RD Sharma solution class 8 chapter 9 Linear Equation In One Variable Exercise 9.3

Exercise 9.3

Page-9.17

Question 1:

Solve the following equation and verify your answer:
$\frac{2x-3}{3x+2}=-\frac{2}{3}$

Answer 1:

$$\begin{gathered} \frac{2\mathrm{x}-3}{3\mathrm{x}+2}=-\frac{2}{3} \\ \mathrm{or} 6\mathrm{x}-9=-6\mathrm{x}-4 (\text{After c}\mathrm{ross} \mathrm{multipl}\text{ication}) \\ \mathrm{or} 6\mathrm{x}+6\mathrm{x}=-4+9 \\ \mathrm{or} \mathrm{x}=\frac{5}{12} \\ \therefore \mathrm{x}=\frac{5}{12}\mathrm{is} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.= \frac{2\times {\displaystyle \frac{5}{12}}-3}{3\times {\displaystyle \frac{5}{12}}+2}=\frac{{\displaystyle \frac{5}{6}}-3}{{\displaystyle \frac{5}{4}}+2}=\frac{{\displaystyle \frac{-13}{6}}}{{\displaystyle \frac{13}{4}}}=\frac{-4}{6}=\frac{-2}{3} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{-2}{3} \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{x}=\frac{5}{12} \end{gathered}$$

Question 2:

Solve the following equation and verify your answer:
$\frac{2-y}{y+7}=\frac{3}{5}$

Answer 2:

$$\begin{gathered} \frac{2-\mathrm{y}}{\mathrm{y}+7}=\frac{3}{5} \\ \mathrm{or} 10-5\mathrm{y}=3\mathrm{y}+21 (\text{After c}\mathrm{ross} \mathrm{multipl}\text{ication}) \\ \mathrm{or} 3\mathrm{y}+5\mathrm{y}=10-21 \\ \mathrm{or} 8\mathrm{y}=-11 \\ \mathrm{or} \mathrm{y}=\frac{-11}{8} \\ \therefore \mathrm{y}=\frac{-11}{8} \mathrm{is} \mathrm{the} \mathrm{solution} \mathrm{of} \text{the} \mathrm{given} \mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting} \mathrm{y}=\frac{-11}{8} \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{2-\frac{-11}{8}}{\frac{-11}{8}+7}=\frac{16+11}{-11+56}=\frac{27}{45}=\frac{3}{5} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{3}{5} \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{y}=\frac{-11}{8} \end{gathered}$$

Question 3:

Solve the following equation and verify your answer:
$\frac{5x-7}{3x}=2$

Answer 3:

$$\begin{gathered} \frac{5\mathrm{x}-7}{3\mathrm{x}}=2 \\ \mathrm{or} 6\mathrm{x}=5\mathrm{x}-7 (\text{After c}\mathrm{ross} \mathrm{multipl}\text{ication}) \\ \text{or} 6\mathrm{x}-5\mathrm{x}=-7 \\ \mathrm{or} \mathrm{x}=-7 \\ \therefore \mathrm{x}=-7 \mathrm{is} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{given} \mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting} \mathrm{x}=-7 \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}=\frac{5\times (-7)-7}{.3(-7)}=\frac{-35-7}{-21}=\frac{-42}{-21}=2 \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=2 \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{x}=-7. \end{gathered}$$

Question 4:

Solve the following equation and verify your answer:
$\frac{3x+5}{2x+7}=4$

Answer 4:

$$\begin{gathered} \frac{3\mathrm{x}+5}{2\mathrm{x}+7}=4 \\ \mathrm{or} 3\mathrm{x}+5=8\mathrm{x}+28 \\ \mathrm{or} 8\mathrm{x}+28=3\mathrm{x}+5 (\text{After c}\mathrm{ross} \mathrm{multipl}\text{ication}) \\ \mathrm{or}8\mathrm{x}-3\mathrm{x}=5-28 \\ \mathrm{or} 5\mathrm{x}=-23 \\ \mathrm{or} \mathrm{x}=\frac{-23}{5} \\ \therefore \mathrm{x}=\frac{-23}{5} \mathrm{is} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{given} \mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting} \mathrm{x}=\frac{-23}{5} \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{3\times {\displaystyle \frac{-23}{5}}+5}{2\times {\displaystyle \frac{-23}{5}}+7}=\frac{-69+25}{-46+35}=\frac{-44}{-11}=4 \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=4 \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{x}=\frac{-23}{5} \end{gathered}$$

