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

RD Sharma Solutions Class 8 Chapter 9 Exercise 9.1

RD Sharma Class 8 Solutions Chapter 9 Ex 9.1 PDF Download

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