## RD Sharma Solutions Class 7 Chapter 8 Exercise 8.2

#### Exercise 8.2

**Q1. x â€“ 3 = 5**

**SOLUTION :**

x – 3 = 5

Adding 3 to both sides, we get

x â€“ 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 â€“ 9 = 13 â€“ 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 – \(\frac{3}{5}\)

Adding 3/5 to both sides, we get

=> x â€“ \(\frac{3}{5}\)

=> x = \(\frac{7}{5}\)

=> x = \(\frac{10}{5}\)

=> x=2

Verification :

Substituting x = 2 in LHS, we get

LHS = 2 â€“ \(\frac{3}{5}\)

LHS = RHS

Hence, verified.

**Q4. 3x = 0**

**SOLUTION :**

3x = 0

Dividing both sides by 3, we get

\(\frac{3x}{3}\)

x = 0

Verification :

Substituting x = 0 in LHS = 3x, we get LHS = 3 \(\times\)

LHS = RHS

Hence, verified.

**Q5 . \(\frac{x}{2}\) = 0**

**SOLUTION :**

\(\frac{x}{2}\)

Multiplying both sides by 2, we get

=> \(\frac{x}{2}\)

=> x = 0

Verification :

Substituting x= 0 in LHS, we get

LHS = \(\frac{0}{2}\)

LHS = 0 and RHS = 0

LHS = RHS

Hence, verified.

**Q6 . x – \(\frac{1}{3}\) = \(\frac{2}{3}\)**

**SOLUTION :**

x – \(\frac{1}{3}\)

Adding \(\frac{1}{3}\)

x – \(\frac{1}{3}\)

=> x = \(\frac{2}{3}\)

=> x = \(\frac{3}{3}\)

x = 1

Verification :

Substituting x= 1 in LHS, we get

LHS = 1 – \(\frac{1}{3}\)

LHS = RHS

Hence, verified.

**Q7. x + \(\frac{1}{2}\) = \(\frac{7}{2}\)**

**SOLUTION :**

x + \(\frac{1}{2}\)

Subtracting \(\frac{1}{2}\)

x + \(\frac{1}{2}\)

x = \(\frac{7}{2}\)

x = 3

Verification :

Substituting x = 3 in LHS, we get LHS = 3 + \(\frac{1}{2}\)

LHS = RHS

Hence, verified.

**Q8. 10 â€“ y = 6**

**SOLUTION :**

10 â€“ y = 6

Subtracting 10 from both sides, we get

10 – y – 10 = 6 â€“ 10

-y = -4

Multiplying both sides by -1, we get

-y Â \(\times\)

y = 4

Verification :

Substituting y = 4 in LHS, we get

LHS = 10 â€“ y = 10 â€“ 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 â€“ 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.Â \(\frac{4}{5}\) – x = \(\frac{3}{5}\)**

**SOLUTION :**

\(\frac{4}{5}\)

Subtracting \(\frac{4}{5}\)

\(\frac{4}{5}\)

– x = \(\frac{3}{5}\)

– x = –\(\frac{1}{5}\)

Multiplying both sides by -1, we get

-x \(\times\)

x = \(\frac{1}{5}\)

Verification :

Substituting x= \(\frac{1}{5}\)

LHS = \(\frac{4}{5}\)

LHS = RHS

Hence, verified.

**Q11. 2y – \(\frac{1}{2}\) = –\(\frac{1}{3}\)**

**SOLUTION :**

2y –\(\frac{1}{2}\)

Adding \(\frac{1}{2}\)

2y – \(\frac{1}{2}\)

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 = \(\frac{1}{12}\)

LHS = 2\(\frac{1}{12}\)

LHS = RHS

Hence, verified.

**Q12. 14 = \(\frac{7x}{10}\) â€“ 8**

**SOLUTION :**

14 = \(\frac{7x}{10}\)

Adding 8 to both sides, we get

14 + 8 = \(\frac{7x}{10}\)

22 = \(\frac{7x}{10}\)

Multiplying both sides by 10, we get

22 \(\times\)

220 = 7x

Dividing both sides by 7, we get

\(\frac{220}{7}\)

x =Â \(\frac{220}{7}\)

Verification:

Substituting x =Â \(\frac{220}{7}\)

LHS = 14. and RHS = \(\frac {\frac {220}{7}}{\frac {7}{10}}\)

LHS = RHS

Hence, verified.

