RD Sharma solution class 8 chapter 9 Linear Equation In One Variable Exercise 9.2

Exercise 9.2

Page-9.11

Question 1:

Solve each of the following equation and also check your result in each case:
$\frac{2x+5}{3}=3x-10$

Answer 1:

$$\begin{gathered} \frac{2\mathrm{x}+5}{3}=3\mathrm{x}-10 \\ \mathrm{or} 2\mathrm{x}+5=9\mathrm{x}-30 \\ \mathrm{or} 9\mathrm{x}-2\mathrm{x}=5+30 \\ \mathrm{or} 7\mathrm{x}=35 \\ \mathrm{or} \mathrm{x}=\frac{35}{7} \\ \mathrm{or} \mathrm{x}=5 \\ \mathrm{Verification}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{10+5}{3}=\frac{15}{3}=5 \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=15-10=5 \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{x}=5. \end{gathered}$$

Question 2:

Solve each of the following equation and also check your result in each case:
$\frac{a-8}{3}=\frac{a-3}{2}$

Answer 2:

$$\begin{gathered} \frac{\mathrm{a}-8}{3}=\frac{\mathrm{a}-3}{2} \\ \mathrm{or} 2\mathrm{a}-16=3\mathrm{a}-9 \\ \mathrm{or} 3\mathrm{a}-2\mathrm{a}=-16+9 \\ \mathrm{or} \mathrm{a}=-7 \\ \mathrm{Verification}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{-7-8}{3}=\frac{-15}{3}=-5 \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{-7-3}{2}=\frac{-10}{2}=-5 \end{gathered}$$

$\therefore$ L.H.S. = R.H.S. for a = $-$7

Question 3:

Solve each of the following equation and also check your result in each case:
$\frac{7y+2}{5}=\frac{6y-5}{11}$

Answer 3:

$$\begin{gathered} \frac{7\mathrm{y}+2}{5}=\frac{6\mathrm{y}-5}{11} \\ \mathrm{or} 77\mathrm{y}+22=30\mathrm{y}-25 \\ \mathrm{or} 77\mathrm{y}-30\mathrm{y}=-25-22 \\ \mathrm{or} 47\mathrm{y}=-47 \\ \mathrm{or} \mathrm{y}=\frac{-47}{47}=-1 \\ \mathrm{Verification}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{-7+2}{5}=\frac{-5}{5}=-1 \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{-6-5}{11}=\frac{-11}{11}=-1 \end{gathered}$$

$\therefore$ L.H.S. = R.H.S. for y = $-$1

Question 4:

Solve each of the following equation and also check your result in each case:
$x-2x+2-\frac{16}{3}x+5=3-\frac{7}{2}x$

Answer 4:

$$\begin{gathered} \mathrm{x}-2\mathrm{x}+2-\frac{16}{3}\mathrm{x}+5=3-\frac{7}{2}\mathrm{x} \\ \mathrm{or}\frac{3\mathrm{x}-6\mathrm{x}+6-16\mathrm{x}+15}{3}=\frac{6-7\mathrm{x}}{2} \\ \mathrm{or}\frac{-19\mathrm{x}+21}{3}=\frac{6-7\mathrm{x}}{2} \\ \mathrm{or} -38\mathrm{x}+42=18-21\mathrm{x} \\ \mathrm{or} -21\mathrm{x}+38\mathrm{x}=42-18 \\ \mathrm{or} 17\mathrm{x}=24 \\ \mathrm{or} \mathrm{x}=\frac{24}{17} \\ \mathrm{Check}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{24}{17}-2\times \frac{24}{17}+7-\frac{16}{3}\times \frac{24}{17}=\frac{-33}{17} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=3-\frac{7}{2}\times \frac{24}{17}=\frac{-33}{17} \end{gathered}$$

$\therefore$ L.H.S. = R.H.S. for x = $\frac{24}{17}$

Question 5:

Solve each of the following equation and also check your result in each case:
$\frac{1}{2}x+7x-6=7x+\frac{1}{4}$