Question 5:

Solve the following equation and verify your answer:
$\frac{2y+5}{y+4}=1$

Answer 5:

$$\begin{gathered} \frac{2\mathrm{y}+5}{\mathrm{y}+4}=1 \\ \mathrm{or} 2\mathrm{y}+5=\mathrm{y}+4 \\ \mathrm{or} 2\mathrm{y}-\mathrm{y}=4-5 \\ \mathrm{or} \mathrm{y}=-1 \\ \mathrm{Thus}, \mathrm{y}=-1 \mathrm{is} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting} \mathrm{y}=-1 \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{2(-1)+5}{-1+4}=\frac{-2+5}{3}=\frac{3}{3}=1 \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=1 \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{y}=-1. \end{gathered}$$

Question 6:

Solve the following equation and verify your answer:
$\frac{2x+1}{3x-2}=\frac{5}{9}$

Answer 6:

$$\begin{gathered} \frac{2\mathrm{x}+1}{3\mathrm{x}-2}=\frac{5}{9} \\ \mathrm{or} 18\mathrm{x}+9=15\mathrm{x}-10 [\text{After c}\mathrm{ross} \mathrm{multipl}\text{ication}] \\ \mathrm{or} 18\mathrm{x}-15\mathrm{x}=-10-9 \\ \mathrm{or} 3\mathrm{x}=-19 \\ \mathrm{or} \mathrm{x}=\frac{-19}{3} \\ \mathrm{Thus}, \mathrm{x}=\frac{-19}{3} \mathrm{is} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting} \mathrm{x}=\frac{-19}{3} \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{2\left({\displaystyle \frac{-19}{3}}\right)+1}{3\left({\displaystyle \frac{-19}{3}}\right)-2}=\frac{-38+3}{-57-6}=\frac{-35}{-63}=\frac{5}{9} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{5}{9} \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{x}=\frac{-19}{3} \end{gathered}$$

Question 7:

Solve the following equation and verify your answer:
$\frac{1-9y}{19-3y}=\frac{5}{8}$

Answer 7:

$$\begin{gathered} \frac{1-9\mathrm{y}}{19-3\mathrm{y}}=\frac{5}{8} \\ \mathrm{or} 8-72\mathrm{y}=95-15\mathrm{y} [\text{After c}\mathrm{ross} \mathrm{multipl}\text{ication}] \\ \mathrm{or} 95-15\mathrm{y}=8-72\mathrm{y} \\ \mathrm{or} 72\mathrm{y}-15\mathrm{y}=8-95 \\ \mathrm{or} 57\mathrm{y}=-87 \\ \mathrm{or} \mathrm{y}=\frac{-87}{57} \\ \mathrm{or} \mathrm{y}=\frac{-29}{19} \\ \mathrm{Thus} \mathrm{y}=\frac{-29}{19} \mathrm{is} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting} \mathrm{y}=\frac{-29}{19} \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{1-9\left({\displaystyle \frac{-29}{19}}\right)}{19-3\left({\displaystyle \frac{-29}{19}}\right)}=\frac{19+261}{361+87}=\frac{280}{448}=\frac{5}{8} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{5}{8} \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{y}=\frac{-29}{19} \end{gathered}$$

Question 8:

Solve the following equation and verify your answer:
$\frac{2x}{3x+1}=-3$

Answer 8:

$$\begin{gathered} \frac{2\mathrm{x}}{3\mathrm{x}+1}=-3 \\ \mathrm{or} 2\mathrm{x}=-9\mathrm{x}-3 [\text{After c}\mathrm{ross} \mathrm{multipl}\text{ication}] \\ \mathrm{or} 2\mathrm{x}+9\mathrm{x}=-3 \\ \mathrm{or} 11\mathrm{x}=-3 \\ \mathrm{or} \mathrm{x}=\frac{-3}{11} \\ \mathrm{Thus}, \mathrm{x}=\frac{-3}{11} \mathrm{is} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting} \mathrm{x}=\frac{-3}{11} \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{2\left(\frac{-3}{11}\right)}{3\left(\frac{-3}{11}\right)+1}=\frac{-6}{-9+11}=\frac{-6}{2}=-3 \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=-3 \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{x}=\frac{-3}{11} \end{gathered}$$

