# RD Sharma Solutions Class 7 Linear Equations In One Variable Exercise 8.2

## RD Sharma Solutions Class 7 Chapter 8 Exercise 8.2

### RD Sharma Class 7 Solutions Chapter 8 Ex 8.2 PDF Free Download

#### Exercise 8.2

Q1. x – 3 = 5

SOLUTION :

x – 3 = 5

Adding 3 to both sides, we get

x – 3 + 3 = 5 + 3

x = 8

Verification :

Substituting x = 8 in LHS, we get

LHS = x – 3 and RHS = 5

LHS = 8 – 3 = 5 and RHS = 5

LHS = RHS

Hence, verified.

Q2. x + 9 = 13

SOLUTION :

x + 9 = 13

Subtracting 9 from both sides, we get

=> x + 9 – 9 = 13 – 9

=> x = 4

Verification :

Substituting x = 4 on LHS, we get

LHS = 4 + 9 = 13 = RHS

LHS = RHS

Hence, verified.

Q3. x – $\frac{3}{5}$ = $\frac{7}{5}$

SOLUTION :

x – $\frac{3}{5}$ = $\frac{7}{5}$

Adding 3/5 to both sides, we get

=> x – $\frac{3}{5}$  + $\frac{3}{5}$  = $\frac{7}{5}$  + $\frac{3}{5}$

=> x = $\frac{7}{5}$  + $\frac{3}{5}$

=> x = $\frac{10}{5}$

=> x=2

Verification :

Substituting x = 2 in LHS, we get

LHS = 2 – $\frac{3}{5}$ = 10 – $\frac{3}{5}$ = $\frac{7}{5}$. and RHS = $\frac{7}{5}$

LHS = RHS

Hence, verified.

Q4. 3x = 0

SOLUTION :

3x = 0

Dividing both sides by 3, we get

$\frac{3x}{3}$ = $\frac{0}{3}$

x = 0

Verification :

Substituting x = 0 in LHS = 3x, we get LHS = 3 $\times$  0 = 0 and RHS = 0

LHS = RHS

Hence, verified.

Q5 . $\frac{x}{2}$ = 0

SOLUTION :

$\frac{x}{2}$ = 0

Multiplying both sides by 2, we get

=> $\frac{x}{2}$ $\times$ 2 = 0$\times$2

=> x = 0

Verification :

Substituting x= 0 in LHS, we get

LHS = $\frac{0}{2}$  = 0 and RHS = 0

LHS = 0 and RHS = 0

LHS = RHS

Hence, verified.

Q6 . x – $\frac{1}{3}$ = $\frac{2}{3}$

SOLUTION :

x – $\frac{1}{3}$ = $\frac{2}{3}$

Adding $\frac{1}{3}$ to both sides, we get

x – $\frac{1}{3}$ + $\frac{1}{3}$ = $\frac{2}{3}$ + $\frac{1}{3}$

=> x = $\frac{2}{3}$ + $\frac{1}{3}$

=> x = $\frac{3}{3}$

x = 1

Verification :

Substituting x= 1 in LHS, we get

LHS = 1 – $\frac{1}{3}$ = $\frac{3-1}{3}$ =  $\frac{2}{3}$, and RHS = $\frac{2}{3}$

LHS = RHS

Hence, verified.

Q7. x + $\frac{1}{2}$ = $\frac{7}{2}$

SOLUTION :

x + $\frac{1}{2}$ = $\frac{7}{2}$

Subtracting $\frac{1}{2}$  from both sides, we get

x + $\frac{1}{2}$  – $\frac{1}{2}$ = $\frac{7}{2}$$\frac{1}{2}$

x = $\frac{7}{2}$$\frac{1}{2}$ =$\frac{6}{2}$

x = 3

Verification :

Substituting x = 3 in LHS, we get LHS = 3 + $\frac{1}{2}$ = $\frac{6+1}{2}$ =72. and RHS = 72

LHS = RHS

Hence, verified.