Answer 5:

$$\begin{gathered} \frac{1}{2}\mathrm{x}+7\mathrm{x}-6=7\mathrm{x}+\frac{1}{4} \\ \mathrm{or} \frac{1}{2}\mathrm{x}+7\mathrm{x}-7\mathrm{x}=\frac{1}{4}+6 \\ \text{or}\frac{x}{2}=\frac{1+24}{4} \\ \mathrm{or} \frac{\mathrm{x}}{2}=\frac{25}{4} \\ \mathrm{or} \mathrm{x}=\frac{25}{2} \\ \mathrm{Check}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{1}{2}\times \frac{25}{2}+7\times \frac{25}{2}-6=\frac{351}{4} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=7\times \frac{25}{2}+\frac{1}{4}=\frac{351}{4} \end{gathered}$$

$\therefore$ L.H.S. = R.H.S. for x = $\frac{25}{2}$

Question 6:

Solve each of the following equation and also check your result in each case:
$\frac{3}{4}x+4x=\frac{7}{8}+6x-6$

Answer 6:

$$\begin{gathered} \frac{3}{4}\mathrm{x}+4\mathrm{x}=\frac{7}{8}+6\mathrm{x}-6 \\ \mathrm{or} \frac{3}{4}\mathrm{x}-2\mathrm{x}=\frac{7}{8}-6 \\ \mathrm{or} \frac{3\mathrm{x}-8\mathrm{x}}{4}=\frac{7-48}{8} \\ \mathrm{or}\frac{-5\mathrm{x}}{4}=\frac{-41}{8} \\ \mathrm{or} -40\mathrm{x}=-164 \\ \mathrm{or} \mathrm{x}=\frac{-164}{-40}=\frac{41}{10} \\ \mathrm{Check}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{3}{4}\times \frac{41}{10}+4\times \frac{41}{10}=\frac{123}{40}+\frac{164}{10}=\frac{123+656}{40}=\frac{779}{40} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{7}{8}+6\times \frac{41}{10}-6=\frac{7}{8}+\frac{246}{10}-6=\frac{35+984-240}{40}=\frac{779}{40} \end{gathered}$$

$\therefore$ L.H.S. = R.H.S. for x = $\frac{41}{10}$

Question 7:

Solve each of the following equation and also check your result in each case:
$\frac{7}{2}x-\frac{5}{2}x=\frac{20}{3}x+10$

Answer 7:

$$\begin{gathered} \frac{7}{2}\mathrm{x}-\frac{5}{2}\mathrm{x}=\frac{20}{3}\mathrm{x}+10 \\ \mathrm{or} \frac{7\mathrm{x}-5\mathrm{x}}{2}=\frac{20\mathrm{x}+30}{3} \\ \mathrm{or} 40\mathrm{x}+60=6\mathrm{x} \\ \mathrm{or} 34\mathrm{x}=-60 \\ \mathrm{or} \mathrm{x}=-\frac{60}{34}=-\frac{30}{17} \\ \mathrm{Check}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{7}{2}\times \frac{-30}{17}-\frac{5}{2}\times -\frac{30}{17}=-\frac{30}{17} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{20}{3}\times \frac{-30}{17}+10=-\frac{30}{17} \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{x}=-\frac{30}{17} \end{gathered}$$

Question 8:

Solve each of the following equation and also check your result in each case:
$\frac{6x+1}{2}+1=\frac{7x-3}{3}$

Answer 8:

$$\begin{gathered} \frac{6\mathrm{x}+1}{2}+1=\frac{7\mathrm{x}-3}{3} \\ \mathrm{or} \frac{6\mathrm{x}+1+2}{2}=\frac{7\mathrm{x}-3}{3} \\ \mathrm{or} 18\mathrm{x}+9=14\mathrm{x}-6 \\ \mathrm{or} 18\mathrm{x}-14\mathrm{x}=-6-9 \\ \mathrm{or} 4\mathrm{x}=-15 \\ \mathrm{or} \mathrm{x}=\frac{-15}{4} \\ \mathrm{Check}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{6\times {\displaystyle \frac{-15}{4}}+1}{2}+1=\frac{-45+2+4}{4}=\frac{-39}{4} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{7\times {\displaystyle \frac{-15}{4}}-3}{3}=\frac{-105-12}{12}=\frac{-39}{4} \end{gathered}$$