Question 9:

Solve the following equation and verify your answer:
$\frac{y-(7-8y)}{9y-(3+4y)}=\frac{2}{3}$

Answer 9:

$$\begin{gathered} \frac{\mathrm{y}-(7-8\mathrm{y})}{9\mathrm{y}-(3+4\mathrm{y})}=\frac{2}{3} \\ \mathrm{or} \frac{9\mathrm{y}-7}{5\mathrm{y}-3}=\frac{2}{3} \\ \text{or} 27\mathrm{y}-21=10\mathrm{y}-6 \text{[After c}\mathrm{ross} \mathrm{multipl}\text{ication}] \\ \mathrm{or} 27\mathrm{y}-10\mathrm{y}=-6+21 \\ \mathrm{or} 17\mathrm{y}=15 \\ \mathrm{or} \mathrm{y}=\frac{15}{17} \\ \mathrm{Thus}, \mathrm{y}=\frac{15}{17} \mathrm{is} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting} \mathrm{y}=\frac{15}{17} \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{9\left({\displaystyle \frac{15}{17}}\right)-7}{5\left({\displaystyle \frac{15}{17}}\right)-3}=\frac{135-119}{75-51}=\frac{16}{24}=\frac{2}{3} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{2}{3} \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{y}=\frac{15}{17} \end{gathered}$$

Question 10:

Solve the following equation and verify your answer:
$\frac{6}{2x-(3-4x)}=\frac{2}{3}$

Answer 10:

$$\begin{gathered} \frac{6}{2\mathrm{x}-(3-4\mathrm{x})}=\frac{2}{3} \\ \mathrm{or} \frac{6}{6\mathrm{x}-3}=\frac{2}{3} \\ \mathrm{or} 12\mathrm{x}-6=18 [\text{After c}\mathrm{ross} \mathrm{multipl}\text{ication}] \\ \mathrm{or} 12\mathrm{x}=18+6 \\ \mathrm{or} \mathrm{x}=\frac{24}{12} \\ \mathrm{or} \mathrm{x}=2 \\ \mathrm{Thus}, \mathrm{x}=2 \mathrm{is} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting} \mathrm{x}=2 \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{6}{2\times 2-(3-4\times 2)}=\frac{6}{9}=\frac{2}{3} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{2}{3} \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{x}=2. \end{gathered}$$

Question 11:

Solve the following equation and verify your answer:
$\frac{2}{3x}-\frac{3}{2x}=\frac{1}{12}$

Answer 11:

$$\begin{gathered} \frac{2}{3\mathrm{x}}-\frac{3}{2\mathrm{x}}=\frac{1}{12} \\ \mathrm{or}, \frac{4-9}{6\mathrm{x}}=\frac{1}{12} \\ \mathrm{or}, \frac{-5}{\mathrm{x}}=\frac{1}{2} \\ \mathrm{or}, \mathrm{x}=-10 [\text{After} c\mathrm{ross} \mathrm{multipl}\text{ication}] \\ \mathrm{Thus}, \mathrm{x}=-10 \mathrm{is} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting} \mathrm{x}=-10 \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{2}{3(-10)}-\frac{3}{2(-10)}=\frac{2}{-30}-\frac{3}{-20}=\frac{4-9}{-60}=\frac{-5}{-60}=\frac{1}{12} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{1}{12} \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{x}=-10. \end{gathered}$$

Question 12:

Solve the following equation and verify your answer:
$\frac{3x+5}{4x+2}=\frac{3x+4}{4x+7}$

