Class 7 Maths - Simple Equations

$$12=\frac{m}{5}$$

Given $$12=\dfrac{m}{5}$$

$$\implies \dfrac{m}{5}=12$$

$$\implies m=5\times 12$$

$$\implies m=60$$

Solve:$$8x=2x+2$$

$$8x=2x+2$$

$$\Rightarrow 8x-2x=2$$

$$\Rightarrow 6x=2$$

$$\Rightarrow x=\dfrac{2}{6}=\dfrac{1}{3}$$

$$\therefore x=\dfrac{1}{3}$$

Solve: $$3y+5=44$$

The given equation is $$3y+5=44$$

So,

$$3y=44-5$$

$$3y=39$$

$$y=\dfrac{39}{3}$$

$$\therefore$$ $$y=13$$

If $$x-12=9$$, then find the value of $$x$$.

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Solve the following equation by elimination method:
$$x+8=15$$

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Solve the following equation by elimination method:
$$15+y=18$$

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Solve the following equation by elimination method:
$$\dfrac{m}{5}=3$$

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Solve the following equation by elimination method:
$$2p=p+12$$

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Solve the following equation by elimination method:
$$10-4x=26$$

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Solve:
$$2\left( {5x - 3} \right) - 3\left( {2x - 1} \right) = 9$$

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Solve the following equation by elimination method:
$$k-\dfrac{1}{2}=\dfrac{3}{4}$$

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Solve the following equation by elimination method:
$$3\left(x-7\right)=24$$

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Solve the following equation by elimination method:
$$16-5x=6$$

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Solve the following equation:
$$2k+6=0$$

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If $$15p=45$$ then $$p^2+5$$

$$15p=45$$

$$\Rightarrow$$$$p=3$$

$$\Rightarrow$$$$p^2+5$$

Put $$p=3$$

$$\Rightarrow$$$$9+5=14$$

Find $$d$$ if $$5x+8y-z+d=0$$ passes through $$(2,3,4)$$

Point is $$(2,3,4)$$

Equation is $$5x+8y-z+d=0$$

$$5(2)+8(3)-4+d=0$$

$$\Rightarrow$$$$30+d=0$$

$$\Rightarrow$$$$d=-30$$

The value of $$x$$ if $$\dfrac{x-1}{x+1}=3$$

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

$$\Rightarrow$$$$x-1=3x+3$$

$$\Rightarrow$$$$2x+4=0$$

$$\Rightarrow$$$$x=-2$$

Solve $$\sqrt { 2 x + 5 } = 7$$

$$\sqrt {2x+5}=7$$

$$\Rightarrow$$$$2x+5=7^2$$

$$\Rightarrow$$$$2x+5=49$$

$$\Rightarrow$$$$2x=49-5$$

$$\Rightarrow$$$$x=\dfrac{44}2=22$$

$$y=\dfrac 1x$$ find $$x$$ if $$y=5$$

$$y=\dfrac 1x\\\implies x=\dfrac 1y\\y=5\\\implies x=\dfrac 15$$

If $$x=3$$ find $$y$$ when $$x^2=y-7$$

$$x^2=y-7$$

put $$x=3$$

$$\Rightarrow$$$$3^2=y-7$$

$$\Rightarrow$$$$y=9+7$$

$$\Rightarrow$$$$y=16$$

A line passes through $$\left(-4,6\right)$$ and it is parallel to $$y-$$axis.Find its equation.

Any line parallel to $$y-$$axis is of the form $$x=k$$

$$\therefore\,$$ the required equation is $$x=-4$$

If $$8x+2=42$$ find $$x^2$$

$$8x+2=42$$

$$\Rightarrow$$$$8x=42-2$$

$$\Rightarrow$$$$8x=40$$

$$\Rightarrow$$$$x=5$$

$$\Rightarrow$$$$x^2=25$$

If $$2x-3=x+2$$, then find $$x$$.

Given $$2 x-3 =x +2$$

$$\implies 2 x-x=2+3$$

$$\implies x=5$$

Convert $$2km$$ by taking $$200 cm$$ as a unit

Let $$200cm =1u$$

$$2km=2000m=200000cm$$$$=1000u$$

Find a number such that when $$5$$ is a subtracted from $$5$$ times, the result is $$4$$ more then twice the number.