$\therefore$ L.H.S. = R.H.S. for x = $\frac{-15}{4}$

Question 9:

Solve each of the following equation and also check your result in each case:
$\frac{3a-2}{3}+\frac{2a+3}{2}=a+\frac{7}{6}$

Answer 9:

$$\begin{gathered} \frac{3\mathrm{a}-2}{3}+\frac{2\mathrm{a}+3}{2}=\mathrm{a}+\frac{7}{6} \\ \mathrm{or} \frac{6\mathrm{a}-4+6\mathrm{a}+9}{6}=\mathrm{a}+\frac{7}{6} \\ \mathrm{or} 12\mathrm{a}+5=6\mathrm{a}+7 \\ \mathrm{or} 6\mathrm{a}=7-5 \\ \mathrm{or} \mathrm{a}=\frac{2}{6}=\frac{1}{3} \\ \mathrm{Check}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{3\times {\displaystyle \frac{1}{3}}-2}{3}+\frac{2\times {\displaystyle \frac{1}{3}}+3}{2}=\frac{-1}{3}+\frac{11}{6}=\frac{9}{6}=\frac{3}{2} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{1}{3}+\frac{7}{6}=\frac{9}{6}=\frac{3}{2} \end{gathered}$$

$\therefore$ L.H.S. = R.H.S. for a = $\frac{1}{3}$

Question 10:

Solve each of the following equation and also check your result in each case:
$x-\frac{(x-1)}{2}=1-\frac{(x-2)}{3}$

Answer 10:

$$\begin{gathered} \mathrm{x}-\frac{\mathrm{x}-1}{2}=1-\frac{\mathrm{x}-2}{3} \\ \mathrm{or} \frac{2\mathrm{x}-\mathrm{x}+1}{2}=\frac{3-\mathrm{x}+2}{3} \\ \mathrm{or} \frac{\mathrm{x}+1}{2}=\frac{5-\mathrm{x}}{3} \\ \mathrm{or} 3\mathrm{x}+3=10-2\mathrm{x} \\ \mathrm{or} 5\mathrm{x}=10-3 \\ \mathrm{or} \mathrm{x}=\frac{7}{5} \\ \mathrm{Check}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{7}{5}-\frac{\frac{7}{5}-1}{2}=\frac{7}{5}-\frac{1}{5}=\frac{6}{5} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=1-\frac{\frac{7}{5}-2}{3}=1-\frac{-3}{15}=\frac{6}{5} \end{gathered}$$

$\therefore$ L.H.S. = R.H.S. for x = $\frac{7}{5}$

Question 11:

Solve each of the following equation and also check your result in each case:
$\frac{3x}{4}-\frac{(x-1)}{2}=\frac{(x-2)}{3}$

Answer 11:

$$\begin{gathered} \frac{3\mathrm{x}}{4}-\frac{\mathrm{x}-1}{2}=\frac{\mathrm{x}-2}{3} \\ \mathrm{or} \frac{3\mathrm{x}-2\mathrm{x}+2}{4}=\frac{\mathrm{x}-2}{3} \\ \mathrm{or} 4\mathrm{x}-8=3\mathrm{x}+6 \\ \mathrm{or} \mathrm{x}=14 \\ \mathrm{Check}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{3\times 14}{4}-\frac{14-1}{2}=\frac{21}{2}-\frac{13}{2}=\frac{8}{2}=4 \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{14-2}{3}=\frac{12}{3}=4 \end{gathered}$$

$\therefore$ L.H.S. = R.H.S. for x = 14

Question 12:

