a variable, identity the variable.
$7=11\times 5-12\times 4$

$7=11\times 5-12\times 4$
Is an equation

$\to L.H.S = R.H.S \to$ . It has no variable.

Draw the graphs of linear equations $y = x$ and $y = -x$ on the same Cartesian plane. What do you observe?

Any point on the graph of

$y = x$ will have

$x$ and

$y$ coordinates same.

The line passes through the point

$(0,0), (1,1)$ and

$(-1,-1).$

Again, any point on the graph of

$y = x$ will have

$x$ and

$y$ coordinates of opposite signs.

The line passes through the points

$(1,-1)$ and

$(-1,1).$

Also,

$(0,0)$ satisfy

$y + x.$
The graph of the linear equations

$y = x$ and

$y = - x$ on the same Cartesian plane is shown in the figure given below.
We observe that the graph of these equation passes through

$(0,0).$

1659931_1794094_ans_cf5a1621921949e2a42a6f364730b81d.png

The line represented by $x= 7$ parallel to $x-axis$ . justify whether the statement is true or not.

The line represented by

$x = 7$ is of the form

$x= a$ .

The graph of this form of equations is a line parallel to the

$y -axis$
Hence the given statement is not true.

Find the solution of the linear equation $x+2y =8$ which represents a point on
(i) x-axis (ii) y-axis

(i) We know that the point which lies on x-axis has its ordinate

So, putting

$y=0$ we get:

$x+2 \times0 =8 \Rightarrow x=8$

hence,

$(8,0)$ lies on the x-axis

(ii) We know that the point which lies on y-axis has its abscissa

So, putting

$x=0$ we get:

$0+2 y =8 \Rightarrow 2y=8 \Rightarrow y=4$

hence,

$(0,4)$ lies on the y-axis

If the point (3,4) lies on the graph of $3y = ax + 7,$ then find the value of a.

The point

$(3,4)$ lies on the graph of

$3y = ax + 7,$
Substituting

$x = 3$ and

$y=4$ in the given equation

$3y = ax + 7,$ we get

$\therefore 3 \times 4 = a \times 3 + 7$

$\Rightarrow 12 = 3a + 7 \Rightarrow 3a = 5 \Rightarrow a = \dfrac{5}{3}$

There are Some students in two examination halls $A$ and $B$ . To make the number of students equal in each hall $10$ students are sent from $A$ to $B$ . But if $20$ students are sent from $B$ to $A$ , the number of students in $A$ becomes double the number of students in $B$ find the numbers of students in the two halls.

Let the number of students initially in hall A be

$x.$

and the number of students initially in hall B be

$y.$

Case I : 10 students of hall A shifted to B

Now, number of students in hall

$A = (x - 10)$

Now, number of students in hall

$B = (y + 10)$

According to the question, number of student in both halls are equal.

$\therefore x - 10 = y + 10$

$\Rightarrow x - y = 20$ ___(i)

Case II :

$20$ students are shifted from hall B to A. then

Number of students in hall A becomes

$= x + 20$

Number of students In hall B becomes

$= y - 20$

According to the question, students in hall A becomes twice of students in hall B.

$\therefore x + 20 = 2(y - 20)$

$\Rightarrow x + 20 = 2y - 40$

$\Rightarrow x - 2y = -60$ __(ii)

$x - y = 20$ [from (i)]

$-$

$+$

$-$

$\overline{-y = -80}$ [Subtracting eqn. (i) from (ii)]

$\Rightarrow y = 80$

Now,

$x - y = 20$ [From (i)]

$\Rightarrow x - 80 = 20$

$\Rightarrow x = 20 + 80$

$\Rightarrow x = 100$

$\therefore$ Number of students initially in hall

$A = 100$

and number of students initially in hall

$B = 80$

Draw the graph of the linear equation whose solutions are represented by the points having the sum of coordinates as $10$ units.

$\textbf{Step-1: Find the points using given lines & plotting them on the graph.}$

$\text{A linear equation whose solutions are represented by the points}$

$\text{having the sum of coordinates as}$

$10$

$\text{units is}$

$x + y = 10.$

$\text{When}$

$x = 0, y = 10$

$\text{and when}$

$x = 10, y = 0.$

$\text{Now, plot these two points}$

$(0,10)$

$\text{and}$

$(10,0)$

$\text{on a graph paper and join them to obtain a straight line.}$

$\text{The graph of}$

$x = y = 10$

$\text{is a straight line as shown in the figure given above.}$

$\textbf{Hence, the graph has been plotted .}$

2170363_1794117_ans_824d778a69f84f8380cfac2576ba337b.png

Draw the graph for the linear equation given below:
$y = -2x$ .

by choosing a couple of values for

$x, -1, 0, 1$

calculate the corresponding

$y$ values.

$x$

$-1$

$0$

$1$

$y$

$2$

$0$

$-2$

1804119_1843587_ans_c922f900ee0a40979022faa386f00ec4.png

Draw the graph for the linear equation given below:
$y = -x$ .

by choosing a couple of values for

$x, -1, 0, 1$

calculate the corresponding

$y$ values.

$x$

$-1$

$0$

$1$

$y$

$1$

$0$

$-1$

1803595_1843586_ans_d60bd616a55a4f27a62b7bae648fc3ef.png

Draw the graph for the linear equation given below:
$5x + y = 0$ .

by choosing a couple of values for

$x, -1, 0, 1$

calculate the corresponding

$y$ values.

$x$

$-1$

$0$

$1$

$y$

$5$

$0$

$-5$

1804144_1843590_ans_d7b3fb9ba8e44a78ad2220d31fda439e.png

Draw the graph for each linear equation given below: $4x - y = 0$ .

by choosing a couple of values for

$x, -1, 0, 1$

calculate the corresponding

$y$ values.

$x$

$-1$

$0$

$1$

$y$

$-4$

$0$

$4$

1804175_1843592_ans_7bf68dc1478243ce871eb7a362f2e64b.png

Draw the graph for each linear equation given below:
$2x - 7 = 0$ .

We can also write the equation as:

$x = \dfrac {7}{2}$

by choosing a couple of values for

$x,$

$\dfrac {7}{2},$

$\dfrac {7}{2}$

calculate the corresponding

$y$ values.

$x$

$\dfrac {7}{2}$

$y$

$-1$

$0$

$1$

With the help of this the required graph is drawn.

1803961_1843503_ans_2f2e7c5f47b2451996774423b311a60a.png

Draw the graph for each linear equation given below: $y = 0$ .

by choosing a couple of values for

$x, -1, 0, 1$

calculate the corresponding

$y$ values.