Let the number be $$x$$

Given

when $$5$$ is subracted from $$5$$ times the number then the result is $$4$$ more than twice the number

$$\implies 5{x}-5=4+2{x}$$

$$\implies 3 x=9\implies x=3$$

The number is $$3$$

Radhika bought $$2\ kg\ 340\ g$$ of sweets. If she distributed the sweets among $$9$$ people, then how many sweets will each get.

Weight of sweets $$2$$ kg and $$340$$g

$$=2340$$ g $$[\because 1$$ kg $$=1000$$ g $$]$$

No. of people $$=9$$

Every one's share $$=\dfrac{2340}9=260$$ g

A line passes through $$\left(4,-3\right)$$ and it is parallel to $$y-$$axis.Find its equation.

Any line parallel to $$y-$$axis is of the form $$x=k$$

$$\therefore\,$$ the required equation is $$x=4$$

The sum of $$3$$ consecutive numbers is $$123$$ find the numbers

let the numbers be $$x-1,x,x+1$$

Their sum is $$3x=123$$

$$\Rightarrow$$$$x=41$$

So the numbers are $$40,41,42$$

By what number should $$2^{ -8 }$$ be multiplied so that the product is $$4096$$

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Find $$x.$$ if $$ \dfrac{x}{2}=\dfrac{x}{3}+1 $$

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

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

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

$$\implies \dfrac{x}{6}=1$$

$$\implies x=6$$

Solve the equation $$\frac {x}{2}=-6$$

Multiplying both sides by 2, we get $$x = -12$$.

Solve the equation $$7x=14$$

Dividing both sides by 7, we get x =2.

Solve $$x + 3 = 11$$ anf find the value of $$x$$

Given, $$x+3=11$$

Subtract $$3$$ from both the sides,

$$x+3-3=11-3$$

$$\therefore x=8$$

Solve $$\dfrac{s}{7}= 4$$ for the value of $$s$$.

Given, $$\dfrac {s}{7}=4$$

Multiply $$7$$ on both the sides,

$$7 \times \dfrac {s}{7}=4\times 7$$

$$\therefore s=28$$

Solve $$y - 9 = 21$$ for the value of $$y$$.

Given, $$y-9=21$$

Add $$9$$ on both the sides,

$$y-9+9=21+9$$

$$\therefore y=30$$

Find the value of $$'x'$$

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

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

$$\Rightarrow \dfrac{x}{2}-\dfrac{x}{3}=6$$

$$\Rightarrow \dfrac{3x-2x}{6}=6$$

$$\Rightarrow x=36$$

Find the value of $$x$$ if :
$$7x - 9 = 16$$

Given: $$7x-9=16$$

$$\Rightarrow 7x=16+9=25 $$

$$\Rightarrow x=\dfrac{25}{7}$$

Simplify and solve the following equation
$$\dfrac{x-4}{2}-\dfrac{x-3}{4}=\dfrac{1}{6}$$

$$\cfrac{x - 4}{2} - \cfrac{x - 3}{4} = \cfrac{1}{6}$$

$$\cfrac{2 \left( x - 4 \right) - \left( x - 3 \right)}{4} = \cfrac{1}{6}$$

$$\cfrac{2x - 8 - x + 3}{4} = \cfrac{1}{6}$$

$$x - 5 = \cfrac{4}{6}$$

$$x = 5 + \cfrac{2}{3}$$

$$x = \cfrac{15 + 2}{3} = \cfrac{17}{3}$$

Solve for x: 2x+24=34

$$2x+24=34$$

$$= 2x=34-24$$

$$= 2x=10$$

$$= x=5$$

The value of $$x$$ if $$x-3=8$$

The given equation is $$x-3=8$$

$$x-3=8\\x=8+3\\x=11$$

Hence, if $$x-3=8$$, then the value of $$x$$ is $$11$$.