Solve each of the following equation and also check your result in each case:
$\frac{5x}{3}-\frac{(x-1)}{4}=\frac{(x-3)}{5}$

Answer 12:

$$\begin{gathered} \frac{5\mathrm{x}}{3}-\frac{\mathrm{x}-1}{4}=\frac{\mathrm{x}-3}{5} \\ \mathrm{or} \frac{20\mathrm{x}-3\mathrm{x}+3}{12}=\frac{\mathrm{x}-3}{5} \\ \mathrm{or} \frac{17\mathrm{x}+3}{12}=\frac{\mathrm{x}-3}{5} \\ \mathrm{or} 85\mathrm{x}+15=12\mathrm{x}-36 \\ \mathrm{or} 73\mathrm{x}=-51 \\ \mathrm{or} \mathrm{x}=\frac{-51}{73} \\ \mathrm{Check}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{5\times \frac{-51}{73}}{3}-\frac{\frac{-51}{73}-1}{4}=\frac{-255}{219}-\frac{-124}{292}=\frac{-54}{73} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{\frac{-51}{73}-3}{5}=\frac{-54}{73} \end{gathered}$$

$\therefore$ L.H.S. = R.H.S. for x = $\frac{-51}{73}$

Question 13:

Solve each of the following equation and also check your result in each case:
$\frac{(3x+1)}{16}+\frac{(2x-3)}{7}=\frac{(x+3)}{8}+\frac{(3x-1)}{14}$

Answer 13:

$$\begin{gathered} \frac{3\mathrm{x}+1}{16}+\frac{2\mathrm{x}-3}{7}=\frac{\mathrm{x}+3}{8}+\frac{3\mathrm{x}-1}{14} \\ \mathrm{or} \frac{3\mathrm{x}+1}{16}-\frac{\mathrm{x}+3}{8}=\frac{3\mathrm{x}-1}{14}-\frac{2\mathrm{x}-3}{7} \\ \mathrm{or} \frac{3\mathrm{x}+1-2\mathrm{x}-6}{16}=\frac{3\mathrm{x}-1-4\mathrm{x}+6}{14} \\ \mathrm{or} \frac{\mathrm{x}-5}{8}=\frac{-\mathrm{x}+5}{7} \\ \mathrm{or} 7\mathrm{x}-35=-8\mathrm{x}+40 \\ \mathrm{or} 15\mathrm{x}=75 \\ \mathrm{or} \mathrm{x}=\frac{75}{15}=5 \\ \mathrm{Check}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{3\times 5+1}{16}+\frac{2\times 5-3}{7}=\frac{16}{16}+\frac{7}{7}=2 \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{5+3}{8}+\frac{3\times 5-1}{14}=\frac{8}{8}+\frac{14}{14}=2 \end{gathered}$$

$\therefore$ L.H.S. = R.H.S. for x = 5

Question 14:

Solve each of the following equation and also check your result in each case:
$\frac{(1-2x)}{7}-\frac{(2-3x)}{8}=\frac{3}{2}+\frac{x}{4}$

Answer 14:

$$\begin{gathered} \frac{1-2\mathrm{x}}{7}-\frac{2-3\mathrm{x}}{8}=\frac{3}{2}+\frac{\mathrm{x}}{4} \\ \mathrm{or} \frac{1-2\mathrm{x}}{7}=\frac{3}{2}+\frac{\mathrm{x}}{4}+\frac{2-3\mathrm{x}}{8} \\ \mathrm{or} \frac{1-2\mathrm{x}}{7}=\frac{12+2\mathrm{x}+2-3\mathrm{x}}{8} \\ \mathrm{or} \frac{1-2\mathrm{x}}{7}=\frac{14-x}{8} \\ \mathrm{or} 8-16\mathrm{x}=98-7x \\ \mathrm{or} -16\mathrm{x}+7\mathrm{x}=98-8 \\ \mathrm{or} \mathrm{x}=\frac{-90}{9} \\ =-10 \\ \mathrm{Check}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{1-2\times \left(-10\right)}{7}-\frac{2-3\times \left(-10\right)}{8}=\frac{1+20}{7}-\frac{2+30}{8}=3-4=-1 \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{3}{2}+\frac{-10}{4}=\frac{3}{2}+\frac{-5}{2}=\frac{3-5}{2}=-1 \end{gathered}$$

