# RD Sharma Solutions Class 8 Linear Equation In One Variable Exercise 9.1

## RD Sharma Solutions Class 8 Chapter 9 Exercise 9.1

Solve each of the following equations and also verify your solutions:

Q1. $9(\frac{1}{4}) = y – 1(\frac{1}{3})$

Sol:

$9(\frac{1}{4}) = y – 1(\frac{1}{3})$

=> $\frac{37}{4}$ + $\frac{4}{3}$ = y

=> y = $\frac{127}{2}$

Verification

L.H.S = 9($\frac{1}{4}$)

R.H.S = $\frac{127}{2}$ – 1($\frac{1}{3}$)

= $\frac{127}{2}$$\frac{4}{3}$

= $\frac{127 – 16}{12}$

= $\frac{111}{12}$

= 9($\frac{1}{4}$)

Hence, L.H.S = R.H.S

Q2. $\frac{5x}{3} + \frac{2}{5} = 1$

Sol:

$\frac{5x}{3} + \frac{2}{5} = 1$

=> $\frac{5x}{3}$ = 1 – $\frac{2}{5}$

=> $\frac{5x}{3}$ = $\frac{3}{5}$

=> x = $\frac{3}{5}$ $\times$ $\frac{3}{5}$

= $\frac{9}{25}$

Q3 $\frac{x}{2} + \frac{x}{3} + \frac{x}{4} = 13$

Sol:

$\frac{x}{2} + \frac{x}{3} + \frac{x}{4} = 13$

=> $\frac{x\times 6 + x\times 4 + x\times 3}{12}$ = 13

=> $\frac{13x}{12}$ = 13

=>   x = 13 $\times$ $\frac{12}{13}$

= 12

Verification

L.H.S = $\frac{12}{2} + \frac{12}{3} + \frac{12}{4}$

= 6 + 4 + 3

= 13

R.H.S = 13

Hence, L.H.S = R.H.S

Q4 $\frac{x}{2} + \frac{x}{8} = \frac{1}{8}$

Sol:

$\frac{x}{2} + \frac{x}{8} = \frac{1}{8}$

=> $\frac{4x + x}{8}$ = $\frac{1}{8}$

=> $\frac{5x}{8}$ = $\frac{1}{8}$

=>       x =  $\frac{1}{8}$ $\times$ $\frac{8}{5}$

= $\frac{1}{5}$

Verification

L.H.S = $\frac{1}{2}$ $\times$ $\frac{1}{5}$ + $\frac{1}{8}$ $\times$ $\frac{1}{5}$

= $\frac{1}{10}$ + $\frac{1}{40}$

= $\frac{5}{40}$

= $\frac{1}{8}$

R.H.S = $\frac{1}{8}$

Hence, L.H.S = R.H.S

Q5 $\frac{2x}{3} – \frac{3x}{8} = \frac{7}{12}$

Sol:

$\frac{2x}{3} – \frac{3x}{8} = \frac{7}{12}$

=> $\frac{16x – 9x}{24}$= $\frac{7}{12}$

=> $\frac{7x}{24}$ = $\frac{7}{12}$

=>          x = $\frac{7}{12}$ $\times$ $\frac{24}{7}$

=>          x = 2

Verification

L.H.S = $\frac{4}{3}$$\frac{6}{8}$

= $\frac{32 – 18}{24}$

= $\frac{7}{12}$

R.H.S = $\frac{7}{12}$

Hence, L.H.S = R.H.S

Q6 (x + 2) (x + 3) + (x – 3) (x – 2) – 2x(x + 1) = 0

Sol:

(x + 2) (x + 3) + (x – 3) (x – 2) – 2x(x + 1) = 0

=> $x^{2} + 5x + 6 + x^{2} – 5x + 6 – 2x^{2} – 2x$ = 0

=> 12 – 2x = 0

=>         2x = 12

=>          x = 6

Verification

L.H.S = (6 + 2) (6 + 3) + (6 – 3) (6 – 2) – 2(6)(6 + 1)