**Q13. 3(x + 2) = 15**

**SOLUTION :**

3 (x + 2) = 15

Dividing both sides by 3, we get

\(\frac{3(x+2)}{3}\)

(x+2) = Â 5

Subtracting 2 from both sides, we get

x + 2 â€“ 2 = 5 â€“ 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 :**

\(\frac{x}{4}\)

Multiplying both sides by 4, we get

\(\frac{x}{4}\)

x = \(\frac{7}{2}\)

Verification :

Substituting x = \(\frac{7}{2}\)

LHS = \(\frac{7}{2}\)

LHS = RHS

Hence, verified.

**Q15. \(\frac{1}{3}\) -2x = 0**

**SOLUTION :**

\(\frac{1}{3}\)

Subtracting \(\frac{1}{3}\)

\(\frac{1}{3}\)

-2x = –\(\frac{1}{3}\)

Multiplying both sides by -1, we get

-2x \(\times\)

2x = \(\frac{1}{3}\)

Dividing both sides by 2, we get \(\frac{2x}{2}\)

x = \(\frac{1}{6}\)

Verification :

Substituting x = \(\frac{1}{6}\)

LHS = \(\frac{1}{3}\)

LHS = RHS

Hence, verified.

**Q16. 3(x + 6) = 24**

**SOLUTION : **

3(x + 6) = 24

Dividing both sides by 3, we get

\(\frac{3 (x + 6)}{3}\)

(x+ 6) =Â 8

Subtracting 6 from both sides, we get

x + 6 â€“ 6 = 8 â€“ 6

x = 2

Verification :

Substituting x = 2 in LHS , we get

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

LHS = RHS

Hence, verified.

**Q17. 3(x + 2)-2(x â€“ 1) = 7**

**SOLUTION :**

3(x + 2) – 2(x – 1) = 7

On expanding the brackets, we get

3\(\times\)

3x + 6 – 2x + 2 = 7

3x – 2x + 6 + 2 = 7

x + 8 = 7

Subtracting 8 from both sides, we get

x + 8 â€“ 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\(\times\)

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\(\times\)

16x – 40 – 18x + 42 = 1

16x – 18x + 42 – 40 = 1

-2x + 2 = 1

Subtracting 2 from both sides, we get

-2x+ 2 â€“ 2 = 1 -2

-2x = -1

Multiplying both sides by -1, we get

-2x \(\times\)

2x = 1

Dividing both sides by 2, we get

\(\frac{2x}{2}\)

x = \(\frac{1}{2}\)

Verification :

Substituting x = \(\frac{1}{2}\)

= 8(2 \(\times\)

= 8(1 – 5) – 6(32 – 7)

= 8\(\times\)

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\(\times\)

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

\(\frac{11x}{11}\)

x = 3

Verification :

Substituting x = 3 in LHS, we get

=6(1 – 4\(\times\)

= 6(1 – 12) + 7(2 + 15)

= 6(-11) + 7(17)

= -66 + 119

= 53 = RHS

LHS = RHS

Hence, verified.

**Q20. 5(2 â€“ 3x)-17(2x – 5) = 16**

**SOLUTION :**

5(2 – 3x) -17(2x – 5) = 16

On expanding the brackets , we get (5\(\times\)

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

\(\frac{-49x}{-49}\)

x = \(\frac{79}{49}\)

Verification :

Substituting x = \(\frac{79}{49}\)

= 5(2 – 3\(\times\)

= (5\(\times\)

= 10 – \(\frac{1185}{49}\)

= \(\frac{490 – 1185 – 2686 + 4165}{49}\)

LHS = RHS

Hence, verified.

**Q21. \(\frac{x – 3}{5}\) – 2 = -1 **

**SOLUTION :**

\(\frac{x – 3}{5}\)

Adding 2 to both sides, we get

\(\frac{x – 3}{5}\)

\(\frac{x – 3}{5}\)

Multiplying both sides by 5, we get

\(\frac{x – 3}{5}\)

x â€“ 3 = 5

Adding 3 to both sides, we get

x – 3 + 3 = 5 +3

x = 8

Verification :

Substituting x = 8 in LHS, we get

=\(\frac{8 – 3}{5}\)

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 \(\times\)

5x â€“ 10 + 3x + 3 = 25

5x + 3x â€“ 10 + 3 = 25

8x – 7 = 25

Adding 7 to both sides, we get

8x â€“ 7 + 7 = 25 + 7

8x = 32

Dividing both sides by 8, we get

\(\frac{8x}{8}\)

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.