$\therefore$ L.H.S. = R.H.S. for x = $-$10

Question 15:

Solve each of the following equation and also check your result in each case:
$\frac{9x+7}{2}-\left(x-\frac{x-2}{7}\right)=36$

Answer 15:

$$\begin{gathered} \frac{9\mathrm{x}+7}{2}-(\mathrm{x}-\frac{\mathrm{x}-2}{7})=36 \\ \mathrm{or} \frac{63\mathrm{x}+49-14\mathrm{x}+2\mathrm{x}-4}{14}=36 \\ \mathrm{or} \frac{51\mathrm{x}+45}{14}=36 \\ \mathrm{or} 51\mathrm{x}+45=504 \\ \mathrm{or} 51\mathrm{x}=504-45 \\ \mathrm{or} \mathrm{x}=\frac{459}{51}=9 \\ \mathrm{Thus}, \mathrm{x}=9 \mathrm{is} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting} \mathrm{x}=9 \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{9\times 9+7}{2}-(9-\frac{9-2}{7})=\frac{88}{2}-9+\frac{7}{7}=44-9+1=36 \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=36 \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{x}=9. \end{gathered}$$

Question 16:

Solve each of the following equation and also check your result in each case:
0.18(5x − 4) = 0.5x + 0.8

Answer 16:

$$\begin{gathered} 0.18(5\mathrm{x}-4)=0.5\mathrm{x}+0.8 \\ \mathrm{or} 0.9\mathrm{x}-0.72=0.5\mathrm{x}+0.8 \\ \mathrm{or} 0.9\mathrm{x}-0.5\mathrm{x}=0.8+0.72 \\ \mathrm{or} 0.4\mathrm{x}=1.52 \\ \mathrm{or} \mathrm{x}=\frac{1.52}{0.4} \\ \text{or} \mathrm{x}=3.8 \\ \mathrm{Thus}, \mathrm{x}=3.8 \mathrm{is} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting} \mathrm{x}=3.8 \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=0.18(5\times 3.8-4)=0.18\times 15=2.7 \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=0.5\times 3.8+0.8=2.7 \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{x}=3.8. \end{gathered}$$

Question 17:

Solve each of the following equation and also check your result in each case:
$\frac{2}{3x}-\frac{3}{2x}=\frac{1}{12}$

Answer 17:

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

Question 18:

Solve each of the following equation and also check your result in each case:
$\frac{4x}{9}+\frac{1}{3}+\frac{13}{108}x=\frac{8x+19}{18}$

Answer 18:

$$\begin{gathered} \frac{4\mathrm{x}}{9}+\frac{1}{3}+\frac{13}{108}\mathrm{x}=\frac{8\mathrm{x}+19}{18} \\ \mathrm{or} \frac{48\mathrm{x}+36+13\mathrm{x}}{108}=\frac{8\mathrm{x}+19}{18} \\ \mathrm{or} \frac{61\mathrm{x}+36}{108}=\frac{8\mathrm{x}+19}{18} \\ \mathrm{or} 61\mathrm{x}+36=6(8\mathrm{x}+19) [\mathrm{Multiply}\text{ing} \mathrm{both} s\mathrm{ides} \mathrm{by} 108] \\ \mathrm{or} 61\mathrm{x}+36=48\mathrm{x}+114 \\ \mathrm{or} 61\mathrm{x}-48\mathrm{x}=114-36 \\ \mathrm{or} 13\mathrm{x}=78 \\ \mathrm{or} \mathrm{x}=\frac{78}{13} \\ \mathrm{or} \mathrm{x}=6 \\ \mathrm{Thus}, \mathrm{x}=6 \mathrm{is} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting} \mathrm{x}=6 \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{4\times 6}{9}+\frac{1}{3}+\frac{13}{108}\times 6=\frac{24}{9}+\frac{1}{3}+\frac{13}{18}=\frac{48+6+13}{18}=\frac{67}{18} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{8\times 6+19}{18}=\frac{67}{18} \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{x}=6. \end{gathered}$$