= 72 + 12 – 84

= 0

R.H.S = 0

Hence, L.H.S = R.H.S

Q7 $\frac{x}{2} – \frac{4}{5} + \frac{x}{5} + \frac{3x}{10} = \frac{1}{5}$

Sol:

$\frac{x}{2} – \frac{4}{5} + \frac{x}{5} + \frac{3x}{10} = \frac{1}{5}$

=> $\frac{x}{2}$ + $\frac{x}{5}$ + $\frac{3x}{5}$ = $\frac{1}{5}$ + $\frac{4}{5}$

=> $\frac{5x + 2x + 2x}{10}$ = $\frac{5}{5}$

=> $\frac{10x}{10}$ = 1

=>                 x = 1

Verification

L.H.S = $\frac{1}{2}$$\frac{4}{5}$ + $\frac{1}{5}$ + $\frac{3}{10}$

= $\frac{5 – 8 + 2 + 3}{10}$

= $\frac{1}{5}$

R.H.S = $\frac{1}{5}$

Q8 $\frac{7}{x} + 35 = \frac{1}{10}$

Sol:

$\frac{7}{x} + 35 = \frac{1}{10}$

=> $\frac{7}{x}$ =$\frac{1}{10}$ – 35

=> $\frac{7}{x}$ = $\frac{1 – 350}{10}$

=> $\frac{x}{7}$ = $\frac{10}{-349}$

=>        x = $\frac{-10 \times 7}{349}$

= $\frac{-70}{349}$

Verification

L.H.S = $\frac{7}{\frac{-70}{349}} + 35$

= 7 $\times$ $\frac{349}{-70}$ + 35

= $\frac{349}{-70}$ + 35

= $\frac{1}{10}$

R.H.S = $\frac{1}{10}$

Hence, L.H.S = R.H.S

Q9 $\frac{2x – 1}{3} – \frac{6x – 2}{5} = \frac{1}{3}$

Sol:

$\frac{2x – 1}{3} – \frac{6x – 2}{5} = \frac{1}{3}$

=> $\frac{10x – 5 – 18x + 6}{15}$ = $\frac{1}{3}$

=> $\frac{-8x + 1}{15}$ = $\frac{1}{3}$

=> -24x + 3 = 15

=>    x = $\frac{-12}{24}$

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

Verification

L.H.S = $\frac{2(\frac{-1}{2}) – 1}{3}$$\frac{6(\frac{-1}{2}) – 2}{5}$

= $\frac{-2}{3}$$\frac{-5}{5}$

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

= $\frac{1}{3}$

R.H.S = $\frac{1}{3}$

Hence, L.H.S = R.H.S

Q10 13(y – 4) – 3(y – 9) – 5(y + 4) = 0

Sol:

13(y – 4) – 3(y – 9) – 5(y + 4) = 0

=> 13y – 52 – 3y + 27 – 5y – 20 = 0

=> 5y = 45

=>    y = $\frac{45}{5}$

= 9

Verification

L.H.S = 13(9 – 4) – 3(9 – 9) – 5(9 + 4)

= 13(5) – 3(0) – 5(13)

=  0

R.H.S = 0

Hence, L.H.S = R.H.S

Q11 $\frac{2}{3}(x – 5) – \frac{1}{4}(x -2) = \frac{9}{2}$

Sol:

$\frac{2}{3}(x – 5) – \frac{1}{4}(x -2) = \frac{9}{2}$

=> $\frac{2x – 10}{3}$$\frac{x – 2}{4}$ = $\frac{9}{2}$

=> $\frac{8x – 40 – 3x + 6}{12}$ = $\frac{9}{2}$

=> $\frac{5x – 34}{12}$ = $\frac{9}{2}$

=> 10x – 68 = 108

=>   10x = 108 + 68

=>     10x = 176

=>         x = $\frac{176}{10}$

=>         x = $\frac{88}{5}$

Verification

L.H.S = $\frac{2}{3}(\frac{88}{5} – 5) – \frac{1}{4}(\frac{88}{5} – 2)$

= $\frac{2}{3}(\frac{63}{5}) – \frac{1}{4}(\frac{78}{5})$

= $\frac{9}{2}$

R.H.S = $\frac{9}{2}$