RD Sharma solution class 8 chapter 9 Linear Equation In One Variable Exercise 9.1

Exercise 9.1

Page-9.5

Question 1:

Solve each of the following equation and also verify your solution:
$9\frac{1}{4}=y-1\frac{1}{3}$

Answer 1:

$$\begin{gathered} 9\frac{1}{4}=\mathrm{y}-1\frac{1}{3} \\ \mathrm{o}\mathrm{r} \frac{37}{4}+\frac{4}{3}=\mathrm{y}\mathrm{or} \mathrm{y}=\frac{127}{12}\therefore \mathrm{y}=\frac{127}{12} \mathrm{for} \mathrm{the} \mathrm{given} \mathrm{equation}.\text{C}\mathrm{heck}:\mathrm{L}.\mathrm{H}.\mathrm{S}=9\frac{1}{4}\mathrm{R}.\mathrm{H}.\mathrm{S}=\frac{127}{12}-1\frac{1}{3}=\frac{127}{12}-\frac{4}{3}=\frac{127-16}{12}=\frac{111}{12}=9\frac{1}{4}\mathrm{So}, \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S} \mathrm{for} \mathrm{y}=\frac{127}{12} \end{gathered}$$

Question 2:

Solve each of the following equation and also verify your solution:
$\frac{5x}{3}+\frac{2}{5}=1$

Answer 2:

$$\begin{gathered} \frac{5\mathrm{x}}{3}+\frac{2}{5}=1 \\ \Rightarrow \frac{5\mathrm{x}}{3}=1-\frac{2}{5} \\ \Rightarrow \frac{5\mathrm{x}}{3}=\frac{3}{5} \\ \Rightarrow \mathrm{x}=\frac{3}{5}\times \frac{3}{5}=\frac{9}{25} \\ \mathrm{Verification}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{5}{3}\times \frac{9}{25}+\frac{2}{5}=\frac{3}{5}+\frac{2}{ 5}=1 \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=1 \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{x}=\frac{9}{25} \end{gathered}$$

Question 3:

Solve each of the following equation and also verify your solution:
$\frac{x}{2}+\frac{x}{3}+\frac{x}{4}=13$

Answer 3:

$$\begin{gathered} \frac{\mathrm{x}}{2}+\frac{\mathrm{x}}{3}+\frac{\mathrm{x}}{4}=13 \\ \Rightarrow \frac{\mathrm{x}\times 6+\mathrm{x}\times 4+\mathrm{x}\times 3}{12}=13 \\ \Rightarrow \frac{13\mathrm{x}}{12}=13 \\ \Rightarrow \mathrm{x}=13\times \frac{12}{13}=12 \\ \mathrm{Verification}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{12}{2}+\frac{12}{3}+\frac{12}{4}=6+4+3=13=\mathrm{R}.\mathrm{H}.\mathrm{S}. \end{gathered}$$

Question 4:

Solve each of the following equation and also verify your solution:
$\frac{x}{2}+\frac{x}{8}=\frac{1}{8}$

Answer 4:

$$\begin{gathered} \frac{\mathrm{x}}{2}+\frac{\mathrm{x}}{8}=\frac{1}{8} \\ \mathrm{or} \frac{4\mathrm{x}+\mathrm{x}}{8}=\frac{1}{8} \\ \mathrm{or} \frac{5\mathrm{x}}{8}=\frac{1}{8} \\ \mathrm{or} \mathrm{x}=\frac{1}{8}\times \frac{8}{5}=\frac{1}{5} \\ \mathrm{Verification}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{1}{2}\times \frac{1}{5}+\frac{1}{8}\times \frac{1}{5}=\frac{1}{10}+\frac{1}{40}=\frac{5}{40}=\frac{1}{8}=\mathrm{R}.\mathrm{H}.\mathrm{S}. \end{gathered}$$

Question 5:

Solve each of the following equation and also verify your solution:
$\frac{2x}{3}-\frac{3x}{8}=\frac{7}{12}$

Answer 5:

$$\begin{gathered} \frac{2\mathrm{x}}{3}-\frac{3\mathrm{x}}{8}=\frac{7}{12} \\ \mathrm{or} \frac{16\mathrm{x}-9\mathrm{x}}{24}=\frac{7}{12} \\ \mathrm{or} \frac{7\mathrm{x}}{24}=\frac{7}{12} \\ \mathrm{or} \mathrm{x}=\frac{7}{12}\times \frac{24}{7}=2 \\ \mathrm{Verification}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{4}{3}-\frac{6}{8}=\frac{32-18}{24}=\frac{7}{12} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{7}{12} \\ \therefore \mathrm{R}.\mathrm{H}.\mathrm{S}.=\mathrm{L}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{x}=2 \end{gathered}$$

Question 6:

Solve each of the following equation and also verify your solution:
(x + 2)(x + 3) + (x − 3)(x − 2) − 2x(x + 1) = 0

Answer 6:

$$\begin{gathered} (\mathrm{x}+2)(\mathrm{x}+3)+(\mathrm{x}-3)(\mathrm{x}-2)-2\mathrm{x}(\mathrm{x}+1)=0 \\ \mathrm{or} {\mathrm{x}}^2+5\mathrm{x}+6+{\mathrm{x}}^2-5\mathrm{x}+6-2{\mathrm{x}}^2-2\mathrm{x}=0 \\ \mathrm{or} 12-2\mathrm{x}=0 \\ \mathrm{or} \mathrm{x}=\frac{12}{2}=6 \\ \mathrm{Verification}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=(6+2)(6+3)+(6-3)(6-2)-2\times 6(6+1) \\ =72+12-84=0=\mathrm{R}.\mathrm{H}.\mathrm{S}. \end{gathered}$$