$x$

$-1$

$0$

$1$

$y$

$0$

1803974_1843522_ans_06db16c6dae2466c96ab7f6fbf36d2d4.png

Draw the graph for the linear equation given below:
$x + 2y = 0$ .

by choosing a couple of values for

$x, -1, 0, 1$

calculate the corresponding

$y$ values.

$x$

$-1$

$0$

$1$

$y$

$\dfrac {1}{2}$

$0$

$-\dfrac {1}{2}$

1804164_1843591_ans_34e6e8d726b74d578dc4dcc60b0ecd80.png

$\Rightarrow y=\dfrac{-2}{3}$

For

$x=-1\Rightarrow 2(-1)-3y=4$

$\Rightarrow y=-2$

$x:$

$-1$

$0$

$1$

$y:$

$-2$

$-\dfrac {4}{3}$

$-\dfrac {2}{3}$

Hence, plotting points

$(-1,-1),\ \left(0,\dfrac{-4}{3}\right)\ and \ \left(1,\dfrac{-2}{3}\right)$ , we will get the required graph, as shown above.

1804860_1843607_ans_9b0561f906614901bc95d83db2fc643c.png

Draw the graph for the each linear equation given below:
$y = 2x + 3$ .

by choosing a couple of values for

$x, -1, 0, 1$

calculate the corresponding

$y$ values.

$x$

$-1$

$0$

$1$

$y$

$1$

$3$

$5$

1804408_1843597_ans_09c1f68484194526a8671d8de363dea7.png

Draw the graph for the each linear equation given below: $y = \dfrac {3x}{2} + \dfrac {2}{3}$ .

by choosing a couple of values for

$x, -1, 0, 1$

calculate the corresponding

$y$ values.

$x$

$-1$

$0$

$1$

$y$

$-\dfrac {5}{6}$

$\dfrac {2}{3}$

$\dfrac {13}{6}$

Draw the graph for the linear equation given below: $\dfrac {x - 1}{3} - \dfrac {y + 2}{2} = 0$ .

The equation can be written as,

$2x - 3y = 8$

by choosing a couple of values for

$x, -1, 0, 1$

calculate the corresponding

$y$ values.

$x$

$-1$

$0$

$1$

$y$

$-\dfrac {10}{3}$

$-\dfrac {8}{3}$

$-2$

1804891_1843610_ans_11cd791abf9048c88e039711e366ae67.png

Draw the graph for the linear equation given below: $x - 3 = \dfrac {2}{5} (y + 1)$ .

The equation can be written as,

$5x - 2y = 17$

by choosing a couple of values for

$x, -1, 0, 1$

calculate the corresponding

$y$ values.

$x$

$-1$

$0$

$1$

$y$

$-11$

$-\dfrac {17}{2}$

$-6$

1804926_1843614_ans_febb701ff037465697b5f536125cc23f.png

Draw the graph of equation $x + 2y - 3 = 0$ . From the graph, find:
(i) $x_{1}$ , the value of $x$ , when $y = 3$
(ii) $x_{2}$ , the value of $x$ , when $y = -2$ .

by choosing a couple of values for

$x, -1, 0, 1$

calculate the corresponding

$y$ values.

$x$

$-1$

$0$

$1$

$y$

$2$

$\dfrac {3}{2}$

$1$

(i) When

$y = 3$ , we get

$x = -3$

(ii) When

$y = -2$ , we get

$x = 7$ .

1805080_1843645_ans_a5857026b90c4e0f9f2e425e4995e457.png

Draw the graph for each equation, given below: $y + 7 = 0$ .

$y + 7 = 0$

$y = -7$ .
1800682_1843760_ans_00fe5f5971534f099bc4f765f590fa3b.png

Draw the graph for each equation, given below: $x + 5 = 0$ .

$x + 5 = 0$

$x = -5$ .
1800677_1843756_ans_6fa8546c0a074e83a3cf44ad55044a04.png

Draw the graph for each equation, given below: $y = 7$ .

$y = 7$ .

1800681_1843758_ans_3ee1ab027a1e4d409d0afe4c0519ad73.png

Draw the graph for the equation given below:
$2x - 5y = 10$
Also, find the co-ordinates of the points where the graph (line) drawn meets the co-ordinates axes.

by choosing a couple of values for

$x, -1, 0, 1$

calculate the corresponding

$y$ values.

$x$

$-1$

$0$

$1$

$y$

$-\dfrac {12}{5}$

$-2$

$-\dfrac {8}{5}$

According to the graph, the line intersect

$x$ axis at

$(5, 0)$ and

$y$ at

$(0, -2)$ .

1804993_1843622_ans_af207441e89a45e494d7c776e66b1f27.png

Draw the graph of the equation $3x - 4y = 12$ .
Use the graph drawn to find:
(i) $y_{1}$ , the value of $y$ , when $x = 4$ .
(ii) $y_{2}$ , the value of $y$ , when $x = 0$ .

by choosing a couple of values for

$x, -1, 0, 1$

calculate the corresponding

$y$ values.

$x$

$-1$

$0$

$1$

$y$

$-\dfrac {15}{4}$

$-3$

$-\dfrac {9}{4}$

According to the graph,

When

$x = 4, y = 0$

When

$x = 0, y = -3$ .

1805057_1843647_ans_fc40bcab5b6d469a942b4e39c78628e0.png

Draw the graph for the equation given below:
$3x + 2y = 6$
Also, find the co-ordinates of the points where the graph (line) drawn meets the co-ordinates axes.

by choosing a couple of values for

$x, -1, 0, 1$

calculate the corresponding

$y$ values.

$x$

$-2$

$0$

$2$

$y$

$6$

$3$

$0$

According to the graph, the line intersect

$x$ axis at

$(2, 0)$ and

$y$ at

$(0, 3)$

1805144_1843621_ans_87bed477a93c43c398f67e862238fade.png

Draw the graph for the linear equation given below: $x + 5y + 2 = 0$ .

by choosing a couple of values for

$x, -1, 0, 1$

calculate the corresponding

$y$ values.

$x$

$-1$

$0$

$1$

$y$

$-\dfrac {1}{5}$

$-\dfrac {2}{5}$

$-\dfrac {3}{5}$

1804968_1843615_ans_7ddf7c39c6984a87bef82572b8462d37.png

Draw the graph for each equation, given below: $3x + 2y = 6$ .

$3x + 2y = 6$

$2y = 6 - 3x$

$\therefore y = \dfrac {6 - 3x}{2}$
When

$x = 0; y = \dfrac {6 - 3\times 0}{2} = \dfrac {6 - 0}{2} = 3$
When

$x = 2; y = \dfrac {6 - 3\times 2}{2} = \dfrac {6 - 6}{2} = 0$
When

$x = 4; y = \dfrac {6 - 3\times 4}{2} = \dfrac {6 - 12}{2} = -3$

$x$	$0$	$2$	$4$
$y$	$3$	$0$	$-3$

1800685_1843761_ans_f3d0a768c7704ffba79a7722066ee57b.png

Draw the graph for each equation, given below: $x = 5$ .