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

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

$$\dfrac{x-6}{2} = \dfrac{15+x}{3}$$

$$3x-18=30+2x$$

$$x=48$$

Solve the following equations .$$4x -\dfrac { 1 }{ 2 } =\dfrac { 9 }{ 2 } $$

$$4x-\dfrac{1}{2}=\dfrac{9}{2}$$

$$\Rightarrow 4x=\dfrac{9}{2}+\dfrac{1}{2}$$

$$\Rightarrow 4x=\dfrac{9+1}{2}=\dfrac{10}{2}=5$$

$$\therefore x=\dfrac{5}{4}$$

Solve for x:
$$3x+42x+42=0$$

3x+42x+42=0

45x+42=0

$$x=\dfrac{-42}{45}=\dfrac{-14}{15}$$

Solve :
$$3x+11=6x-13$$

$$3x+11=6x-13$$

$$3x=24$$

$$x=8$$

Solve for $$x:$$ $$2x+36=48$$

$$2x+36=48$$

$$\Rightarrow$$$$2x=48-36$$

$$\Rightarrow$$$$2x=12$$

$$\Rightarrow$$$$x=6$$

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

Sum of $$-2a+b-5c$$ and $$3a-2b+c$$ = $$-2a+b-5c+3a-2b+c$$

= $$a-b-4c$$

Subtracting $$3a-2b+4c$$ from $$a-b-4c$$

= $$a-b-4c-\left(3a-2b+4c\right)$$

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

$$=-2a+b-8c$$

Solve:
$$6x-1=3x+8$$

$$6x-1=3x+8$$

$$\Rightarrow 6x-3x=8+1$$

$$\Rightarrow 3x=9$$

$$\Rightarrow x=\dfrac{9}{3}=3$$

$$\therefore x=3$$

Construct 3 equations starting with $$x = 2$$.

$${\textbf{Step -1: Form equations.}}$$

$${\text{Given }}x = 2,{\text{ to construct equations starting with this value of }}x.$$

$${\text{Add}}$$ $${\mathrm{2}}$$ $${\text{on both sides}}{\text{.}}$$

$$\Rightarrow x + 2 = 2 + 2$$

$$\Rightarrow x + 2 = 4....(1)$$

$${\text{Subtract}}$$ $${\mathrm{2}}$$ $${\text{on both sides}}{\text{.}}$$

$$\Rightarrow x - 2 = 2 - 2$$

$$\Rightarrow x - 2 = 0$$

$${\text{Multiply}}$$ $${\mathrm{2}}$$ $${\text{on both sides}}{\text{.}}$$

$$\Rightarrow x \times 2 = 2 \times 2$$

$$\Rightarrow 2x = 4$$

$${\textbf{Hence, }}$$$${\mathbf{3}}$$$${\textbf{ equations starting with }}$$ $${\mathbf{x = 2}}$$ $${\textbf{ are }}$$ $${\mathbf{x + 2 = 4,x - 2 = 0,2x = 4.}}$$

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

Given,

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

or, $$x=\dfrac{3}{2}+2$$

or, $$x=\dfrac{3+4}{2}$$

or, $$x=\dfrac{7}{2}$$.

A number is added to 3 ,5 times to give 88 Find the number

Let the number be $$x$$

It is added to $$3$$, $$ 5$$ times so $$3+x+x+x+x+x=88\\5x+3=88\\5x=88-3\\5x=85\\x=\dfrac{85}{5}\\x=17$$

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

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

$$\Rightarrow \dfrac{{x}^{2}-\left({x}^{2}+3x-2\right)}{5x+1}=6$$

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

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

$$\Rightarrow -3x-30x=6-2$$

$$\Rightarrow -33x=4$$

$$\therefore x=\dfrac{-4}{33}$$

Find the value of $$a$$ and the value of $$b$$ when $$x^{2} - 16x + a = (x + b)^{2} $$.

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Make $$x$$ the subject of the formula
$$A - x = \dfrac{xr}{t} $$

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$$\textbf{Fill in the blanks to make the statements true.}$$
Any term of an equation may be transposed from on side of the equation to the other side of the equation by changing the ________ of the term.

Any term of an equation may be transposed from on side of the equation to the other side of the equation by changing the $$\text{sign}$$ of the term.

What value of $$n$$ satisfies the equation $$\displaystyle \dfrac { 7 }{ 8 } \left( n-6 \right) =\dfrac { 21 }{ 2 } $$ ?