Q8. 10 – y = 6

SOLUTION :

10 – y = 6

Subtracting 10 from both sides, we get

10 – y – 10 = 6 – 10

-y = -4

Multiplying both sides by -1, we get

-y  $\times$ -1 = -4  $\times$-1

y = 4

Verification :

Substituting y = 4 in LHS, we get

LHS = 10 – y = 10 – 4 = 6 and RHS = 6

LHS = RHS

Hence, verified.

Q9. 7 + 4y = -5

SOLUTION :

7 + 4y = -5

Subtracting 7 from both sides, we get

7 + 4y – 7 = -5 -7

4y = -12

Dividing both sides by 4, we get

y = -12/ 4

y = -3

Verification :

Substituting y = -3 in LHS, we get

LHS = 7 + 4y = 7 + 4(-3) = 7 – 12 = -5, and RHS = -5

LHS = RHS

Hence, verified.

Q10.  $\frac{4}{5}$ – x = $\frac{3}{5}$

SOLUTION :

$\frac{4}{5}$ – x = $\frac{3}{5}$

Subtracting $\frac{4}{5}$  from both sides, we get

$\frac{4}{5}$ – x – $\frac{4}{5}$ = $\frac{3}{5}$ –  $\frac{4}{5}$

– x = $\frac{3}{5}$$\frac{4}{5}$

– x = –$\frac{1}{5}$

Multiplying both sides by -1, we get

-x $\times$ (-1) = –$\frac{1}{5}$ $\times$ (-1)

x = $\frac{1}{5}$

Verification :

Substituting x= $\frac{1}{5}$ in LHS, we get

LHS = $\frac{4}{5}$  – $\frac{1}{5}$ = $\frac{4-1}{5}$ = $\frac{3}{5}$ , and RHS = $\frac{3}{5}$

LHS = RHS

Hence, verified.

Q11. 2y – $\frac{1}{2}$ = –$\frac{1}{3}$

SOLUTION :

2y –$\frac{1}{2}$ = –$\frac{1}{3}$

Adding $\frac{1}{2}$  to both sides, we get

2y – $\frac{1}{2}$ + $\frac{1}{2}$ =- $\frac{1}{3}$  + $\frac{1}{2}$

2y = $\frac{-2+3}{6}$

2y = $\frac{1}{6}$

Dividing both sides by 2, we get

2y/2 = 16/2

y =$\frac{1}{12}$

Verification :

Substituting y = $\frac{1}{12}$  in LHS, we get

LHS = 2$\frac{1}{12}$  – $\frac{1}{2}$  = $\frac{1}{6}$  – $\frac{1}{2}$  = $\frac{1-3}{6}$ = $\frac{-2}{6}$  = –$\frac{1}{3}$, and RHS = –$\frac{1}{3}$

LHS = RHS

Hence, verified.

Q12. 14 = $\frac{7x}{10}$ – 8

SOLUTION :

14 = $\frac{7x}{10}$ – 8

Adding 8 to both sides, we get

14 + 8 = $\frac{7x}{10}$– 8 + 8

22 = $\frac{7x}{10}$

Multiplying both sides by 10, we get

22 $\times$  10 = $\frac{7x}{10}$ $\times$  10

220 = 7x

Dividing both sides by 7, we get

$\frac{220}{7}$ = $\frac{7x}{7}$

x =  $\frac{220}{7}$

Verification:

Substituting x =  $\frac{220}{7}$  in RHS, we get

LHS = 14. and RHS = $\frac {\frac {220}{7}}{\frac {7}{10}}$10 – 8 = $\frac{220}{10}$   – 8 = 22 – 8 = 14

LHS = RHS

Hence, verified.

Q13. 3(x + 2) = 15

SOLUTION :

3 (x + 2) = 15

Dividing both sides by 3, we get

$\frac{3(x+2)}{3}$  = $\frac{15}{3}$

(x+2) =  5

Subtracting 2 from both sides, we get

x + 2 – 2 = 5 – 2

x = 3

Verification :

Substituting x = 3 in LHS, we get

LHS = 3 (x + 2) = 3 (3+2) = 3(5) = 15, and RHS = 15

LHS = RHS

Hence, verified.