$x = 5$

1800676_1843755_ans_c75fc165cc8045f6958de1bc46191036.png

Draw the graph of the equation $2x - 3y - 5 = 0$
(i) $x_{1}$ , the value of $x$ , when $y = 7$
(ii) $x_{2}$ , the value of $x$ , when $y = -5$ .

$2x - 3y - 5 = 0$

$\Rightarrow 2x = 3y + 5$

$\Rightarrow x = \dfrac {3x + 5}{2}$
When

$y = 1, x = \dfrac {3(1) + 5}{2} = \dfrac {8}{2} = 4$
When

$y = 3, x = \dfrac {3(3) + 5}{2} = \dfrac {9 + 5}{2} = 7$
When

$y = -1, x = \dfrac {3(-1) + 5}{2} = \dfrac {5 - 3}{2} = 1$

$x$	$4$	$7$	$1$
$y$	$1$	$3$	$-1$

The value of

$x$ , when

$y = 7$ ;
We have the equation of the line as

$x = \dfrac {3y + 5}{2}$
Now substitute

$y = 7$ and

$x = x_{1}$ ;
Now substitute

$y = 7$ and

$x = x_{1}$ :

$x_{1} = \dfrac {3(7) + 5}{2} = \dfrac {21 + 5}{2} = \dfrac {26}{2} = 13$
The value of

$x$ , when

$y = -5$
Now substitute

$y = -5$ and

$x = x_{2}$ :

$x_{2} = \dfrac {3(-5) + 5}{2} = \dfrac {-15 + 5}{2} = \dfrac {-10}{2} = -5$ .
1800696_1843767_ans_14fc26cbc8a04f34bc6cd231cbb92029.png

Draw the graph of the equation $4x + 3y + 6 = 0$
From the graph, find:
(i) $y_{1}$ , the value of $y$ , when $x = 12$
(ii) $y_{2}$ , the value of $y$ , when $x = -6$ .

$4x + 3y + 6 = 0$

$\Rightarrow 3y = -4x - 6$

$\Rightarrow y = \dfrac {-4x - 6}{3}$
When

$x = 0, y = \dfrac {-4(0) - 6}{3} = \dfrac {-6}{3} = -2$
When

$x = 3, y = \dfrac {-4(3) - 6}{3} = \dfrac {-12 - 6}{3} = -6$
When

$x = -3, y = \dfrac {-4(-3) - 6}{3} = \dfrac {12 - 6}{2} = 2$

$x$	$0$	$3$	$-3$
$y$	$-2$	$-6$	$2$

The value of

$y$ , when

$x = 12$ :
We have the equation of the line as

$y = \dfrac {-4x - 6}{3}$
Now substitute

$x = 12$ and

$y = y_{1}$ :

$y_{1} = \dfrac {-4(12) - 6}{3} = \dfrac {-48 - 6}{3} = \dfrac {-54}{4} = -18$
The value of

$y$ , when

$x = -6$ :
Now substitute

$x = -6$ and

$y = y_{2}$ :

$y_{2} = \dfrac {-4(-6) - 6}{3} = \dfrac {24 - 6}{3} = \dfrac {18}{3} = 6$ .
1800697_1843768_ans_dd993fb2362e4d73af9250d36f3131a1.jpg

Draw the graph for each equation, given below: $x - 5y + 4 = 0$ .

$x - 5y + 4 = 0$

$5y = 4 + x$

$\therefore y = \dfrac {x + 4}{5}$
When

$x = 1; y = \dfrac {1 + 4}{5} = \dfrac {5}{5} = 1$
When

$x = 6; y = \dfrac {6 + 4}{5} = \dfrac {10}{5} = 2$
When

$x = -4; y = \dfrac {-4 + 4}{5} = \dfrac {0}{5} = 0$

$x$	$1$	$6$	$-4$
$y$	$1$	$2$	$0$

1800689_1843762_ans_a91999019f214f9db5ea2df481a6fd8b.png

Draw the graph of the equation given below
$x + y = 2$

The equation of the given line is x + y = 2.

x + y = 2

⇒ y = 2 − x .....(1)

Putting x = 0 in (1), we get

y = 2 − 0 = 2

Putting x = 2 in (1), we get

y = 2 − 2 = 0

Putting x = −1 in (1), we get

y = 2 − (−1) = 2 + 1 = 3

Putting x = 5 in (1), we get

y = 2 − 5 = −3

1807971_1855758_ans_63532dc4cd5d40f9bc500310961bf528.png

Draw the graph of the equation given below
$2x + y = 1$

These values can be represented in the table in the form of ordered pairs as follows:

Plot these points on the graph paper.

The equation of the given line is 2x + y = 1.

2x + y = 1

⇒ y = 1 − 2x .....(1)

Putting x = 0 in (1), we get

y = 1 − 2 × 0 = 1 − 0 = 1

Putting x = 1 in (1), we get

y = 1 − 2 × 1 = 1 − 2 = −1

Putting x = −1 in (1), we get

y = 1 − 2 × (−1) = 1 + 2 = 3

Putting x = 2 in (1), we get

y = 1 − 2 × 2 = 1 − 4 = −3

1808025_1855762_ans_3827734e02d44457a5d916257bf92fd7.png

Complete the following table to draw the graph of $2x - 6y = 3$
$\textbf{x}$ -5 x
$\textbf{y}$ x 0
$\textbf{(x,y)}$ (-5,x) (x,0)

$2x - 6y = 3$

$\implies y=\dfrac{2x-3}{6}$

$\textbf{x}$	$-5$	$\frac{3}{2}$
$\textbf{y}$	$-\frac{13}{6}$	$0$
$\textbf{(x,y)}$	$\left( -5, -\dfrac{13}{6}\right)$	$\left( \dfrac{3}{2}, 0\right)$

1815198_1850803_ans_d5e37866741c4f84999a79743223aaf8.png

Draw the graph (straight line) given by equation $x - 3y = 18$ . If the straight line drawn passes through the points $(m, -5)$ and $(6, n)$ ; find the values of $m$ and $n$ .

According to the question,

$x - 3y = 18$

$\Rightarrow - 3y = 18 - x$

$\Rightarrow 3y = x - 18$

$\Rightarrow y = \dfrac {x - 18}{3}$
The table for

$x - 3y = 18$ is

$x$	$9$	$0$	$6$	$3$
$y$	$-3$	$-6$	$-4$	$-5$

Hence, from the graph, we can conclude that

$m = 3$ and

$n = -4$ .
1800713_1843775_ans_e737521549f74e3faa9524ba996214e1.png

Draw the graph of the equation given below
$3x - y = 0$

The equation of the given line is 3x − y = 0.