Given,

$$\dfrac{7}{8}(n-6)=\dfrac{21}{2}$$

$$\Rightarrow$$ $$\dfrac{7n}{8}-\dfrac{21}{4}=\dfrac{21}{2}$$

Multiplying both sides by $$8$$ we get,

$$\Rightarrow$$ $$7n-42=84$$

$$\Rightarrow$$ $$7n=84+42$$

$$\Rightarrow$$ $$7n=126$$

$$\Rightarrow$$ $$n=\dfrac{126}{7}$$

$$\Rightarrow$$ $$n=18$$

$$\therefore$$ The value of $$n$$ is $$18.$$

Solve $$10x = 30$$ for the value of $$x$$.

Given, $$10x=30$$

Divide $$10$$ on both the sides,

$$\Rightarrow \dfrac {10x}{10}=\dfrac {30}{10}$$

$$\Rightarrow x=3$$

Solve the following equations
$$2 + y = 7$$

$$2 + y = 7$$
So we get
$$y = 7 - 2 = 5$$.

A number subtracted by $$30$$ gives $$14$$ subtracted by $$3$$ times the number. Find the number.

Let the number be $$x$$.

As per the question, we have

$$30-x=3x-14$$

Transpose $$x$$ on one side,

$$30+14=3x+x$$

$$4x=44$$

$$x=11$$

Therefore, the number is $$11$$.

Solve the equation: $$3y+39=8$$

Given equation : $$3y+39=8$$
$$3y=8-39$$ (Transporting $$39$$ to RHS)
$$3y=-31$$
$$y=\cfrac{-31}{3}$$ (Transporting $$3$$ to RHS)
$$\therefore$$ The solution of $$3y+39=8$$ is $$y=\cfrac{-31}{3}$$

Leela is Radha's younger sister. Leela is $$4$$ years younger than Radha. Can you write Leela's age in terms of Radha's age? Take Radha's age to be $$x$$ years.

$$\textbf{Step 1: Writing Leela's age in terms of Radha's}$$
$$\text{It is given that, Leela is 4 years younger than Radha}$$

$$\text{If we assume Radha's age to be }x \text{ years.}$$

$$\text{Then Leela's age is found to be }(x - 4) \text{ years.}$$

$$\textbf{Hence, Leela's age in terms of Radha's age is }$$$$(x-4)$$ $$\textbf{ if Radha's age is }$$$$x.$$

The sum of three number isIf the second number is subtracted from the sum of first and third number then we get 5 an if twice of first number is added to second and if the third number is subtracted from the sum we get 4.

Let the three numbers be $$a,b,c$$

The sum of three numbers is $$15$$

$$\implies a+b+c=15\cdots(1)$$

The second number is subtracted from the sum of first and third number to geet $$5$$

$$\implies a-b+c=5\cdots(2)$$

twice of first number is added to second and if the third number is subtracted from sum we get $$4$$

$$\implies 2a+b-c=4\cdots(3)$$

$$(1)-(2)$$

$$\implies 2b=10\\b=5$$

put it in $$(1)$$

$$\implies a+c=10\\c=10-a$$

put it in $$(3)$$

$$2a+5-10+a=4\\3a-5=4\\3a=9\\a=3\\\implies c=10- 3\\c=7\\a=3;b=5;c=7$$

The sum of the digits of a 2-digit number isWhen 18 is added to the number, its digits are reversed. Find the number

Let the ten's and unit digit be x and $$8/x$$ respectively.

Then, $$10x + \dfrac{8}{x} + 18 = 10 \times \dfrac{8}{x} + x$$

$$\Rightarrow 10x^2 + 8 + 18x = \dfrac{80}{x} + x$$

$$\Rightarrow \dfrac{10x^2 + 18x + 8}{x} = \dfrac{80 + x^2}{x}$$

$$\Rightarrow 10x^2 + 18x + 8 = 80 + x^2$$

$$\Rightarrow 10x^2 - x^2 + 18 x + 8 - 80 = 0$$

$$\Rightarrow 9x^2 + 18x - 72 = 0$$

$$\Rightarrow x^2 + 2x - 8 = 0$$

$$\Rightarrow x^2 + 4x - 2x - 8 = 0$$

$$\Rightarrow x (x + 4) -2(x + 4) = 0$$

$$\Rightarrow x = -4, 2$$ $$\therefore x = 2 \sin \theta x > 0$$

$$\Rightarrow 8/x = 8/2 = 4$$.

$$\therefore $$ The numbers is $$24$$.

