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.

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