3x − y = 0

⇒ y = 3x .....(1)

Putting x = 0 in (1), we get

y = 3 × 0 = 0

Putting x = 1 in (1), we get

y = 3 × 1 = 3

Putting x = −1 in (1), we get

y = 3 × (−1) = −3

Putting x = 2 in (1), we get

y = 3 × 2 = 6

1807984_1855760_ans_1ada4038ca234f7984d8f5344b19288d.png

Draw the graphs of the following straight lines.
$4y=-x+3$

$4y=-x+3$

$x$	$-1$	$3$	$-5$	$-9$
$y$	$1$	$0$	$2$	$3$
$(x,y)$	$(-1,1)$	$(3,0)$	$(-5,2)$	$(-9,3)$

When

$x=-1\quad \quad$ When

$x=3\quad \quad when\,x=-5$

$4y=-(-1+3)\quad 4y=-3+3\quad 4y=-(-5)+3$

$4y=1+3\quad y=\dfrac 04 \quad y=\dfrac 84 =\boxed {2}$

$y=\dfrac 44 =\boxed {1}\quad \boxed {y=0}\quad when\, x=-9$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad 4y=-(-9)-3$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\dfrac {12}{4}=\boxed {3}$
1956445_1871001_ans_c68d9ae7a9d2487d8224a9016c5aa89e.png

Draw the graphs of the following straight lines.
$3y=1$

$3y=1$

$3y+0x=1$

$x$	$0$	$1$	$2$	$3$
$y$	$1/3$	$1/3$	$1/3$	$1/3$
$(x,y)$	$\left(0,\dfrac 13 \right)$	$\left(1,\dfrac 13 \right)$	$\left(2,\dfrac 13 \right)$	$\left(3,\dfrac 13 \right)$

$P(0,0.3)\ Q(1,0.3)\ R(2,0.3)\ S(3,0.3)$

1956450_1871004_ans_fafd7bcf4d4647ef960bed6f1879432e.png

Draw the graphs of the following straight lines.
$y=3-x$

$y=3-x$

$x=$	$0,$	$1,$	$2$
$y=$	$3,$	$2,$	$1$

When

$x=0, \quad y=3-0=(3)$
When

$x=1, \quad y=3-1=(2)$
When

$x=2, \quad y=3-2=(1)$
1956437_1870995_ans_699f5ea9f1f04718bcde1bac1806ebf9.png

Draw the graphs of the following straight lines.
$y=x-3$

$y=x-3$

$x=$	$0$	$1$	$2$
$y=$	$-3$	$-2$	$-1$

When

$x=0, \quad y=0-3=(-3)$
When

$x=1, \quad y=1-3=(-2)$
When

$x=2, \quad y=2-3=(-1)$
1955414_1870996_ans_194e2cf1344542c7ad0465b871698880.png

Draw the graphs of the following straight lines.
$3y=4x+1$

$3y=4x+1$

$x$	$-1$	$2$	$5$	$8$
$y$	$-1$	$3$	$7$	$11$
$(x,y)$	$(-1,0)$	$(2,3)$	$(5,7)$	$(7,11)$

When

$x=-1\quad$ When

$x=3\quad \quad x=-5$

$3y=4(-1+1)\quad 3y=4(2)+1\quad 3y=4(5)+1$

$=\dfrac {-3}{3}=\boxed {-1}\quad y=\dfrac {9}{3}=\boxed {3}\quad =\dfrac {21}{3}=\boxed {7}$
When

$x=8$

$3y=4(8)+1$

$=\dfrac {33}{3}=\boxed {11}$
1956447_1871002_ans_c036489679b84cdf952bb5a1703e844a.png

Draw the graphs of the following straight lines.
$y=3x-2$

$y=3x-2$

$x=$	$0,$	$1,$	$2$
$y=$	$-2,$	$1,$	$4$

When

$x=0, \quad y=3(0)-2=(-2)$
When

$x=1, \quad y=3(1)-2=(1)$
When

$x=2, \quad y=3(2)-2=(4)$
1956443_1870998_ans_886bb926a5874d70bc6b10820f7bc945.png

Give the geometric representations of $2x + 9 = 0$ as an equation.
(ii) in two variable

$2x + 9 = 0$ , this equation in
ii)

$2x + 9 = 0$ in two variables,

$2x + 9 = 0$ .

$2x = -9$

$x = - \dfrac{9}{2} = -4.5$

This equation passes through

$(-4.5, 0)$ and this is parallel ot y-axis. Coordinate of

$x = - 4.5$

Draw the graph of each of the following linear equations in two variables:
$3 = 2x + y$

$3 = 2x + y$

$2x + y = 3$

$y = 3 - 2x$

1830481_1869796_ans_45599d96c26a401aa183ac98bbe2cf8e.png

Draw the graphs of the following straight lines.
$y=5-3x$

$y=5-3x$

$x=$	$0,$	$1,$	$2$
$y=$	$5,$	$2,$	$-1$

When

$x=0, \quad y=5-3(0)=5$
When

$x=1, \quad y=5-3(1)=2$
When

$x=2, \quad y=5-3(2)=5-6=-1$
1956444_1871000_ans_49da7b1e98e34cb883fe30851030ede9.png

Draw the graphs of the following straight lines.
$x=4$

$x=4$

$x-0y=4$

$x$	$4$	$4$	$4$	$4$
$y$	$0$	$1$	$2$	$3$
$(x,y)$	$(4,0)$	$(4,0)$	$(4,2)$	$(4,3)$

1956449_1871003_ans_045cdd0ba17247b0acb4a86b88c89dff.png

By taking the values of $x = -2$ to $x = 2$ and also included value of x for the equation $y = 2x + 1.$ Form the and draw the graph of the equation.

The given equation is

$y = 2x + 1$
When

$x = -2$

$y = 2 \times (-2) + 1 = -3$
When

$x = -1$

$y = 2 \times (-1) + 1 = -2 + 1 = -1$
When

$x = 0$

$y = 2 \times 0 + 1 = 1$
When

$x = 1$

$y = 2 \times 1 + 1 = 3$
When

$x = 2$

$y = 2 \times 2 + 1 = 5$

$|x=-2,-1,0,1,2|$

$|y=-3,-1,1,3,5|$

1867650_1879013_ans_a8f518fe790e4c3dbf6c0aa2f7ca3860.png

Construct $3$ equations starting with $x=-2$ .

