Class 8 Maths Linear Equations In One Variable Worksheets

1. What is the value of y, if y-2=7

1. 8

2. 9

3. 10

4. 7

2. What is the value of a, if 6a= 18

1. 6

2. 3

3. 3

4. 4

3. What is the value of X if $X+\dfrac{3}{7}=\dfrac{17}{7}$?

1. 2

2. 3

3. 6

4. 8

4. What is the value of $A$, If $7 A-16=9$

1. $\dfrac{25}{7}$

2. $\dfrac{17}{7}$

3. $\dfrac{22}{7}$

4. $\dfrac{1}{7}$

5. What is the value of $z$, If $\dfrac{z}{1.6}=1.5$

1. $2.4$

2. $1.6$

3. $3.2$

4. $4.2$

Fill in the blanks, For Question (6-10)

6. If $\dfrac{8 z-3}{3 z}=\mathbf{2}$, so $\mathbf{z}=\ldots \ldots$

7. If $\mathrm{y}=\dfrac{1}{2}$, then $\dfrac{5}{4}=\dfrac{y}{2}=\ldots \ldots$

8. $\dfrac{1}{4} \times\left(\dfrac{1}{4}+\dfrac{1}{2}\right)=\ldots \ldots$

9. If $5-\dfrac{3}{5}=\dfrac{2}{5} y-2$,then $y=\ldots \ldots$

10. If 6 is subtracted from the product of $x$ and 5, the result is 12. The value of $x$ is.....

11. Six times a number is 48. What is the number?

12. Sum of the two numbers is 96. If one exceeds the other by 16. Find the number.

13. Simplify $5(1+2 y)=(y-3) 3$

14. A positive number is 4  times another number. If 16  is added to both the numbers, then one of the new numbers becomes thrice  the other new number. What are the numbers?

15. Aman’s age is three times his son’s age. Ten years ago he was five times his son’s age. Find their present ages

16. Two numbers are the ratio 10:6. If they differ by 36. What are the number?

17. Two equal sides of a triangle are each 4m less than three times the third side. Find the dimensions of the triangle, if its perimeter is 55m.

True / False( For question 18 and 19)

18. $\mathbf{y}=10$ for equation $\dfrac{y}{6}-\dfrac{y}{2}+\dfrac{y}{4}=\dfrac{3}{4}$

19. $\left(6-5 x^2\right)$ is a binomial?

Solve:

20. Solve the equation: $y+3=10$.

21. Solve the equation: $6=z+2$

22. Solve the equations: $\dfrac{3}{7}+x=\dfrac{17}{7}$

23. Solve the equation $6 x=12$

24. Solve the equation $\dfrac{t}{5}=10$.

25. Solve the equation $\dfrac{2 x}{3}=18$.

Answers to the Worksheet:

1. Correct option is (b)

If $y-2=7$

$\Rightarrow y=7+2$

$\Rightarrow y=9$

2. Correct option is (b)

If $6 a=18$

$\Rightarrow a=\dfrac{18}{6}$

$\Rightarrow a=3$

3. Correct option is (a)

If $X+\dfrac{3}{7}=\dfrac{17}{7}$

$X=\dfrac{17}{7}-\dfrac{3}{7}$

$\Rightarrow X=\dfrac{17-3}{7}$

$\Rightarrow X=\dfrac{14}{7}$

$\Rightarrow X=2$

4. Correct option is (a)

If $7 A-16=9$

$\Rightarrow 7 A=9+16$

$\Rightarrow 7 A=25$

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

5. Correct option is (a)

If $\dfrac{z}{1.6}=1.5$

$\Rightarrow z=1.6 \times 1.5$

$\Rightarrow z=2.4$

6. If $\dfrac{8 z-3}{3 z}=2$

So, $8 z-3=2 \times 3 z$

$\Rightarrow 8 z-3=6 z$

$\Rightarrow 8 z-6 z=3$

$\Rightarrow 2 z=3$

$\Rightarrow \mathrm{z}=\dfrac{3}{2}$

7. If $y=\dfrac{1}{2}$

Put the value on then $\dfrac{5}{4}-\dfrac{y}{2}$

$= \dfrac{5}{4}-\dfrac{\dfrac{1}{2}}{2}$

$= \dfrac{5}{4}-\dfrac{1}{2 \times 2}$

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

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

$= \dfrac{4}{4}$

= 1

8.  $\dfrac{1}{4} \times\left(\dfrac{1}{4}+\dfrac{1}{2}\right)$

$= \dfrac{1}{4} \times\left(\dfrac{1}{4}+\dfrac{2}{4}\right)$

$= \dfrac{1}{4} \times\left(\dfrac{1+2}{4}\right)$

$= \dfrac{1}{4} \times\left(\dfrac{3}{4}\right)$

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

$= \dfrac{3}{16}$

9. If $5-\dfrac{3}{5}=\dfrac{2}{5} y-2$

$\Rightarrow \dfrac{5 \times 5}{1 \times 5}-\dfrac{3}{5}=\dfrac{2}{5} y-\dfrac{2 \times 5}{1 \times 5}$

