## RD Sharma Solutions Class 7 Chapter 8 Exercise 8.3

#### Exercise 8.3

Q1. 6x + 5 = 2x + 17

SOLUTION :

We have

6x + 5 = 2x+ 17

Transposing 2x to LHS and 5 to RHS, we get

6x – 2x = 17 – 5

4x = 12

Dividing both sides by 4, we get

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

x= 3

Verification :

Substituting x =3 in the given equation, we get

6\(\times\)

18 + 5 = 6 + 17

23 = 23

LHS = RHS

Hence, verified.

Q2. 2(5x – 3)-3(2x – 1)= 9

SOLUTION :

We have

2(5x – 3) – 3(2x – 1) = 9

Expanding the brackets, we get

2\(\times\)

10x – 6 -6x + 3 = 9

10x – 6x – 6 + 3 = 9

4x – 3 = 9

Adding 3 to both sides, we get

4x – 3 + 3 = 9 + 3

4x = 12

Dividing both sides by 4, we get

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

Thus, x = 3.

Verification :

Substituting x = 3 in LHS, we get

= 2(5\(\times\)

= 2\(\times\)

= 24 – 15

= 9

LHS = RHS

Hence, verified.

Q3 .Â \(\frac{x}{2}\)

SOLUTION :

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

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

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

\(\frac{3x-2x}{6}\)

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

Multiplying both sides by 6, we get

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

x=6

Verification :

Substituting x = 6 in the given equation, we get

\(\frac{6}{2}\)

3 = 2 + 1

3 = 3

LHS = RHS

Hence, verified.

Q4. \(\frac{x}{2}\)

SOLUTION :

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

Transposing \(\frac{2x}{5}\)

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

= \(\frac{5x-4x}{10}\)

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

Multiplying both sides by 10, we get

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

= x = -25

Verification :

Substituting x = -25 in the given equation, we get

\(\frac{-25}{2}\)

\(\frac{-22}{2}\)

-11 = -11

LHS = RHS

Hence, verified.

Q5. \(\frac{3}{4}\)

SOLUTION :

\(\frac{3}{4}\)

On expanding the brackets on both sides, we get

=\(\frac{3x}{4}\)

TransposingÂ \(\frac{3x}{4}\)

= 3 â€“ \(\frac{3}{4}\)

= \(\frac{12-3}{4}\)

= \(\frac{9}{4}\)

Multiplying both sides by 4, we get

= x = 9

Verification :

Substituting x = 9 on both sides, we get

\(\frac{3}{4}\)

\(\frac{3}{4}\)

6 = 6

LHS = RHS

Hence, verified.

Q6. 3(x – 3) = 5(2x + 1)

SOLUTION :

3(x – 3) = 5(2x + 1)

On expanding the brackets on both sides, we get

= 3\(\times\)

= 3x – 9 = 10x + 5

Transposing 10x to LHS and 9 to RHS, we get

= 3x – 10x = 9 + 5

= -7x = 14

Dividing both sides by 7, we get

= –\(\frac{7x}{7}\)

=x = -2

Verification :

Substituting x = -2 on both sides, we get

3( -2 â€“ 3) = 5{2(-2) +1}

3(-5) = 5(-3)

-15 = -15

LHS = RHS

Hence, verified.

Q7. 3x â€“ 2(2x – 5) = 2(x + 3) – 8

SOLUTION :

3x – 2(2x – 5) = 2(x + 3) – 8

On expanding the brackets on both sides, we get

= 3x – 2\(\times\)

= 3x – 4x + 10 = 2x + 6 – 8

= -x + 10 = 2x – 2

Transposing x to RHS and 2 to LHS, we get

= 10 + 2 = 2x + x

= 3x = 12

Dividing both sides by 3, we get

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

= x = 4

Verification :

Substituting x = 4 on both sides, we get

3(4) â€“ 2{2(4) – 5} = 2(4 + 3) – 8

12 – 2(8 – 5) = 14 – 8

12 – 6 = 6

6 = 6

LHS = RHS

Hence, verified.

Q8. x – \(\frac{x}{4}\)

SOLUTION :

x – \(\frac{x}{4}\)

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

= x – \(\frac{x}{4}\)

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

= \(\frac{2x}{4}\)

Multiplying both sides by 4, we get

= \(\frac{2x}{4}\)

= 2x = 14

Dividing both sides by 2, we get

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

= x = 7

Verification :

Substituting x = 7 on both sides, we get

7 – \(\frac{7}{4}\)

\(\frac{28-7-2}{4}\)

\(\frac{19}{4}\)

LHS = RHS

Hence, verified.