$\begin{array}{l} x=-2 \\ \\\left( 1 \right) x=-2 \\ x+2=0 \\ \\ \left( 2 \right) x-2=-2-2 \\ x-2=-4 \\\\ \left( 3 \right) 2x=2\times \left( { -2 } \right) \\ 2x=-4 \end{array}$

Two candidates A and B contest an election. A gets 46% of the valid votes and is defeated by 1600 votes. Find the total number of valid votes cast in the election ?

Let total number of vaild votes = x

$\Rightarrow$ votes A got = 46 % of x

$= \dfrac{46x}{100}$

votes B get = 54% of x

$= \dfrac{54x}{100}$

$\Rightarrow$ B won the election by

votes

$= \dfrac{54x}{100}-\dfrac{46x}{100}$

$= \dfrac{8x}{100}$

Difference of votes =

$1600$

$\Rightarrow \dfrac{8x}{100} = 1600$

$\Rightarrow x = \dfrac{1600\times 100}{8}$

$= 20000$

$\Rightarrow$ Total votes cast =

$20000$

Express the following linear equations in the form ax + by + c = 0 and indicate the value of a, b and c in each case :
i) $2x + 3y = 9.3 \bar{5}$
ii) $x - \dfrac{y}{5} - 10 = 0$
iii) -2x + 3y = 6

$2x+3y=9.3\bar 5\implies 2x+3y-9.3\bar 5$

where

$a=2,b=3$ and

$c=-9.3\bar 5$

ii)

$x-\dfrac 15 y-10=0$

where

$a=1,b=\dfrac 15$ and

$c=-10$

iii)

$-2x+3y=6\implies 2x-3y+6=0$

where

$a=2,b=-3$ and

$c=6$

What is the sum of $2x-y$ and $3y -2x$ ?

Sum of

$2x-y$ and

$3y-2x$

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

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

$=2y$

Hence, the answer is

$2y.$

Express the following linear equation in the form $ax+by+c=0$ and indicate the values of $a,b$ and $c$ in each case :
$2x+3y=9.35$

Given

$: 2x+3y=9.35$

we have to express this in the form of a linear equation of the type,

$ax+by+c=0 \dots\left(1\right)$

$2x+3y=9.35$

$2x+3y-9.35=0\dots \left(2\right)$

Comparing

$\left( 1\right)\;\&\;\left(2\right)$ we get:

Hence, the values of

$a, b$ and

$c$ are:

$a=2$

$b=3$

$c=-9.35$

Find the solution of given problem $\dfrac{3}{4}(7x-1)-\left(2x-\dfrac{1-x}{2}\right)=x+\dfrac{3}{2}$ .

$\cfrac { 3 }{ 4 } \left( 7x-1 \right) -\left( 2x-\cfrac { 1-x }{ 2 } \right) =x+\cfrac { 3 }{ 2 } \\ =>\cfrac { 21x }{ 4 } -\cfrac { 3 }{ 4 } -2x+\cfrac { 1-x }{ 2 } =x+\cfrac { 3 }{ 2 } \\ =>21x-3-8x+2\left( 1-x \right) =4x+6\\ =>21x-3-8x+2-2x=4x+6\\ =>11x-1=4x+6\\ =>7x=7\\ =>x=1$

Solve the following equations.
$\displaystyle\, 32^{\frac{x + 5}{x - 7}} = 0.25 \cdot 128^{\frac{x + 17}{x - 3}}$

$32^{\dfrac{x+5}{x-7}} = 0.25.128^{\dfrac{x+17}{x-3}}$

$(2^{5})^{\dfrac{x+5}{x-7}} = 1/4.(2^{7})^{\dfrac{x+17}{x-3}}$

$2^{2}.2^{5(\dfrac{x+5}{x-7})} = 2^{7.(\dfrac{x+17}{x-3})}$

$2^{2+5.(\dfrac{x+5}{x-7})} = 2^{7(\dfrac{x+17}{x-3})}$

$2+5(\dfrac{x+5}{x-7}) = 7(\dfrac{x+17}{x-3})$

$2+\dfrac{5x+25}{x-7} = \dfrac{7x+119}{x-3}$

$\dfrac{2x-14+5x+25}{x-7} = \dfrac{7x+119}{x-3}$

$\dfrac{7x+11}{x-7} = \dfrac{7x+119}{x-3}$

$(x-3)(7x+11) = (x-7)(74+119)$

$7x^{2}+11x-21x-33 = 7x^{2}+119x-49x-833$

$-10x-33 = 70x-833$

$800 = 80x$

$\boxed{x = 10}$

1131842_886911_ans_42db43cb692b469083a9251fba3b8755.jpg

Express the following linear equation in the form $ax+by+c=0$ and indicate the values of $a,b$ and $c$ in each case :
$5=2x$

Given :

$5=2x$

Expressing this in the form of

$\Rightarrow ax+by+c=0$

we get

$2x+0.y-5=0$

comparing both we get,

$\Rightarrow a=2$

$b=0$ and

$c=-5$

Hence, solved.

$A$ and $B$ can dig a trench in $20$ days. If $B$ is twice as good as $A$ in his work, in how many days can A alone complete the work?

(

$A's$ one day work

$= \mathrm { B } 's \dfrac { 1 } { 2 }$ days work )

$A$ is twice as good as $B$ .

Let $B=x,A=2x$

$A$ and $B$ can complete the work together in $20$ days.

So, $(A+B)’s\ 20$ days’ work $=1$

$(A+B)’s\ 1$ day’s work $=1/20$

So,

$x+2x=\dfrac{1}{20}$

$3x=\dfrac{1}{20}$

$x=\dfrac{1}{60}$

$2x=\dfrac{1}{30}$

A’s one day’s work $=1/30$

A can complete the work alone in $30$ days.

$\therefore answer \ is \ 30$

If $x + y = 3,\,\,\,3x - 2y - 4 = 0$ , then find coefficient determinant(D).

1308838_1451270_ans_0a5d8a6e508c4347adea1444464b2d03.jpg

Solve:
$\dfrac{3}{x+y}+\dfrac{2}{x-y}=2$ ; $\dfrac{9}{x+y}-\dfrac{4}{x-y}=1$

1311455_1586294_ans_1f542ac1640343dd93e9a7fb442d554e.jpg

Solve : $3 x = 27$

$3x = 27$

$x = \cfrac{27}{3}$

$x = 9$

Draw the graph of each of the following equations.
(i) x = 4
(ii) x+4 = 0
(ii) y = 3
(iv) y = -3
(v) x =-2
(vi) x=5
(vii) y+5 = 0
(viii) y = 4

1671154_1715163_ans_b74e8961f60d4ab1bf6fcc09cbc8d7b2.jpg

Solve : $4 x - 30 = 0$

$4x - 30 = 0$

$4x = 30$

$x = \cfrac{30}{4}$

$x = \cfrac{15}{2}$

For fig. (i)	For fig. (ii)
(i) $y=x$	(i) $y=x+2$
(ii) $x+y=0$	(ii) $y=x-2$
(iii) $y=2x$	(iii) $y=-x+2$
(iv) $2+3y=7x$	(iv) $x+2y=6$

Linear Equations In Two Variable - Class 9 Maths - Extra Questions

y−2=0y - 2 = 0 is parallel to xx-axis.If true enter 1 else 0.