Answer 12:

$$\begin{gathered} \frac{3\mathrm{x}+5}{4\mathrm{x}+2}=\frac{3\mathrm{x}+4}{4\mathrm{x}+7} \\ \mathrm{or}, 12{\mathrm{x}}^2+20\mathrm{x}+21\mathrm{x}+35=12{\mathrm{x}}^2+16\mathrm{x}+6\mathrm{x}+8 [\mathrm{Cross} \mathrm{multiply}] \\ \mathrm{or}, 12{\mathrm{x}}^2-12{\mathrm{x}}^2+41\mathrm{x}-22\mathrm{x}=8-35 \\ \mathrm{or}, 19\mathrm{x}=-27 \\ \mathrm{or}, \mathrm{x}=\frac{-27}{19} \\ \mathrm{Thus}, \mathrm{x}=\frac{-27}{19} \mathrm{is} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{equation} \\ \mathrm{Check}: \\ \mathrm{Substituting} \mathrm{x}=\frac{-27}{19} \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{3\left({\displaystyle \frac{-27}{19}}\right)+5}{4\left({\displaystyle \frac{-27}{19}}\right)+2}=\frac{-81+95}{-108+38}=\frac{14}{-70}=\frac{-1}{5} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{3\left({\displaystyle \frac{-27}{19}}\right)+4}{4\left({\displaystyle \frac{-27}{19}}\right)+7}=\frac{-81+76}{-108+133}=\frac{-5}{25}=\frac{-1}{5} \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{x}=\frac{-27}{19} \end{gathered}$$

Question 13:

Solve the following equation and verify your answer:
$\frac{7x-2}{5x-1}=\frac{7x+3}{5x+4}$

Answer 13:

$$\begin{gathered} \frac{7\mathrm{x}-2}{5\mathrm{x}-1}=\frac{7\mathrm{x}+3}{5\mathrm{x}+4} \\ \mathrm{or} 35{\mathrm{x}}^2+28\mathrm{x}-10\mathrm{x}-8=35{\mathrm{x}}^2+15\mathrm{x}-7\mathrm{x}-3 [\text{After c}\mathrm{ross} \mathrm{multipl}\text{ication}] \\ \mathrm{or} 35{\mathrm{x}}^2-35{\mathrm{x}}^2+18\mathrm{x}-8\mathrm{x}=-3+8 \\ \mathrm{or} 10\mathrm{x}=5 \\ \mathrm{or} \mathrm{x}=\frac{5}{10}\mathrm{or} \mathrm{x}=\frac{1}{2} \\ \mathrm{Thus}, \mathrm{x}=\frac{1}{2} \mathrm{is} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting} \mathrm{x}=\frac{1}{2} \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{7\left(\frac{1}{2}\right)-2}{5\left(\frac{1}{2}\right)-1}=\frac{7-4}{5-2}=\frac{3}{3}=1 \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{7\left(\frac{1}{2}\right)+3}{5\left(\frac{1}{2}\right)+4}=\frac{7+6}{5+8}=\frac{13}{13}=1 \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{x}=\frac{1}{2} \end{gathered}$$

Question 14:

Solve the following equation and verify your answer:
${\left(\frac{x+1}{x+2}\right)}^2=\frac{x+2}{x+4}$

Answer 14:

$$\begin{gathered} {\left(\frac{\mathrm{x}+1}{\mathrm{x}+2}\right)}^2=\frac{\mathrm{x}+2}{\mathrm{x}+4} \\ \mathrm{or} \frac{{\mathrm{x}}^2+2\mathrm{x}+1}{{\mathrm{x}}^2+4\mathrm{x}+4}=\frac{\mathrm{x}+2}{\mathrm{x}+4} \\ \mathrm{or} {\mathrm{x}}^3+2{\mathrm{x}}^2+\mathrm{x}+4{\mathrm{x}}^2+8\mathrm{x}+4={\mathrm{x}}^3+4{\mathrm{x}}^2+4\mathrm{x}+2{\mathrm{x}}^2+8\mathrm{x}+8 [\text{After c}\mathrm{ross} \mathrm{multipl}\text{ication}] \\ \mathrm{or} {\mathrm{x}}^3-{\mathrm{x}}^3+6{\mathrm{x}}^2-6{\mathrm{x}}^2+9\mathrm{x}-12\mathrm{x}=8-4 \\ \mathrm{or} -3\mathrm{x}=4 \\ \mathrm{or} \mathrm{x}=\frac{4}{-3}=\frac{-4}{3} \\ \mathrm{Thus}, \mathrm{x}=\frac{-4}{3} \mathrm{is} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting} \mathrm{x}=\frac{-4}{3} \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.={\left(\frac{\frac{-4}{3}+1}{\frac{-4}{3}+2}\right)}^2={\left(\frac{-4+3}{-4+6}\right)}^2=\frac{1}{4} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{\frac{-4}{3}+2}{\frac{-4}{3}+4}=\frac{-4+6}{-4+12}=\frac{2}{8}=\frac{1}{4} \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{x}=\frac{-4}{3} \end{gathered}$$