Page-9.12

Question 19:

Solve each of the following equation and also check your result in each case:
$\frac{(45-2x)}{15}-\frac{(4x+10)}{5}=\frac{(15-14x)}{9}$

Answer 19:

$$\begin{gathered} \frac{45-2\mathrm{x}}{15}-\frac{4\mathrm{x}+10}{5}=\frac{15-14\mathrm{x}}{9} \\ \mathrm{or} \frac{45-2\mathrm{x}-12\mathrm{x}-30}{15}=\frac{15-14\mathrm{x}}{9} \\ \mathrm{or} \frac{15-14\mathrm{x}}{5}=\frac{15-14\mathrm{x}}{3} [\mathrm{Multiply}\text{ing} \mathrm{both} \mathrm{sides} \mathrm{by} 3] \\ \mathrm{or} 45-42\mathrm{x}=75-70\mathrm{x} [\text{After c}\mathrm{ross} \mathrm{multipl}\text{ication}] \\ \mathrm{or} 70\mathrm{x}-42\mathrm{x}=75-45 \\ \mathrm{or} 28\mathrm{x}=30 \\ \mathrm{or} \mathrm{x}=\frac{30}{28} \\ \mathrm{or} \mathrm{x}=\frac{15}{14} \\ \mathrm{Thus}, \mathrm{x}=\frac{15}{14} \mathrm{is} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting} \mathrm{x}=\frac{15}{14} \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{45-2\times {\displaystyle \frac{15}{14}}}{15}-\frac{4\times {\displaystyle \frac{15}{14}}+10}{5}=\frac{45\times 7-15}{105}-\frac{30+70}{35}=\frac{300}{105}-\frac{100}{35}=0 \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{15-14\times {\displaystyle \frac{15}{14}}}{9}=0 \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{x}=\frac{15}{14} \end{gathered}$$

Question 20:

Solve each of the following equation and also check your result in each case:
$5\left(\frac{7x+5}{3}\right)-\frac{23}{3}=13-\frac{4x-2}{3}$

Answer 20:

$$\begin{gathered} 5\left(\frac{7\mathrm{x}+5}{3}\right)-\frac{23}{3}=13-\frac{4\mathrm{x}-2}{3} \\ \mathrm{or} \frac{35\mathrm{x}+25}{3}+\frac{4\mathrm{x}-2}{3}=13+\frac{23}{3} \\ \mathrm{or} \frac{35\mathrm{x}+25+4\mathrm{x}-2}{3}=\frac{39+23}{3} \\ \mathrm{or} 39\mathrm{x}+23=62 [\mathrm{Multiply}\text{ing} \mathrm{both} \mathrm{sides} \mathrm{by} 3] \\ \mathrm{or} 39\mathrm{x}=62-23 \\ \mathrm{or} \mathrm{x}=\frac{39}{39} \\ \mathrm{or} \mathrm{x}=1 \\ \mathrm{Thus}, \mathrm{x}=1 \mathrm{is} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting} \mathrm{x}=1 \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=5\left(\frac{7\times 1+5}{3}\right)-\frac{23}{3}=\frac{60}{3}-\frac{23}{3}=\frac{37}{3} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=13-\frac{4\times 1-2}{3}=\frac{39-2}{3}=\frac{37}{3} \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{x}=1. \end{gathered}$$

Question 21:

Solve each of the following equation and also check your result in each case:
$\frac{7x-1}{4}-\frac{1}{3}\left(2x-\frac{1-x}{2}\right)=\frac{10}{3}$