A truck runs $$8$$ km using $$1$$ litre of petrol. How much distance will it cover using $$10\dfrac { 2 } { 3 }$$ litre of petrol.

The question can be solved using unitary method.

Distance covered using $$1\ \text{L}$$ of petrol = $$8\ \text{km}$$

Distance covered using $$10\dfrac{2}{3}\ \text{L}$$ of petrol $$= 8\times 10\dfrac{2}{3}\ \text{km}= 8\times \dfrac{32}{3}\ \text{km}=\dfrac{256}{3}\ \text{km}$$

$$=85\dfrac{1}{3}\ \text{km}$$

Find the value of $$x:-$$
$$\displaystyle \frac{{x + 1}}{{x + 2}} = \frac{8}{2}$$

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Identify the operations ( addition, subtraction, division, multiplication) in forming the following expressions and tell how the expressions have been formed.
$$2y+17,2y-17$$

$$2y + 17, 2y – 17$$

Multiplication and addition

y is multiplied with $$2$$, and $$17$$ is added to the result

y is multiplied with $$2$$, and $$17$$ is subtracted from the result

Identify the operations ( addition, subtraction, division, multiplication) in forming the following expressions and tell how the expressions have been formed.
$$7m, -7m+3, -7m-3$$

$$7m \rightarrow \ multiplication$$

$$ -7m + 3 \rightarrow \ multiplication \ and \ then \ addition$$

$$ -7m – 3 \rightarrow \ multiplication \ and \ then \ subtraction$$

The present age of woman is $$3$$ years more than three times the age of her daughter. Three years hence, the woman age will be $$10$$ years more than twice the age of her daughter. Find their present ages.

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Identify the operations ( addition, subtraction, division, multiplication) in forming the following expressions and tell how the expressions have been formed.
$$z+1, z-1, y+17,y-17$$

Addition as $$1$$ is added to z

Subtraction as $$1$$ is subtracted from z

Addition as $$17$$ is added to y

Subtraction as $$17$$ is subtracted from y

If $$29$$ workers were paid $$Rs\ 3335$$ as wages, then how many workers can be paid with $$Rs4830$$ at the same rate?

Amount paid to $$29$$ workers $$= Rs. 3335$$

Amount paid to one worker $$= \cfrac{3335}{29} = Rs. 115$$

Therefore,

No. of workers can be paid with $$Rs. 4830 = \cfrac{4830}{115} = 42$$

Hence $$42$$ workers can be paid with $$Rs 4830$ at the same rate.

When Raju multiplies a certain number by $$17$$ and adds $$4$$ to the product, he gets $$225$$. Find that number.

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The cost of $$2\dfrac{3}{4}$$ metres of cloth is Rs. $$583$$. Find the cost of the cloth per meter.

Cost of $$2\dfrac{3}{4}$$ metres of cloth $$=$$Rs.$$583$$

Cost of $$1$$ metre of cloth,

$$=\dfrac{1\times 583}{2\dfrac{3}{4}}$$

$$=\dfrac{583}{\dfrac{11}{4}}$$

$$=583\times\dfrac{4}{11}=53\times 4=212$$

Hence Cost of $$1$$ metre of cloth is Rs.$$212$$

Solve for x: 2x+54=4x-46

2x+54=4x-46

= 54+46=4x-2x

= 100=2x

= x=50

The cost of 5 kilogram of sugar is 72 rupees. Find out the cost of 15 kilogram of sugar

Cost of $$5$$kg of sugar$$=$$Rs.$$72$$

Cost of $$1$$kg of sugar$$=$$Rs.$$\dfrac{72}{5}$$

Cost of $$15$$kg of sugar$$=$$Rs.$$\dfrac{72}{5}\times 15=72\times 3=$$Rs.$$216$$

Some plants have to be planted in a plantation program. If four plants are planted into each row, then find the number of plants in terms of rows.