The graph for the linear equation 2y−5=02y - 5 = 0 is parallel to xx-axis.If true enter 1 else 0.

y+6=0y + 6 = 0 is parallel to xx-axis.If true enter 1 else 0.

The cost of notebook is twice the cost of a pen. Write a linear equation in two variable to represent this statement.

x−5=0x - 5 = 0 is parallel to y−y-axis. Find the value of xx co-ordinate where it crosses x−axisx-axis.

Give the geometric representation of y=3y = 3 as an equation in one variable.

Give the geometric representation of x=−92x = -\dfrac{9}{2} in one variable.

A greeting card costs Rs.33 and a ream of photocopy paper costs Rs.88. Amir paid Rs.3232 for xx cards and yy reams of paper. Write a linear equation based on the information above.

The perimeter of △PQR\triangle PQR is 30cm30 cm.Write a linear equation in two variables based on the information given above.

The length and width of a rectangle is 4xcm4x cm and ycmy cm respectively. If its length is 3cm3 cm longer than its width, write an equation involving xx and yy.

Express the following linear equation in the form of ax + by + c = 0 and indicate the value of a, b and c.y=−32xy = \dfrac{-3}{2} x

Given the function hh such that h(x)=x2+kx+5h(x)=x^2+kx+5 and h(3)=2h(3)=2. Find the value of kk.

Write each of the following in the form of ax + by + c = 0 and find the value of a, b and c(i) x = - 5(ii) y = 2(iii) 2x = 3(iv) 5y = -3

Express the following linear equation in the form of ax + by + c = 0 and indicate the value of a, b and c.2x=−5y2x = - 5y

Express each of the following equations in the form of ax+by+c=0ax + by + c= 0 and write the values of a, b and c.x2+y2=16\dfrac{x}{2} + \dfrac{y}{2} = \dfrac{1}{6}

Hema's age is 4 times the age of Mary. Write a linear equation in two variables to represent this information.

Express each of the following equations in the form of ax+by+c=0ax + by + c= 0 and write the values of a, b and c.3x=y3x = y

Express the following linear equation in the form of ax + by + c = 0 and indicate the value of a, b and c.x3+y4=7\dfrac{x}{3} + \dfrac{y}{4} = 7

Express the following linear equation in the form of ax + by + c = 0 and indicate the value of a, b and c.8x+5y−3=08x + 5y - 3 = 0

Sachin and Sehwag scored 137137 runs together. Express the information in the form of an equation.

Express the following linear equation in the form of ax + by + c = 0 and indicate the value of a, b and c.y7=3\dfrac{y}{7} = 3

Express the following statements as a linear equation in two variable.The cost of a pencil is Rs. 2 and one ball point pen costs Rs.Sheela pays Rs. 100 for the pencils and pens she purchased.

Express the following statements as a linear equation in two variable.The cost of a ball pen is Rs. 5 less than half the cost of a fountain pen.

Express the following statements as a linear equation in two variableThe sum of two numbers is 34.

Express the following linear equation in the form of ax + by + c = 0ax + by + c = 0 and indicate the values of a, ba, b and cc .y - 2 = 0y - 2 = 0

Express the following statements as a linear equation in two variable.Bhargavi got 10 more marks than double of the marks of Sindhu.

Express the following statements as a linear equation in two variable.Yamini and Fatima of class IX together contributed Rs. 200/- towards the Prime Minister's Relief Fund.

Express the following linear equation in the form of ax + by + c = 0 and indicate the value of a, b and c.2x = 5

Express the following linear equation in the form of ax + by + c = 0 and indicate the value of a, b and c.3x + 5y = 123x + 5y = 12

Write the linear equations in two variables using 'xx' and 'yy' to represent the following information :"The sum of the two numbers is 3030."

Determine whether (4, 6)(4, 6) is the solution of linear equation x + y = 10x + y = 10.

By using two variables, write the following statement in mathematical form:"The cost of two tables and five chairs is Rs. 2,200.2,200."

The sum of two numbers is 2828. Write this information in linear equation of two variable. (Use variable xx and yy)

Draw the graph of the line y=-\dfrac{5}{3}x+2y=-\dfrac{5}{3}x+2.

Write any one linear equation using the variables 'xx' and 'yy'.

Write any one linear equation using the variables xx and yy.

In linear equation x + y = 5, if x = 3, then find the value of y.

Write the following equation in the general form of a linear equation in two variables:7x = 3y + 237x = 3y + 23

Write L.H.S and R.H.S of the following simple equations:4z+1=84z+1=8

Find the limits within which cc must lie in order that the equation 19x+14y=c19x+14y=c may have six solutions, zero solutions being excluded.

Prove that the coefficient of x^r x^r in the expansion of (1-4x)^{- \frac{1}{2} } (1-4x)^{- \frac{1}{2} } in \dfrac {|\underline{2r}|}{(|\underline{r}|)^2} \dfrac {|\underline{2r}|}{(|\underline{r}|)^2}

Write L.H.S and R.H.S of the following simple equations:7m=147m=14

Draw the graph of y=4x-1y=4x-1.

Solve the following equation without transposing:x+5=9x+5=9

Write L.H.S and R.H.S of the following simple equations:5p+3=2p+95p+3=2p+9

Write L.H.S and R.H.S of the following simple equations:8=q+58=q+5

Write L.H.S and R.H.S of the following simple equations:2x=102x=10

Write L.H.S and R.H.S of the following simple equations:2x-3=92x-3=9

Solve the following equation.\sqrt[4]{1+5x}+\sqrt[4]{5-y}=3, 5x-y=11\sqrt[4]{1+5x}+\sqrt[4]{5-y}=3, 5x-y=11.

Romila went to a stationary stall and purchased 22 pencils and 33 erasers for Rs. 99. Her friend Sonali saw the new variety of pencils and erasers with Romila, and she also bought 44 pencils and 66 erasers of the same kind for Rs. 1818. Represent this situation algebraically.

The diameter of front and rear wheels of a tractor are 80cm80cm and 2m2m respectively. Find the number of revolutions that rear wheel will make in covering distance in which the front wheel makes 14001400 revolutions.

Solve the following linear equations for xx and yy3x+2y=53x+2y=52x-3y=72x-3y=7

The temperature of certain place fell by 2^02^0C ( a charge of -2^02^0C ) each day for a week. What is the change of temperature over the whole week?