Answer 21:

$$\begin{gathered} \frac{7\mathrm{x}-1}{4}-\frac{1}{3}(2\mathrm{x}-\frac{1-\mathrm{x}}{2})=\frac{10}{3} \\ \mathrm{or} \frac{7\mathrm{x}-1}{4}-\frac{2\mathrm{x}}{3}+\frac{1-\mathrm{x}}{6}=\frac{10}{3} \\ \mathrm{or} \frac{21\mathrm{x}-3-8\mathrm{x}+2-2\mathrm{x}}{12}=\frac{10}{3} \\ \mathrm{or} 11\mathrm{x}-1=40 [\mathrm{Multiply}\text{ing} \mathrm{both} \mathrm{sides} \mathrm{by} 12] \\ \mathrm{or} 11\mathrm{x}=40+1 \\ \mathrm{or} \mathrm{x}=\frac{41}{11} \\ \mathrm{Thus}, \mathrm{x}=\frac{41}{11} \mathrm{is} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting} \mathrm{x}=\frac{41}{11} \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{7\times {\displaystyle \frac{41}{11}}-1}{4}-\frac{1}{3}(2\times \frac{41}{11}-\frac{1-{\displaystyle \frac{41}{11}}}{2})=\frac{276}{44}-\frac{82}{33}+\frac{-30}{66}=\frac{10}{3} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{10}{3} \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{x}=\frac{41}{11} \end{gathered}$$

Question 22:

Solve each of the following equation and also check your result in each case:
$\frac{0.5(x-0.4)}{0.35}-\frac{0.6(x-2.71)}{0.42}=x+6.1$

Answer 22:

$$\begin{gathered} \frac{0.5(\mathrm{x}-0.4)}{0.35}-\frac{0.6(\mathrm{x}-2.71)}{0.42}=\mathrm{x}+6.1 \\ \text{or} \frac{(\mathrm{x}-0.4)}{0.7}-\frac{(\mathrm{x}-2.71)}{0.7}=\mathrm{x}+6.1 \\ \text{or}\frac{\mathrm{x}-0.4-\mathrm{x}+2.71}{0.7}=\mathrm{x}+6.1 \\ \mathrm{or} -0.4+2.71=0.7x+4.27 \\ \mathrm{or} 0.7\mathrm{x}=2.71-0.4-4.27 \\ \mathrm{or} \mathrm{x}=\frac{-1.96}{0.7}=-2.8 \\ \mathrm{Thus}, \mathrm{x}=-2.8 \mathrm{is} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting} \mathrm{x}=-2.8 \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{0.5(-2.8-0.4)}{0.35}-\frac{0.6(-2.8-2.71)}{0.42}=\frac{-1.6}{0.35}+\frac{3.306}{0.42}=-4.571+7.871=3.3 \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=-2.8+6.1=-3.3 \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{x}=-2.8 \end{gathered}$$

Question 23:

Solve each of the following equation and also check your result in each case:
$6.5x+\frac{19.5x-32.5}{2}=6.5x+13+\left(\frac{13x-26}{2}\right)$

Answer 23:

$$\begin{gathered} 6.5\mathrm{x}+\frac{19.5\mathrm{x}-32.5}{2}=6.5\mathrm{x}+13+\frac{13\mathrm{x}-26}{2} \\ \mathrm{or} \frac{19.5\mathrm{x}-32.5}{2}-\frac{13\mathrm{x}-26}{2}=13 \\ \mathrm{or} \frac{19.5\mathrm{x}-32.5-13\mathrm{x}+26}{2}=13 \\ \mathrm{or} 6.5\mathrm{x}-6.5=26 [\text{After} c\mathrm{ross} \mathrm{multipl}\text{ication}] \\ \mathrm{or} 6.5\mathrm{x}=26+6.5 \\ \mathrm{or} \mathrm{x}=\frac{32.5}{6.5}=5 \\ \mathrm{Thus}, \mathrm{x}=5 \mathrm{is} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting} \mathrm{x}=5 \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=6.5\times 5+\frac{19.5\times 5-32.5}{2}=65 \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=6.5\times 5+13+\frac{13\times 5-26}{2}=65 \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{x}=5. \end{gathered}$$