Q9. \(\frac{6x-2}{9}\)

SOLUTION :

\(\frac{6x-2}{9}\)

= \(\frac{6x(2)-2(2)+ 3x+5}{18}\)

= \(\frac{12x-4+ 3x+5}{18}\)

=Â \(\frac{15x+1}{18}\)

Multiplying both sides by 18, we get

= \(\frac{15x+1}{18}\)

= 15x + 1 = 6

Transposing 1 to RHS, we get

= 15x = 6 – 1

= 15x = 5

Dividing both sides by 15, we get

= \(\frac{15x}{15}\)

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

Verification :

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

\(\frac{6(\frac{1}{3})-2}{9}\)

\(\frac{2-2}{9}\)

0 + \(\frac{6}{18}\)

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

LHS = RHS

Hence, verified.

Q10. m – \(\frac{m – 1}{2}\)

SOLUTION :

m – \(\frac{m – 1}{2}\)

= \(\frac{2m -m + 1}{2}\)

= \(\frac{m + 1}{2}\)

= \(\frac{m + 1}{2}\)

= \(\frac{m }{2}\)

Transposing Â \(\frac{m}{3}\)

= \(\frac{m }{2}\)

= \(\frac{ 3m+2m}{6}\)

Multiplying both sides by 6, we get

= \(\frac{ 5m}{6}\)

= 5m = 7

Dividing both sides by 5, we get

= \(\frac{5m}{5}\)

= m = \(\frac{7}{5}\)

Verification :

Substituting m =\(\frac{7}{5}\)

\(\frac{7}{5}\)

\(\frac{7}{5}-\frac{7-5}{10}=1-\frac{7-10}{15}\)

\(\frac{7}{5}-\frac{2}{10}=\frac{15+3}{15}\)

\(\frac{14-2}{10}=\frac{15+3}{15}\)

\(\frac{12}{10}=\frac{18}{15}\)

\(\frac{6}{5}=\frac{6}{5}\)

LHS = RHS

Hence, verified.

Â

Q11. \(\frac{5x-1}{3}\)

SOLUTION :

\(\frac{5x-1}{3}\)

\(\frac{5x-1-2x+2}{3}\)

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

Multiplying both sides by 3, we get 3

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

= 3x + 1 = 3

Subtracting 1 from both sides, we get

= 3x + 1 – 1 = 3 – 1

= 3x = 2

Dividing both sides by 3, we get

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

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

Verification:

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

\(\frac{5(\(\frac{2}{3}\)

= \(\frac{5\left ( \frac{2}{3} \right )-1}{3}-\frac{2\left ( \frac{2}{3} \right )-2}{3}\)

= \(\frac{\left ( \frac{10}{3} \right )-1}{3}-\frac{\left ( \frac{4}{3} \right )-2}{3}\)

= \(\frac{\frac{10-3}{3}}{3}-\frac{ \frac{4-6}{3} }{3}\)

= \(\frac{10-3}{9}-\frac{ 4-6 }{9}\)

= \(\frac{7}{9}+\frac{ 2 }{9}\)

= \(\frac{ 9 }{9}\)

= RHS

LHS = RHS Hence, verified.

Q12. 0.6x + \(\frac{4}{5}\)

SOLUTION :

0.6x + \(\frac{4}{5}\)

Transposing 0.28x to LHS and \(\frac{4}{5}\)

= 0.6x – 0.28x = 1.16 –Â \(\frac{4}{5}\)

= 0.32x = 1.16 – 0.8

= 0.32x = 0.36

Dividing both sides by 0.32, we get

= \(\frac{0.32x}{0.32}\)

= x = \(\frac{9}{8}\)

Verification:

Substituting x = \(\frac{9}{8}\)

0.6(\(\frac{9}{8}\)

\(\frac{5.4}{8}\)

0.675 + 0.8 = 0.315 + 1.16

1.475 = 1.475

LHS = RHS

Hence, verified.

Q13. 0.5x + \(\frac{x}{3}\)

SOLUTION :

0.5x + \(\frac{x}{3}\)

\(\frac{5}{10}\)

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

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

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

\(\frac{6x+4x-3x}{12}\)

\(\frac{7x}{12}\)

Multiplying both sides by 12, we get

\(\frac{7x}{12}\)

= 7x = 84

Dividing both sides by 7, we get

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

= x = 12

Verification:

Substituting x = 12 on both sides, we get

0.5(12) + (12)3 = 0.25(12) + 7

6 + 4 = 3 + 7

10 = 10

LHS = RHS

Hence, verified.