Solve the following equations :\cfrac{x-ab}{a+b} + \cfrac{x-ac}{a+c} + \cfrac{x-bc}{b+c}\cfrac{x-ab}{a+b} + \cfrac{x-ac}{a+c} + \cfrac{x-bc}{b+c} = a+ b + c

There are 'a' trees in the village Lat. If the number of trees increases every year by 'b', then how many trees will there be after 'x' years?

After 1212 years, Kanwar shall be 33 times as old as I was 44 years ago. Find my present age.

A man is twice as old as his son. Twelve years ago, the man was thrice as old as his son. Find their present ages.

Given A= 2(a +b)h A= 2(a +b)h find aa if A=144, b=1, h=12A=144, b=1, h=12

Find the value of kk for which the system of equations kx+y=2, x+ky=1kx+y=2, x+ky=1 is undefined.

State which of the following are equation (with variable). Given reasons for your an mention which variable is used from the equation with a variable.9x+6=129x+6=12

Express the given statements in the form of linear equations in two variables.11 is added to the numerator and 44 is subtracted from the denominator, the fraction becomes 1.

State which of the following are equation (with variable). Given reasons for your an mention which variable is used from the equation with a variable.5x=05x=0

A train 100 m100 m long is moving with a velocity of 60 km /hr60 km /hr. Find the time it takes tocross the bridge one km long

Verify a-(-b)=a+ba-(-b)=a+b for the following value of aa and bb.(i). a=28,b=1a=28,b=1(ii). a=118,b=125a=118,b=125(iii). a=75,b=84a=75,b=84(iv). a=28,b=1a=28,b=1

The sum of the two digits of a 2-2-digit number is 99. When the digits are reversed, the number deceases by 99. Find the original number.

What will be the ratio of the ages after 5 years?Rina lost her weights inn the ratio 5:Her original wights was 80 kg. What is her new weight.

A works twice as fast as B. They work together and complete a job in 8 days. How long would A take to complete the job?

State which of the following are equation (with variable). Given reasons for your an mention which variable is used from the equation with a variable.17q+3=5417q+3=54

Find the values of xx and yy if:(x - 1, y + 3) = (4, 4)(x - 1, y + 3) = (4, 4).

Solve \dfrac{{2x - 3}}{5} + \dfrac{{x + 3}}{4} = \dfrac{{4x + 1}}{7}\dfrac{{2x - 3}}{5} + \dfrac{{x + 3}}{4} = \dfrac{{4x + 1}}{7}

Solve 10p\, - \,\left( {3p - 4} \right)\, = \,4\left( {p + 1} \right)\, + \,910p\, - \,\left( {3p - 4} \right)\, = \,4\left( {p + 1} \right)\, + \,9

Solve:8x+4=3(3c-1)+78x+4=3(3c-1)+7

Solve \dfrac{p}{4} + \dfrac{p}{3} = 55 - \dfrac{{p + 40}}{{50}}\dfrac{p}{4} + \dfrac{p}{3} = 55 - \dfrac{{p + 40}}{{50}}

Express the following statement as a linear equation in two variables."Seetha got 88 mark more than double of the marks of Ramya"

In the given expression -2y+3x=14-2y+3x=14, express yy is terms of xx.

If four is subtracted from 33 times the square of a number to get 66. Write this in the form of a linear equation.

Solve for xx and yy:\dfrac{6}{x -1} - \dfrac{3}{y - 2} = 1\dfrac{6}{x -1} - \dfrac{3}{y - 2} = 1\dfrac{5}{x-1} + \dfrac{1}{y-2} =2\dfrac{5}{x-1} + \dfrac{1}{y-2} =2, where x \neq 1, y \neq 2x \neq 1, y \neq 2.

Solve the equation \dfrac {x}{3}+2=2x-3y\dfrac {x}{3}+2=2x-3y

Factorize the following using appropriate identities:(i)9x^2+6xy+y^29x^2+6xy+y^2(ii)4y^2-4y+14y^2-4y+1(iii)x^2-\dfrac{y^2}{100}x^2-\dfrac{y^2}{100}

$y - 2 = 0$ is parallel to $x$ -axis.
If true enter 1 else 0.

$2y - 5 = 0$ is parallel to $x$ -axis.
If true enter 1 else 0.

$y + 6 = 0$ is parallel to $x$ -axis.
If true enter 1 else 0.

$x - 5 = 0$ is parallel to $y-$ axis. Find the value of $x$ co-ordinate where it crosses $x-axis$ .

Give the geometric representation of $y = 3$ as an equation in one variable.

Give the geometric representation of $x = -\dfrac{9}{2}$ in one variable.

A greeting card costs Rs. $3$ and a ream of photocopy paper costs Rs. $8$ . Amir paid Rs. $32$ for $x$ cards and $y$ reams of paper. Write a linear equation based on the information above.

The perimeter of $\triangle PQR$ is $30 cm$ .
Write a linear equation in two variables based on the information given above.

The length and width of a rectangle is $4x cm$ and $y cm$ respectively. If its length is $3 cm$ longer than its width, write an equation involving $x$ and $y$ .

Express the following linear equation in the form of ax + by + c = 0 and indicate the value of a, b and c.
$y = \dfrac{-3}{2} x$

Given the function $h$ such that $h(x)=x^2+kx+5$ and $h(3)=2$ . Find the value of $k$ .

Write each of the following in the form of ax + by + c = 0 and find the value of a, b and c
(i) x = - 5
(ii) y = 2
(iii) 2x = 3
(iv) 5y = -3

Express the following linear equation in the form of ax + by + c = 0 and indicate the value of a, b and c.
$2x = - 5y$

Express each of the following equations in the form of $ax + by + c= 0$ and write the values of a, b and c.
$\dfrac{x}{2} + \dfrac{y}{2} = \dfrac{1}{6}$

Express each of the following equations in the form of $ax + by + c= 0$ and write the values of a, b and c.
$3x = y$

Express the following linear equation in the form of ax + by + c = 0 and indicate the value of a, b and c.
$\dfrac{x}{3} + \dfrac{y}{4} = 7$

Express the following linear equation in the form of ax + by + c = 0 and indicate the value of a, b and c.
$8x + 5y - 3 = 0$

Sachin and Sehwag scored $137$ runs together. Express the information in the form of an equation.

Express the following linear equation in the form of ax + by + c = 0 and indicate the value of a, b and c.
$\dfrac{y}{7} = 3$

Express the following statements as a linear equation in two variable.
The cost of a pencil is Rs. 2 and one ball point pen costs Rs.Sheela pays Rs. 100 for the pencils and pens she purchased.

Express the following statements as a linear equation in two variable.
The cost of a ball pen is Rs. 5 less than half the cost of a fountain pen.