Q14. $\frac{x}{4}$ = $\frac{7}{8}$

SOLUTION :

$\frac{x}{4}$ = $\frac{7}{8}$

Multiplying both sides by 4, we get

$\frac{x}{4}$ $\times$  4 = $\frac{7}{8}$ $\times$  4

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

Verification :

Substituting x = $\frac{7}{2}$ in LHS, we get

LHS = $\frac{7}{2}$/4 = $\frac{7}{8}$, and RHS = $\frac{7}{8}$

LHS = RHS

Hence, verified.

Q15. $\frac{1}{3}$ -2x = 0

SOLUTION :

$\frac{1}{3}$  – 2x = 0

Subtracting $\frac{1}{3}$  from both sides, we get

$\frac{1}{3}$  – 2x – $\frac{1}{3}$  = 0 – $\frac{1}{3}$

-2x = –$\frac{1}{3}$

Multiplying both sides by -1, we get

-2x $\times$   (-1) = –$\frac{1}{3}$ $\times$  (-1)

2x = $\frac{1}{3}$

Dividing both sides by 2, we get $\frac{2x}{2}$ = $\frac{1}{3}$/2

x = $\frac{1}{6}$

Verification :

Substituting x = $\frac{1}{6}$ in LHS, we get

LHS = $\frac{1}{3}$ – 2 $\times$   $\frac{1}{6}$ = $\frac{1}{3}$ –  $\frac{1}{3}$ = 0 , and RHS = 0

LHS = RHS

Hence, verified.

Q16. 3(x + 6) = 24

SOLUTION :

3(x + 6) = 24

Dividing both sides by 3, we get

$\frac{3 (x + 6)}{3}$ = $\frac{24}{3}$

(x+ 6) =  8

Subtracting 6 from both sides, we get

x + 6 – 6 = 8 – 6

x = 2

Verification :

Substituting x = 2 in LHS , we get

LHS = 3 (x + 6) = 3 (2 + 6)  = 24 , and RHS = 24

LHS = RHS

Hence, verified.

Q17. 3(x + 2)-2(x – 1) = 7

SOLUTION :

3(x + 2) – 2(x – 1) = 7

On expanding the brackets, we get

3$\times$ x + 3 $\times$  2 – 2 $\times$ x + 2 $\times$ 1 = 7

3x + 6 – 2x + 2 = 7

3x – 2x + 6 + 2 = 7

x + 8 = 7

Subtracting 8 from both sides, we get

x + 8 – 8 = 7 – 8

x = -1

Verification :

Substituting x = -1 in LHS, we get

LHS = 3 (x + 2) -2(x -1) = 3 (-1 + 2) -2(-1-1) = (3$\times$1) – (2$\times$-2) = 3 + 4 = 7, and RHS = 7

LHS = RHS

Hence, verified .

Q18. 8(2x – 5)-6(3x – 7) = 1

SOLUTION :

8(2x – 5) – 6(3x – 7) = 1

On expanding the brackets, we get (8$\times$2x) – (8 $\times$ 5) – (6 $\times$ 3x) + (-6) $\times$ (-7) = 1

16x – 40 – 18x + 42 = 1

16x – 18x + 42 – 40 = 1

-2x + 2 = 1

Subtracting 2 from both sides, we get

-2x+ 2 – 2 = 1 -2

-2x = -1

Multiplying both sides by -1, we get

-2x $\times$ (-1) = -1$\times$ (-1)

2x = 1

Dividing both sides by 2, we get

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

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

Verification :

Substituting x = $\frac{1}{2}$ in LHS, we get

= 8(2 $\times$ $\frac{1}{2}$ – 5) – 6(3 $\times$ $\frac{1}{2}$ – 7)

= 8(1 – 5) – 6(32 – 7)

= 8$\times$ (-4) – (6 $\times$ 32) +(6 $\times$ 7) = -32 – 9 + 42 =- 41 + 42 = 1 =RHS

LHS = RHS

Hence, verified .