Let the number of rows be $$x$$ .

And number of plants in each row$$=4$$

Then,number of plants $$=4x$$

Transform $$x-2y+3=0$$ to the parallel axes through the point $$\left(2,-5\right)$$

Let the co-ordinates $$\left(x,y\right)$$ of any point on the given line change to $$\left(X,Y\right)$$ on shifting the origin to the point $$\left(2,-5\right)$$, the new axes remaining parallel to the original axes.Then, we get

$$x=X+2$$ and $$y=Y-5$$

Substituting for $$x=X+2$$ and $$y=Y-5$$ we get

$$\left(X+2\right)-2\left(Y-5\right)+3=0$$

$$\Rightarrow\,X+2-2Y+10+3=0$$

$$\Rightarrow\,X-2Y+15=0$$ is the required transformation.

Evaluate:$$3$$ times $$4$$ is divided by $$2$$ times $$y$$

$$3\times 4=12$$

$$2\times y=2y$$

Given:$$3\times 4\div 2\times y=\dfrac{12}{2y}=\dfrac{6}{y}$$

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

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

$$8(x+1) = 3(2x+3)$$

$$8x+8=6x+9$$

$$2x=1$$

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

A train travelling at a uniform speed $$360$$ km would have taken $$48$$ minutes less to travel the same distance if its speed were $$5$$ km/h more. Find the original speed of the train.

Let the original speed of train $$= x \ km/hr$$

So, the new increased speed of train $$ = (x + 5) km/hr$$

Time taken by train in covering $$360 km$$ with original speed $$= \dfrac{Distance}{Speed} = \dfrac{360}{x} hr$$

Time taken by train in covering $$360 km$$ with new speed $$ = \dfrac{360}{x + 5} hr$$

According to the question,

$$\dfrac{360}{x} - \dfrac{360}{x + 5} = \dfrac{48}{60} hr$$

$$\Rightarrow 360 \left[\dfrac{1}{x} - \dfrac{1}{(x + 5)} \right] = \dfrac{4}{5}$$

$$\Rightarrow 360 \left[\dfrac{x + 5 - x}{x (x + 5)} \right] = \dfrac{4}{5}$$

$$\Rightarrow \dfrac{90 [5]}{x^2 + 5x } = \dfrac{1}{5}$$

$$\Rightarrow x^2 + 5x = 90 \times 25$$

$$\Rightarrow x^2 + 5x - 90 \times 25 = 0$$

$$\Rightarrow x^2 + 50 x - 45x - 90 \times 25 = 0$$

$$\Rightarrow x (x + 50) - 45[x + 10 \times 5] = 0$$

$$\Rightarrow (x + 50) (x - 45) = 0$$

$$\Rightarrow x + 50 = 0$$ or $$x - 45 = 0$$

$$\Rightarrow x = -50 $$ or $$ x = 45$$

$$x = -50 $$ is rejected as it is negative.

Hence, the original speed of train is $$45 km/hr$$

Form the mathematical expression:
If the breadth is subtracted from twice the length, the answer is $$27$$.

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Solve
$$2x+3x=36$$

Given $$2 x+3 x=36$$

$$\implies 5 x=36$$

$$\implies x=\dfrac{36}{5}=7.2$$

Find the value of $$x$$
$$9\dfrac { 1 }{ 3 } +x=19$$

$$9\dfrac{1}{3}+x=19$$

$$\Rightarrow \dfrac{28}{3}+x=19$$

$$\Rightarrow x=19-\dfrac{28}{3}=\dfrac{57-28}{3}=\dfrac{29}{3}$$

$$\therefore x=\dfrac{29}{3}$$

A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.

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From the choices given below, choose the equation whose graph is given in Fig.
(i) $$y = x$$
(ii) $$x + y = 0$$
(iii) $$y = 2x$$
(iv) $$2 + 3y = 7x$$.

If we put the values $$(-1, 1)$$ and $$(1, -1)$$ in equation $$x + y = 0$$ it satisfies the equation.

for $$(-1,1)$$ $$\Rightarrow$$ Let $$x=-1$$ then $$y=1$$

for $$(1,-1)$$ $$\Rightarrow$$ Let $$x=1$$ then $$y=-1$$

Other equations don't satisfy the given coordinates.