Question 24:

Solve each of the following equation and also check your result in each case:
(3x − 8)(3x + 2) − (4x − 11)(2x + 1) = (x − 3)(x + 7)

Answer 24:

$$\begin{gathered} (3\mathrm{x}-8)(3\mathrm{x}+2)-(4\mathrm{x}-11)(2\mathrm{x}+1)=(\mathrm{x}-3)(\mathrm{x}+7) \\ \mathrm{or} 9{\mathrm{x}}^2+6\mathrm{x}-24\mathrm{x}-16-8{\mathrm{x}}^2-4\mathrm{x}+22\mathrm{x}+11={\mathrm{x}}^2+7\mathrm{x}-3\mathrm{x}-21 \\ \mathrm{or} {\mathrm{x}}^2-5={\mathrm{x}}^2+4\mathrm{x}-21 \\ \mathrm{or} 4\mathrm{x}=-5+21 \\ \mathrm{or} \mathrm{x}=\frac{16}{4}=4 \\ \mathrm{Thus}, \mathrm{x}=4 \mathrm{is} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting} \mathrm{x}=4 \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=(3\times 4-8)(3\times 4+2)-(4\times 4-11)(2\times 4+1)=4\times 14-5\times 9=11 \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=(4-3)(4+7)=11 \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{x}=4. \end{gathered}$$

Question 25:

Solve each of the following equation and also check your result in each case:
[(2x + 3) + (x + 5)]² + [(2x + 3) − (x + 5)]² = 10x² + 92

Answer 25:

$$\begin{gathered} \left[\right(2\mathrm{x}+3)+(\mathrm{x}+5){]}^2+[(2\mathrm{x}+3)-(\mathrm{x}+5){]}^2=10{\mathrm{x}}^2+92 \\ \mathrm{or} (3\mathrm{x}+8{)}^2+(\mathrm{x}-2{)}^2=10{\mathrm{x}}^2+92 \\ \mathrm{or} 9{\mathrm{x}}^2+48\mathrm{x}+64+{\mathrm{x}}^2-4\mathrm{x}+4=10{\mathrm{x}}^2+92 [ (\mathrm{a}+\mathrm{b}{)}^2={\mathrm{a}}^2+{\mathrm{b}}^2+2\mathrm{ab} \mathrm{and} (\mathrm{a}-\mathrm{b}{)}^2={\mathrm{a}}^2+{\mathrm{b}}^2-2\mathrm{ab} ] \\ \mathrm{or} 10{\mathrm{x}}^2-10{\mathrm{x}}^2+44\mathrm{x}=92-68 \\ \mathrm{or} \mathrm{x}=\frac{24}{44} \\ \mathrm{or} \mathrm{x}=\frac{6}{11} \\ \mathrm{Thus}, \mathrm{x}=\frac{6}{11} \mathrm{is} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting} \mathrm{x}=\frac{6}{11} \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.={\left[\left(2\times \frac{6}{11}+3\right)+\left(\frac{6}{11}+5\right)\right]}^2+{\left[\left(2\times \frac{6}{11}+3\right)-\left(\frac{6}{11}+5\right)\right]}^2 \\ = {\left[\left(\frac{45}{11}\right)+\left(\frac{61}{11}\right)\right]}^2+{\left[\left(\frac{45}{11}\right)-\left(\frac{61}{11}\right)\right]}^2 \\ ={\left({\displaystyle \frac{106}{11}}\right)}^2+{\left({\displaystyle \frac{-16}{11}}\right)}^2 \\ =\frac{11492}{121} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=10\times {\left(\frac{6}{11}\right)}^2+92=\frac{360}{121}+92=\frac{11492}{121} \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{for} \mathrm{x}=\frac{6}{11} \end{gathered}$$