$\Rightarrow \dfrac{25}{5}-\dfrac{3}{5}=\dfrac{2}{5} y-\dfrac{10}{5}$

$\Rightarrow \dfrac{25-3}{5}=\dfrac{2 y-10}{5}$

$\Rightarrow 22=2 y-10$

$\Rightarrow 2 y=22+10$

$\Rightarrow 2 y=32$

$\Rightarrow y=\dfrac{32}{2}$

$\Rightarrow y=16$

10. If 6 is subtracted from the product of $x$ and 5, the result is 12.

$5x - 6 =12$

$5x = 12 - 6$

$5x = 6$

$x=\dfrac{6}{5}$

11. Let us know the number is $y$.

Then, $6 y=48$

$y=\dfrac{48}{6}$

$y=8$

12. According to the question,

Let the smaller number be $\mathrm{X}$.

The other number is $X+96$

According to the question,

$\Rightarrow X+X+96=16$

$\Rightarrow 2 X=96-16$

$\Rightarrow X=40$

The other number is $40+16=56$

So, the two numbers are $40$ and $56$.

13. Simplify $5(1+2 y)=(y-3) 3$

$\Rightarrow 5 \times 1+5 \times 2 y=y \times 3-3 \times 3$

$\Rightarrow 5+10 y=3 y-9$

$\Rightarrow 10 y-3 y=-9-5$

$\Rightarrow 7 y=-14$

$\Rightarrow y=\dfrac{-14}{7}$

$\Rightarrow y=-2$

14. Let the first number be $x$,then another number be $4 x$. 

After adding 16, the new numbers are $x+16$ and $4 x+16$.

According to the question,

$4 x+16=3(x+16)$

$4 x+16=3 x+48$

$4 x-3 x=48-16$

$x=32$

Other number $=4 x=4 \times 32=128$

15. Let Aman’s son’s age be x years.

Therefore, Aman’s age will be 3x years.

Ten years ago, their age was (x - 10) years and (3x - 10) years respectively.

According to the question,10 years ago,

Aman’s age = 5 × Aman’s son’s age 10 years ago

3x - 10 = 5(x - 10)

3x - 10 = 5x - 50

2x = 40

x = 20 and 3x = 60

Therefore, Aman and his son's age are 60 and 20 years respectively.

16 . Let the two numbers be $10 x$ and $6 x$.

According to the condition,

$10 x-6 x=36$

$4 x=36$

$x=9$

Two numbers are

$10 x=10 \times 9=90$

And $6 x=6 \times 9=54$

So, the two numbers are $90$ and $54$.

17. Given, two equal sides of a triangle are each 4 m less than three times the third side.

The perimeter of the triangle is 55 m.

We have to find the dimensions of the triangle.

Let the third side of the triangle be x m

Two equal sides = 3x - 4 m

We know, Perimeter = sum of all sides.

According to the question,

3x - 4 + 3x - 4 + x = 55

7x - 8 = 55

By transposing,

7x = 55 + 8

7x = 63

$x = \dfrac{63}{7}$

x = 9

Now, 3x - 4 = 3(9) - 4

= 27 - 4

= 23 m

Therefore, the dimensions of the triangle are 23 m, 23 m and 9 m.