Question 7:

Solve each of the following equation and also verify your solution:
$\frac{x}{2}-\frac{4}{5}+\frac{x}{5}+\frac{3x}{10}=\frac{1}{5}$

Answer 7:

$$\begin{gathered} \frac{\mathrm{x}}{2}-\frac{4}{5}+\frac{\mathrm{x}}{5}+\frac{3\mathrm{x}}{10}=\frac{1}{5} \\ \mathrm{or} \frac{\mathrm{x}}{2}+\frac{\mathrm{x}}{5}+\frac{3\mathrm{x}}{10}=\frac{1}{5}+\frac{4}{5} \\ \mathrm{or} \frac{5\mathrm{x}+2\mathrm{x}+3\mathrm{x}}{10}=\frac{5}{5} \\ \mathrm{or} \frac{10\mathrm{x}}{10}=1 \\ \text{or} \mathrm{x}=1 \\ \mathrm{Verification}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{1}{2}-\frac{4}{5}+\frac{1}{5}+\frac{3}{10} \\ =\frac{5-8+2+3}{10}=\frac{1}{5}=\mathrm{R}.\mathrm{H}.\mathrm{S}. \end{gathered}$$

Question 8:

Solve each of the following equation and also verify your solution:
$\frac{7}{x}+35=\frac{1}{10}$

Answer 8:

$$\begin{gathered} \frac{7}{\mathrm{x}}+35=\frac{1}{10} \\ \mathrm{or} \frac{7}{\mathrm{x}}=\frac{1}{10}-35 \\ \mathrm{or} \frac{7}{\mathrm{x}}=\frac{1-350}{10} \\ \mathrm{or} \frac{\mathrm{x}}{7}=\frac{10}{-349} \\ \mathrm{or} \mathrm{x}=\frac{-10\times 7}{349}=\frac{-70}{349} \\ \mathrm{Verification}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{7}{{\displaystyle \frac{-70}{349}}}+35 \\ =7\times \frac{349}{-70}+35 \\ =\frac{349}{-10}+35=\frac{1}{10}=\mathrm{R}.\mathrm{H}.\mathrm{S}. \end{gathered}$$

Question 9:

Solve each of the following equation and also verify your solution:
$\frac{2x-1}{3}-\frac{6x-2}{5}=\frac{1}{3}$

Answer 9:

$$\begin{gathered} \frac{2\mathrm{x}-1}{3}-\frac{6\mathrm{x}-2}{5}=\frac{1}{3} \\ \mathrm{or} \frac{10\mathrm{x}-5-18\mathrm{x}+6}{15}=\frac{1}{3} \\ \mathrm{or} \frac{-8\mathrm{x}+1}{15}=\frac{1}{3} \\ \mathrm{or} -24\mathrm{x}+3=15 \\ \mathrm{or} 24\mathrm{x}=3-15 \\ \mathrm{or} \mathrm{x}=\frac{-12}{24}=\frac{-1}{2} \\ \mathrm{Verification}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{2\times {\displaystyle \frac{-1}{2}}-1}{3}-\frac{6\times {\displaystyle \frac{-1}{2}}-2}{5} \\ =\frac{-2}{3}-\frac{-5}{5} \\ =\frac{-2+3}{3}=\frac{1}{3}=\mathrm{R}.\mathrm{H}.\mathrm{S}. \end{gathered}$$

Question 10:

Solve each of the following equation and also verify your solution:
13(y − 4) − 3(y − 9) − 5(y + 4) = 0

Answer 10:

$$\begin{gathered} 13(\mathrm{y}-4)-3(\mathrm{y}-9)-5(\mathrm{y}+4)=0 \\ \mathrm{or} 13\mathrm{y}-52-3\mathrm{y}+27-5\mathrm{y}-20=0 \\ \mathrm{or} 5\mathrm{y}=45 \\ \mathrm{or} \mathrm{y}=\frac{45}{5}=9 \\ \mathrm{Verification}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=13(9-4)-3(9-9)-5(9+4) \\ =13\times 5-3\times 0-5\times 13=0=\mathrm{R}.\mathrm{H}.\mathrm{S}. \end{gathered}$$

Question 11:

Solve each of the following equation and also verify your solution:
$\frac{2}{3}(x-5)-\frac{1}{4}(x-2)=\frac{9}{2}$

Answer 11:

$$\begin{gathered} \frac{2}{3}(\mathrm{x}-5)-\frac{1}{4}(\mathrm{x}-2)=\frac{9}{2} \\ \mathrm{or} \frac{2\mathrm{x}-10}{3}-\frac{\mathrm{x}-2}{4}=\frac{9}{2} \\ \mathrm{or} \frac{8\mathrm{x}-40-3\mathrm{x}+6}{12}=\frac{9}{2} \\ \mathrm{or} \frac{5\mathrm{x}-34}{12}=\frac{9}{2} \\ \mathrm{or} 10\mathrm{x}-68=108 \\ \mathrm{or} 10\mathrm{x}=108+68 \\ \mathrm{or} \mathrm{x}=\frac{176}{10}=\frac{88}{5} \\ \mathrm{Verification}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{2}{3}(\frac{88}{5}-5)-\frac{1}{4}(\frac{88}{5}-2) \\ =\frac{2}{3}\times \frac{63}{5}-\frac{1}{4}\times \frac{78}{5}=\frac{9}{2}=\mathrm{R}.\mathrm{H}.\mathrm{S}. \end{gathered}$$