Question 15:

Solve the following equation and verify your answer:
${\left(\frac{x+1}{x-4}\right)}^2=\frac{x+8}{x-2}$

Answer 15:

$$\begin{gathered} {\left(\frac{\mathrm{x}+1}{\mathrm{x}-4}\right)}^2=\frac{\mathrm{x}+8}{\mathrm{x}-2} \\ \mathrm{or} \frac{{\mathrm{x}}^2+2\mathrm{x}+1}{{\mathrm{x}}^2-8\mathrm{x}+16}=\frac{\mathrm{x}+8}{\mathrm{x}-2} \left[\right(a+b{)}^2=a^2+b^2+2ab and (a-b{)}^2=a^2+b^2-2ab ] \\ \mathrm{or} {\mathrm{x}}^3+2{\mathrm{x}}^2+\mathrm{x}-2{\mathrm{x}}^2-4\mathrm{x}-2={\mathrm{x}}^3-8{\mathrm{x}}^2+16\mathrm{x}+8{\mathrm{x}}^2-64\mathrm{x}+128 [\text{After c}\mathrm{ross} \mathrm{multipl}\text{ication}] \\ \mathrm{or} {\mathrm{x}}^3-{\mathrm{x}}^3-3\mathrm{x}+48\mathrm{x}=128+2 \\ \mathrm{or} 45\mathrm{x}=130 \\ \mathrm{or} \mathrm{x}=\frac{130}{45}=\frac{26}{9} \\ \mathrm{Thus} \mathrm{x}=\frac{26}{9} \mathrm{is} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting} \mathrm{x}=\frac{26}{9} \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.={\left(\frac{\frac{26}{9}+1}{\frac{26}{9}-4}\right)}^2={\left(\frac{26+9}{26-36}\right)}^2=\frac{1225}{100}=\frac{49}{4} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\left(\frac{\frac{26}{9}+8}{\frac{26}{9}-2}\right)=\left(\frac{26+72}{26-18}\right)=\frac{98}{8}=\frac{49}{4} \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{x}=\frac{26}{9} \end{gathered}$$

Question 16:

Solve the following equation and verify your answer:
$\frac{9x-7}{3x+5}=\frac{3x-4}{x+6}$

Answer 16:

$$\begin{gathered} \frac{9\mathrm{x}-7}{3\mathrm{x}+5}=\frac{3\mathrm{x}-4}{\mathrm{x}+6} \\ \mathrm{or} 9{\mathrm{x}}^2-7\mathrm{x}+54\mathrm{x}-42=9{\mathrm{x}}^2-12\mathrm{x}+15\mathrm{x}-20 [\text{After} c\mathrm{ross} \mathrm{multipl}\text{ication}] \\ \mathrm{or} 9{\mathrm{x}}^2-9{\mathrm{x}}^2+47\mathrm{x}-3\mathrm{x}=-20+42 \\ \mathrm{or} 44\mathrm{x}=22 \\ \mathrm{or} \mathrm{x}=\frac{22}{44} \\ \mathrm{or} \mathrm{x}=\frac{1}{2} \\ \mathrm{Thus}, \mathrm{x}=\frac{1}{2} \mathrm{is} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting} \mathrm{x}=\frac{1}{2} \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{9\left(\frac{1}{2}\right)-7}{3\left(\frac{1}{2}\right)+5}=\frac{9-14}{3+10}=\frac{-5}{13} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{3\left(\frac{1}{2}\right)-4}{\frac{1}{2}+6}=\frac{3-8}{1+12}=\frac{-5}{13} \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{x}=\frac{1}{2} \end{gathered}$$