Q19. 6(1 – 4x)+7(2 + 5x) = 53

SOLUTION :

6(1 – 4x) + 7(2 + 5x) = 53

On expanding the brackets, we get (6$\times$1) – (6 $\times$ 4x) + (7 $\times$2) + (7$\times$5x) = 53

6 – 24x + 14 + 35x = 53

6 + 14 + 35x – 24x = 53

20 + 11x = 53

Subtracting 20 from both sides, we get 20 + 11x – 20 = 53 – 20

11x = 33

Dividing both sides by 11, we get

$\frac{11x}{11}$ =$\frac{33}{11}$

x = 3

Verification :

Substituting x = 3 in LHS, we get

=6(1 – 4$\times$3) + 7(2 + 5$\times$3)

= 6(1 – 12) + 7(2 + 15)

= 6(-11) + 7(17)

= -66 + 119

= 53 = RHS

LHS = RHS

Hence, verified.

Q20. 5(2 – 3x)-17(2x – 5) = 16

SOLUTION :

5(2 – 3x) -17(2x – 5) = 16

On expanding the brackets , we get (5$\times$2) – (5 $\times$ 3x) – (17 $\times$2x ) + (17$\times$5) = 16

10 – 15x – 34x + 85 = 16

10 + 85 – 34x – 15x = 16

95 – 49x = 16

Subtracting 95 from both sides, we get – 49x + 95 – 95 = 16 – 95

-49x = -79

Dividing both sides by -49, we get

$\frac{-49x}{-49}$ = $\frac{-79}{-49}$

x = $\frac{79}{49}$

Verification :

Substituting x = $\frac{79}{49}$ in LHS, we get

= 5(2 – 3$\times$ $\frac{79}{49}$) – 17(2$\times$ $\frac{79}{49}$ – 5)

= (5$\times$2) – (5 $\times$ 3$\times$ $\frac{79}{49}$ )- (17 $\times$2$\times$ $\frac{79}{49}$ )+(17$\times$5)

= 10 – $\frac{1185}{49}$$\frac{2686}{49}$ + 85

= $\frac{490 – 1185 – 2686 + 4165}{49}$= $\frac{784}{49}$= 16 = RHS

LHS = RHS

Hence, verified.

Q21. $\frac{x – 3}{5}$ – 2 = -1

SOLUTION :

$\frac{x – 3}{5}$ – 2 = -1

Adding 2 to both sides, we get

$\frac{x – 3}{5}$ – 2  + 2 = -1 + 2

$\frac{x – 3}{5}$  = 1

Multiplying both sides by 5, we get

$\frac{x – 3}{5}$  $\times$ 5 = 1 $\times$ 5

x – 3 = 5

Adding 3 to both sides, we get

x – 3 + 3 = 5 +3

x = 8

Verification :

Substituting x = 8 in LHS, we get

=$\frac{8 – 3}{5}$ -2 = $\frac{5}{5}$  – 2 = 1 – 2 = -1 = RHS

LHS = RHS

Hence, verified.

Q22. 5(x – 2)+3(x + 1) = 25

SOLUTION :

5(x – 2) + 3(x + 1) = 25

On expanding the brackets, we get

(5 $\times$ x) – (5 $\times$ 2) +3$\times$  x + 3$\times$ 1 = 25

5x – 10 + 3x + 3 = 25

5x + 3x – 10 + 3 = 25

8x – 7 = 25

Adding 7 to both sides, we get

8x – 7 + 7 = 25 + 7

8x = 32

Dividing both sides by 8, we get

$\frac{8x}{8}$ = $\frac{32}{8}$

x = 4

Verification :

Substituting x= 4 in LHS, we get

= 5(4 – 2) + 3(4 + 1) = 5(2) + 3(5) = 10 + 15 = 25 = RHS

LHS = RHS

Hence, verified.