Mohini is $$x$$ years old. Express the following in algebraic form:
Four times Mohini's age $$3$$ years ago.

Mohini is $$x$$ years old

Mohini's age 3 years ago $$=x-3$$

Four times Mohini's age 3 years ago $$=4(x-3)$$

Raju's father's age is $$2$$ years more than $$3$$ times Raju's age. If Raju's present age is $$y$$ years, then what is his father's age?

Raju's present age is $$y$$ years
Raju's father's age $$=$$ 2 years more than 3 times Raju's age $$=2+3y$$

Find $$x$$ ? $$10=\frac{x}{1.3}$$

Given $$10=\dfrac{x}{1.3}$$

$$\implies 10=\dfrac{x}{\frac{13}{10}}$$

$$\implies \dfrac{10 x}{13}=10$$

$$\implies x=\dfrac{10\times 13}{10}$$

$$\implies x=13$$

Draw the graph of the equation $$2x + 3y = 12$$. From the graph, find the coordinates of the point:
(i) whose y-coordinates is $$3$$.
(ii) whose x-coordinate is $$-3$$.

$$2 x+3 y=12\cdots(1)$$

To draw the graph ,we need at least two solutions of the equation

Put $$x=0$$ in $$(1)$$

$$2(0)+3 y=12$$

$$\implies 3 y=12$$

$$\implies y=4$$

So $$\left(0,4\right)$$ is a solution of the equation

Put $$y=0$$ in $$(1)$$

$$2 x+3(0)=12$$

$$\implies 2 x= 12$$

$$\implies x=6$$

So $$(6,0)$$ is a solution of the equation

With these two equations we can draw graph of $$2x+3y=12$$

from the graph

$$\text{(i)} $$ when $$y=3$$ then $$x=1.5$$

So, point is $$(1.5,3)$$

$$\text{(ii)} $$ when $$x=-3$$ then $$y=6$$

So, point is $$(-3,6)$$

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From the choices given below, choose the equation whose graph is given in Fig.
(i) $$y = x + 2$$
(ii) $$y = x - 2$$
(iii) $$y = -x +2$$
(iv) $$x + 2y = 6$$.

If we put the values $$(2, 0)$$ and $$(-1, 3)$$ in equation $$y = -x + 2$$ it satisfies the equation.

for $$(2,0)$$ $$\Rightarrow$$ Let $$x=2$$ then $$y=0$$

for $$(-1,3)$$ $$\Rightarrow$$ Let $$x=-1$$ then $$y=3$$

Other equations don't satisfy the given coordinates.

$$\textbf{choose a letter x,y,z,p etc...., wherever necessary, for the unknown (variable) and}$$ $$\textbf{write the corresponding expressions:choose a letter x,y,z,p etc...., wherever}$$ $$\textbf{necessary, for the unknown (variable) and write the corresponding expressions:}$$
Write an equation for which $$0$$ is a solution.

There can be many equations having solution $$=0$$
The following equation has a solution $$0$$,
$$x+5=5,$$
This equation is satisfied at $$x=0$$ as $$0+5 =5$$
$$\textbf{Therefore the required equation is x+5=5}$$

Write two equations for which $$2$$ is the solution.

There can be many equations whose solution is $$2$$.
Below are two examples,
$$x+2=4$$ and $$x-2=0$$,
as, $$2+2=4$$ and $$2-2=0$$. ($$2$$ is the solution of both the equations)

$$\textbf{Therefore the required equations are} \ (x+2=4) \textbf{ and } (x-2=0).$$

If $$$$ is an operation such that for integers $$a$$ and $$b$$ we have $$a^{} b=a \times b+(a \times a+b \times b)$$ then find
(i) (-3)$$(-5)$$
(ii) (-6)$$ 2$$

Given, $$a^{*} b=a \times b+(a \times a+b \times b)$$

So.

$$(i)$$

we put $$a=(-3)$$ and $$b=(-5)$$

$$(-3) *(-5)=(-3) \times(-5)+[(-3) \times(-3)+(-5) \times(-5)]$$

$$=15+(9+25)=15+34=49$$

$$(ii)$$

Similarly, $$a=-6, b=2$$

$$(-6) * 2=(-6) \times 2+[(-6) \times(-6)+2 \times 2]$$

$$=-12+(36+4)$$

$$=-12+40=28$$

Write an equation whose solution is not a whole number.