Question 17:

Solve the following equation and verify your answer:
$\frac{x+2}{x+5}=\frac{x}{x+6}$

Answer 17:

$$\begin{gathered} \frac{\mathrm{x}+2}{\mathrm{x}+5}=\frac{\mathrm{x}}{\mathrm{x}+6} \\ \mathrm{or} {\mathrm{x}}^2+2\mathrm{x}+6\mathrm{x}+12={\mathrm{x}}^2+5\mathrm{x} [\text{After} c\mathrm{ross} \mathrm{multipl}\text{ication}] \\ \mathrm{or} {\mathrm{x}}^2-{\mathrm{x}}^2+8\mathrm{x}-5\mathrm{x}=-12 \\ \mathrm{or} 3\mathrm{x}=-12 \\ \mathrm{or} \mathrm{x}=\frac{-12}{3} \\ \mathrm{or} \mathrm{x}=-4 \\ \mathrm{Thus}, \mathrm{x}=-4 \mathrm{is} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{given} \mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting} \mathrm{x}=-4 \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{-4+2}{-4+5}=-2 \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{-4}{-4+6}=-2 \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{x}=-4. \end{gathered}$$

Question 18:

Solve the following equation and verify your answer:
$\frac{2x-(7-5x)}{9x-(3+4x)}=\frac{7}{6}$

Answer 18:

$$\begin{gathered} \frac{2\mathrm{x}-(7-5\mathrm{x})}{9\mathrm{x}-(3+4\mathrm{x})}=\frac{7}{6} \\ \mathrm{or} \frac{7\mathrm{x}-7}{5\mathrm{x}-3}=\frac{7}{6} \\ \mathrm{or} 42\mathrm{x}-42=35\mathrm{x}-21 [\text{After c}\mathrm{ross} \mathrm{multipl}\text{ication}] \\ \mathrm{or} 42\mathrm{x}-35\mathrm{x}=-21+42 \\ \mathrm{or} 7\mathrm{x}=21 \\ \mathrm{or} \mathrm{x}=\frac{21}{7} \\ \mathrm{or} \mathrm{x}=3 \\ \mathrm{Thus}, \mathrm{x}=3 \mathrm{is} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting} \mathrm{x}=3 \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{2\times 3-(7-5\times 3)}{9\times 3-(3+4\times 3)}=\frac{6-(7-15)}{27-(3+12)}=\frac{6+8}{27-15}=\frac{14}{12}=\frac{7}{6} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{7}{6} \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{x}=3. \end{gathered}$$

Question 19:

Solve the following equation and verify your answer:
$\frac{15(2-x)-5(x+6)}{1-3x}=10$

Answer 19:

$$\begin{gathered} \frac{15(2-\mathrm{x})-5(\mathrm{x}+6)}{1-3\mathrm{x}}=10 \\ \mathrm{or} \frac{30-15\mathrm{x}-5\mathrm{x}-30}{1-3\mathrm{x}}=10 \\ \mathrm{or} \frac{-20\mathrm{x}}{1-3\mathrm{x}}=10 \\ \mathrm{or} 10-30\mathrm{x}=-20\mathrm{x} [\text{After c}\mathrm{ross} \mathrm{multipl}\text{ication}] \\ \mathrm{or} -20\mathrm{x}+30\mathrm{x}=10 \\ \mathrm{or} 10\mathrm{x}=10 \\ \mathrm{or} \mathrm{x}=1 \\ \mathrm{Thus}, \mathrm{x}=1 \mathrm{is} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting} \mathrm{x}=1 \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{15(2-1)-5(1+6)}{1-3\left(1\right)}=\frac{15-35}{-2}=\frac{-20}{-2}=10 \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=10 \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{x}=1. \end{gathered}$$

Question 20:

Solve the following equation and verify your answer:
$\frac{x+3}{x-3}+\frac{x+2}{x-2}=2$