18. False

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

$\Rightarrow \dfrac{2 \times y}{2 \times 6}-\dfrac{6 \times y}{6 \times 2}+\dfrac{3 \times y}{3 \times 4}=\dfrac{3}{4}$

$\Rightarrow \dfrac{2 y}{12}-\dfrac{6 y}{12}+\dfrac{3 y}{12}=\dfrac{3}{4}$

$\Rightarrow \dfrac{2 y-6 y+3 y}{12}=\dfrac{3}{4}$

$\Rightarrow \dfrac{-y}{12}=\dfrac{3}{4}$

$\Rightarrow -y=\dfrac{12 \times 3}{4}$

$\Rightarrow -\mathrm{y}=3 \times 3$

$\Rightarrow -\mathrm{y}=9$

$\Rightarrow \mathrm{y}=-9$

19. True, A polynomial or algebraic expression with just two terms is called a binomial.

20. Given: $y+3=10$

$\Rightarrow y+3-3=10-3$ (subtracting 3 from each side)

$\Rightarrow y=7$ (Required solution)

21. We have $6=z+2$

$\Rightarrow 6-2=z+2-2$ (subtracting 2 from each side)

$\Rightarrow 4=z$

Thus, $z=4$ is the required solution.

22. We have $\dfrac{3}{7}+x=\dfrac{17}{7}$

$\Rightarrow \quad \dfrac{3}{7}-\dfrac{3}{7}+x=\dfrac{17}{7}-\dfrac{3}{7}$

(subtracting $\dfrac{3}{7}$ from each side)

$\Rightarrow x=\dfrac{17-3}{7}$

$\Rightarrow x=\dfrac{14}{7}$

$\Rightarrow x=2$

Thus, $x=2$ is the required solution.

23. We have $6 x=12$

$\Rightarrow 6 x \div 6=12 \div 6$ (dividing each side by 6 )

$\Rightarrow x=2$

Thus, $x=2$ is the required solution.

24. Given $\dfrac{t}{5}=10$

$\Rightarrow \dfrac{t}{5} \times 5=10 \times 5$ (multiplying both sides by 5 )

$\Rightarrow t=50$

Thus, $t=50$ is the required solution.

25. We have $\dfrac{2 x}{3}=18$

$\Rightarrow \dfrac{2 x}{3} \times 3=18 \times 3$ (multiplying both sides by 3 )

$\Rightarrow 2 x=54$

$\Rightarrow 2 x \div 2=54 \div 2$ (dividing both sides by 2 )

$\Rightarrow x=27$

Thus, $x=27$ is the required solution.

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Examples of Class 8 Linear Equations in One Variable

Here are some common examples of Class 8 Linear Equations questions:

Q. A rational number has an eight times larger denominator than its numerator. The result is 3/2 if the numerator is raised by 17 and the denominator is lowered by 1. Find the rational number.

Solution:

The denominator will be (x + 8) if the numerator is x.

As given,

(x + 17)/(x + 8 - 1) = 3/2

⇒ (x + 17)/(x + 7) = 3/2

⇒ 2(x + 17) = 3(x + 7)

⇒ 2x + 34 = 3x + 21

⇒ 34 - 21 = 3x - 2x

⇒ 13 = x

The rational number is 13/21 (or x/(x + 8)).

Q.(3y + 4)/(2 – 6y) = -2/5

Solution:

(3y + 4)/(2 – 6y) = -2/5

⇒ 3y + 4 = -2/5 (2 – 6y)

⇒ 5(3y + 4) = -2(2 – 6y)

⇒ 15y + 20 = -4 + 12y

⇒ 15y – 12y = -4 – 20

⇒ 3y = -24

⇒ y = -8

Q. (7y + 4)/(y + 2) = -4/3

Solution:

(7y + 4)/(y + 2) = -4/3

⇒ 7y + 4 = -4/3 (y + 2)

⇒ 3(7y + 4) = -4(y + 2)

⇒ 21y + 12 = -4y – 8

⇒ 21y + 4y = -8 – 12

⇒ 25y = -20

⇒ y = -20/25 = -4/5

Q. The ratio between Hari and Harry's ages is 5:7. Their ages will be proportionately 3:4 in four years. What is their age right now?

Solution:

Let Harry's age be seven times that of Hari. After four years,

Age of Hari = 5 times + 4 Age of Harry =7 x + 4

As given,

(5x + 4)/(7x + 4) = ¾

⇒ 4(5x + 4) = 3(7x + 4)

⇒ 20x + 16 = 21x + 12

⇒ 21x - 20x = 16 - 12

⇒ x = 4

Hari age is 5x, or i.e., 5* 4 years = 20 years

Age of Harry = 7x = 7*4 = 28 years.

Q. 3(t – 3) = 5(2t + 1)

Solution:

3(t – 3) = 5(2t + 1)

⇒ 3t – 9 = 10t + 5

⇒ 3t – 10t = 5 + 9

⇒ -7t = 14

⇒ t = 14/-7

⇒ t = -2

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