Let us consider the equation $$4x=1$$.
Clearly if we multiply $$4$$ to $$\frac{1}{4}$$ we get $$4 \times \frac{1}{4}=1$$ which matches the RHS.
Hence $$x=\frac{1}{4}$$ is the solution of the above equation,
But we know that $$\frac{1}{4}$$ is not a whole number.
$$\textbf{Hence, the required equation is} \ 4 x=1$$

If the temperature of a liquid can be measured in Kelvin units as $$x$$ K or in Fahrenheit units as y $$F$$, the relation between the two systems of measurement of temperature is given by the linear equation
(i) Find the temperature of the liquid in Fahrenheit if the temperature of the liquid is $$313$$ $$K.$$
(ii) If the temperature is $$158$$ $$F$$, then find the temperature in Kelvin.

The linear equation that converts Kelvin (x) to Fahrenheit (y) is given by the relation:

$$y = \frac{9}{5}\left ( x-273 \right ) + 32$$

(i) When the temperature of the liquid is x = $$313^{\circ }$$ K

$$Y =\frac{9}{5} \left ( 313-273 \right ) + 32 = \frac{9}{5}\times 40 + 32 = 72^{\circ } + 32^{\circ } = 104^{\circ }$$ F

(ii) When the temperature of the liquid is y = $$158^{\circ }$$ F

$$158 = \frac{9}{5}\left ( x-273 \right ) + 32$$ $$\Rightarrow \frac{9}{5} \left ( x-273 \right ) = 158 - 32$$

$$\Rightarrow x - 273 = 126 \times \frac{5}{9} = 70$$

$$\Rightarrow x - 273 = 70 = 273 + 70 = 343 K$$

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

From the given equation- $$\dfrac{2x-3}{4x+5}=\dfrac{1}{3}$$
$$\Rightarrow 3(3x-3)=4x+5$$
$$\Rightarrow 6x-9=4x+5$$
$$\Rightarrow 6x-4x=9+5$$
$$\Rightarrow 2x=14$$
$$\Rightarrow x=7$$.

Baichung's father is $$26$$ years younger than Baichung's grandfather and $$29$$ years older than Baichung. The sum of the ages of all the three is $$135$$ years. What is the age of each one of them?

1304555_1581660_ans_7400e31db51949b3b476449d4a0ceada.jpg

Huma, Hubna and Seema received a total of Rs. 2,016 as monthly allowance from their mother such that Seema gets $$\frac{1}{2}$$ of what Huma gets and Hubna gets $$2\frac{1}{3}$$ times Seema's share. How much money do the three sister get individually?

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If you subtract $$\dfrac{1}{2}$$ from a number and if the obtained number is multiplied by the by $$\dfrac{1}{2}$$, you get $$\dfrac{1}{8}$$ What is the number. ?

Let be original number be $$x$$

According to the given condition

$$(x-\dfrac{1}{2})(\dfrac{1}{2})=\dfrac{1}{8}$$

$$(x-\dfrac{1}{2})=\dfrac{1}{4}$$

$$x=\dfrac{1}{2}+\dfrac{1}{4}$$

$$=\dfrac{2+4}{8}$$

$$=\dfrac{3}{4}$$

If 105 kg corn is enough to feed 9 houses in a day, how much corn will be needed to feed 12 horses ?

1323263_1588420_ans_2229bc500b4a4c90afa34b087817d436.jpg

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

1311936_1589557_ans_5ff4e4bf9de7402686abee575caad170.jpg

$$2(x-5)=10$$.

1323308_1588493_ans_5aa06d0060504e77b0d24ebd63e6b36d.jpg

The length of the minute hand of a clock is $$14\ cm$$. Find the area swept by the minute hand is $$5$$ minutes.

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Simple Equations - Class 7 Maths - Extra Questions