Answer 20:

$$\begin{gathered} \frac{\mathrm{x}+3}{\mathrm{x}-3}+\frac{\mathrm{x}+2}{\mathrm{x}-2}=2 \\ \mathrm{or} \frac{\mathrm{x}+3}{\mathrm{x}-3}=2-\frac{\mathrm{x}+2}{\mathrm{x}-2} \\ \mathrm{or} \frac{\mathrm{x}+3}{\mathrm{x}-3}=\frac{2\mathrm{x}-4-\mathrm{x}-2}{\mathrm{x}-2} \\ \mathrm{or} \frac{\mathrm{x}+3}{\mathrm{x}-3}=\frac{\mathrm{x}-6}{\mathrm{x}-2} \\ \mathrm{or} {\mathrm{x}}^2-2\mathrm{x}+3\mathrm{x}-6={\mathrm{x}}^2-3\mathrm{x}-6\mathrm{x}+18 [\text{After c}\mathrm{ross} \mathrm{multipl}\text{ication}] \\ \mathrm{or} {\mathrm{x}}^2-{\mathrm{x}}^2+\mathrm{x}+9\mathrm{x}=18+6 \\ \mathrm{or} 10\mathrm{x}=24 \\ \mathrm{or} \mathrm{x}=\frac{24}{10} \\ \mathrm{or} \mathrm{x}=\frac{12}{5} \\ \mathrm{Thus}, \mathrm{x}=\frac{12}{5} \mathrm{is} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting} \mathrm{x}=\frac{12}{5} \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{\frac{12}{5}+3}{\frac{12}{5}-3}+\frac{\frac{12}{5}+2}{\frac{12}{5}-2}=\frac{12+15}{12-15}+\frac{12+10}{12-10}=\frac{27}{-3}+\frac{22}{2}=\frac{54-66}{-6}=\frac{-12}{-6}=2 \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=2 \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{x}=\frac{12}{5} \end{gathered}$$

Question 21:

Solve the following equation and verify your answer:
$\frac{(x+2)(2x-3)-2x^2+6}{x-5}=2$

Answer 21:

$$\begin{gathered} \frac{(\mathrm{x}+2)(2\mathrm{x}-3)-2{\mathrm{x}}^2+6}{\mathrm{x}-5}=2 \\ \mathrm{or} \frac{2{\mathrm{x}}^2+\mathrm{x}-6-2{\mathrm{x}}^2+6}{\mathrm{x}-5}=2 \\ \mathrm{or} \frac{\mathrm{x}}{\mathrm{x}-5}=2 \\ \mathrm{or} 2\mathrm{x}-10=\mathrm{x} [\text{After c}\mathrm{ross} \mathrm{multipl}\text{ication}] \\ \mathrm{or} 2\mathrm{x}-\mathrm{x}=10 \\ \mathrm{or} \mathrm{x}=10 \\ \mathrm{Thus}, \mathrm{x}=10 \mathrm{is} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting} \mathrm{x}=10 \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{(10+2)(2\times 10-3)-2\times {10}^2+6}{10-5}=\frac{12\times 17-200+6}{5}=\frac{10}{5}=2 \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=2 \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{x}=10. \end{gathered}$$

Question 22:

Solve the following equation and verify your answer:
$\frac{x^2-(x+1)(x+2)}{5x+1}=6$

Answer 22:

$$\begin{gathered} \frac{{\mathrm{x}}^2-(\mathrm{x}+1)(\mathrm{x}+2)}{5\mathrm{x}+1}=6 \\ \mathrm{or} \frac{{\mathrm{x}}^2-{\mathrm{x}}^2-2\mathrm{x}-\mathrm{x}-2}{5\mathrm{x}+1}=6 \\ \mathrm{or} \frac{-3\mathrm{x}-2}{5\mathrm{x}+1}=6 \\ \mathrm{or} 30\mathrm{x}+6=-3\mathrm{x}-2 [\text{After c}\mathrm{ross} \mathrm{multipl}\text{ication}] \\ \mathrm{or} 30\mathrm{x}+3\mathrm{x}=-2-6 \\ \mathrm{or} 33\mathrm{x}=-8\mathrm{or} \mathrm{x}=\frac{-8}{33} \\ \mathrm{Thus}, \mathrm{x}=\frac{-8}{33} \mathrm{is} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting} \mathrm{x}=\frac{-8}{33} \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{(\frac{-8}{33}{)}^2-(\frac{-8}{33}+1\left)\right(\frac{-8}{33}+2)}{5\left(\frac{-8}{33}\right)+1}=\frac{{\displaystyle \frac{64}{1089}}-{\displaystyle \frac{25}{33}}\times {\displaystyle \frac{58}{33}}}{{\displaystyle \frac{-40}{33}}+1}=\frac{{\displaystyle \frac{64}{1089}}-{\displaystyle \frac{1450}{1089}}}{{\displaystyle \frac{-7}{33}}}=\frac{{\displaystyle \frac{-1386}{1089}}}{{\displaystyle \frac{-7}{33}}}=\frac{42}{7}=\mathrm{R}.\mathrm{H}.\mathrm{S}.=6 \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{x}=\frac{-8}{33} \end{gathered}$$