Express the following statements as a linear equation in two variable
The sum of two numbers is 34.

Express the following linear equation in the form of $ax + by + c = 0$ and indicate the values of $a, b$ and $c$ .
$y - 2 = 0$

Express the following statements as a linear equation in two variable.
Bhargavi got 10 more marks than double of the marks of Sindhu.

Express the following statements as a linear equation in two variable.
Yamini and Fatima of class IX together contributed Rs. 200/- towards the Prime Minister's Relief Fund.

Express the following linear equation in the form of ax + by + c = 0 and indicate the value of a, b and c.
2x = 5

Express the following linear equation in the form of ax + by + c = 0 and indicate the value of a, b and c.
$3x + 5y = 12$

Write the linear equations in two variables using ' $x$ ' and ' $y$ ' to represent the following information :
"The sum of the two numbers is $30$ ."

Determine whether $(4, 6)$ is the solution of linear equation $x + y = 10$ .

By using two variables, write the following statement in mathematical form:
"The cost of two tables and five chairs is Rs. $2,200.$ "

The sum of two numbers is $28$ . Write this information in linear equation of two variable. (Use variable $x$ and $y$ )

Draw the graph of the line $y=-\dfrac{5}{3}x+2$ .

Write any one linear equation using the variables ' $x$ ' and ' $y$ '.

Write any one linear equation using the variables $x$ and $y$ .

Write the following equation in the general form of a linear equation in two variables:
$7x = 3y + 23$

Write L.H.S and R.H.S of the following simple equations:
$4z+1=8$

Find the limits within which $c$ must lie in order that the equation $19x+14y=c$ may have six solutions, zero solutions being excluded.

Prove that the coefficient of $x^r$ in the expansion of $(1-4x)^{- \frac{1}{2} }$ in $\dfrac {|\underline{2r}|}{(|\underline{r}|)^2}$

Write L.H.S and R.H.S of the following simple equations:
$7m=14$

Draw the graph of $y=4x-1$ .

Solve the following equation without transposing:
$x+5=9$

Write L.H.S and R.H.S of the following simple equations:
$5p+3=2p+9$

Write L.H.S and R.H.S of the following simple equations:
$8=q+5$

Write L.H.S and R.H.S of the following simple equations:
$2x=10$

Write L.H.S and R.H.S of the following simple equations:
$2x-3=9$

Solve the following equation.
$\sqrt[4]{1+5x}+\sqrt[4]{5-y}=3, 5x-y=11$ .

Romila went to a stationary stall and purchased $2$ pencils and $3$ erasers for Rs. $9$ . Her friend Sonali saw the new variety of pencils and erasers with Romila, and she also bought $4$ pencils and $6$ erasers of the same kind for Rs. $18$ . Represent this situation algebraically.

The diameter of front and rear wheels of a tractor are $80cm$ and $2m$ respectively. Find the number of revolutions that rear wheel will make in covering distance in which the front wheel makes $1400$ revolutions.

Solve the following linear equations for $x$ and $y$
$3x+2y=5$
$2x-3y=7$

The temperature of certain place fell by $2^0$ C ( a charge of - $2^0$ C ) each day for a week. What is the change of temperature over the whole week?

Solve the following equations :
$\cfrac{x-ab}{a+b} + \cfrac{x-ac}{a+c} + \cfrac{x-bc}{b+c}$ = a+ b + c

After $12$ years, Kanwar shall be $3$ times as old as I was $4$ years ago. Find my present age.

Given $A= 2(a +b)h$ find $a$ if $A=144, b=1, h=12$

Find the value of $k$ for which the system of equations $kx+y=2, x+ky=1$ is undefined.

State which of the following are equation (with variable). Given reasons for your an mention which variable is used from the equation with a variable.
$9x+6=12$

Express the given statements in the form of linear equations in two variables.

$1$ is added to the numerator and $4$ is subtracted from the denominator, the fraction becomes 1.

State which of the following are equation (with variable). Given reasons for your an mention which variable is used from the equation with a variable.
$5x=0$

A train $100 m$ long is moving with a velocity of $60 km /hr$ . Find the time it takes to
cross the bridge one km long

Verify $a-(-b)=a+b$ for the following value of $a$ and $b$ .
(i). $a=28,b=1$
(ii). $a=118,b=125$
(iii). $a=75,b=84$
(iv). $a=28,b=1$

The sum of the two digits of a $2-$ digit number is $9$ . When the digits are reversed, the number deceases by $9$ . Find the original number.

What will be the ratio of the ages after 5 years?
Rina lost her weights inn the ratio 5:Her original wights was 80 kg. What is her new weight.

State which of the following are equation (with variable). Given reasons for your an mention which variable is used from the equation with a variable.
$17q+3=54$

Find the values of $x$ and $y$ if:
$(x - 1, y + 3) = (4, 4)$ .

Solve $\dfrac{{2x - 3}}{5} + \dfrac{{x + 3}}{4} = \dfrac{{4x + 1}}{7}$

Solve $10p\, - \,\left( {3p - 4} \right)\, = \,4\left( {p + 1} \right)\, + \,9$

Solve:
$8x+4=3(3c-1)+7$

Solve $\dfrac{p}{4} + \dfrac{p}{3} = 55 - \dfrac{{p + 40}}{{50}}$

Express the following statement as a linear equation in two variables.
"Seetha got $8$ mark more than double of the marks of Ramya"

In the given expression $-2y+3x=14$ , express $y$ is terms of $x$ .

If four is subtracted from $3$ times the square of a number to get $6$ . Write this in the form of a linear equation.

Solve for $x$ and $y$ :

$\dfrac{6}{x -1} - \dfrac{3}{y - 2} = 1$

$\dfrac{5}{x-1} + \dfrac{1}{y-2} =2$ , where $x \neq 1, y \neq 2$ .

Solve the equation $\dfrac {x}{3}+2=2x-3y$

Factorize the following using appropriate identities:
(i) $9x^2+6xy+y^2$
(ii) $4y^2-4y+1$
(iii) $x^2-\dfrac{y^2}{100}$

Solve $2x+3=5x+6$ & represent it as
i) One variable
ii) Two variable

The marks of Urmila of $4$ subjects are $52,70,66,60$ respectively. If the total average marks is equal to $60$ . How many marks should urmila obtain in fifth subject.

Find $k$ , if $kx+3y=5$ passes through $(1,5)$ .

Solve: $\dfrac{2x}{7} + \dfrac{x - 1}{5} = \dfrac{x}{3} + \dfrac{x}{7}$ .

Solve following equation:
$x^2 + \dfrac{x}{\sqrt{2}} + 1 = 0$

Find the value of $x$ in terms of $y$ $4x-4y-20=0$

Give two equations passing through $(3,10)$