Question 23:

Solve the following equation and verify your answer:
$\frac{(2x+3)-(5x-7)}{6x+11}=-\frac{8}{3}$

Answer 23:

$$\begin{gathered} \frac{(2\mathrm{x}+3)-(5\mathrm{x}-7)}{6\mathrm{x}+11}=\frac{-8}{3} \\ \mathrm{or} \frac{-3\mathrm{x}+10}{6\mathrm{x}+11}=\frac{-8}{3} \\ \mathrm{or} -9\mathrm{x}+30=-48\mathrm{x}-88 [\text{After c}\mathrm{ross} \mathrm{multipl}\text{ication}] \\ \mathrm{or} -9\mathrm{x}+48\mathrm{x}=-88-30 \\ \mathrm{or} 39\mathrm{x}=-118\mathrm{or} \mathrm{x}=\frac{-118}{39} \\ \mathrm{Thus}, \mathrm{x}=\frac{-118}{39} \mathrm{is} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting} \mathrm{x}=\frac{-118}{39} \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{-3\left(\frac{-118}{39}\right)+10}{6\left(\frac{-118}{39}\right)+11}=\frac{354+390}{-708+429}=\frac{744}{-279}=\frac{-8}{-3} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{-8}{3} \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{x}=\frac{-118}{39} \end{gathered}$$

Question 24:

Find a positive value of x for which the given equation is satisfied:
(i) $\frac{x^2-9}{5+x^2}=-\frac{5}{9}$
(ii) $\frac{y^2+4}{3y^2+7}=\frac{1}{2}$

Answer 24:

$$\begin{gathered} \left(\mathrm{i}\right) \frac{{\mathrm{x}}^2-9}{5+{\mathrm{x}}^2}=\frac{-5}{9} \\ \mathrm{or} 9{\mathrm{x}}^2-81=-25-5{\mathrm{x}}^2 [\text{After c}\mathrm{ross} \mathrm{multipl}\text{ication}] \\ \mathrm{or} 9{\mathrm{x}}^2+5{\mathrm{x}}^2=-25+81 \\ \mathrm{or} 14{\mathrm{x}}^2=56 \\ \mathrm{or} {\mathrm{x}}^2=\frac{56}{14} \\ \mathrm{or} {\mathrm{x}}^2=4=2^2 \\ \mathrm{or} \mathrm{x}=2 \\ \mathrm{Thus}, \mathrm{x}=2 \mathrm{is} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting} \mathrm{x}=2 \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{2^2-9}{5+2^2}=\frac{4-9}{5+4}=\frac{-5}{9} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{-5}{9} \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{x}=2. \\ \left(\mathrm{ii}\right) \frac{{\mathrm{y}}^2+4}{3{\mathrm{y}}^2+7}=\frac{1}{2} \\ \mathrm{or} 3{\mathrm{y}}^2+7=2{\mathrm{y}}^2+8 [\text{After} c\mathrm{ross} \mathrm{multipl}\text{ication}] \\ \mathrm{or} 3{\mathrm{y}}^2-2{\mathrm{y}}^2=8-7 \\ \mathrm{or} {\mathrm{y}}^2=1 \\ \mathrm{or} \mathrm{y}=1 \\ \mathrm{Thus}, \mathrm{y}=1 \mathrm{is} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting} \mathrm{y}=1 \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{1^2+4}{3(1{)}^2+7}=\frac{5}{10}=\frac{1}{2} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{1}{2} \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{y}=1. \end{gathered}$$