Exercise 2.2 (Revised) - Chapter 2 - Linear Equations In One Variable - Ncert Solutions class 8 - Maths

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NCERT Solutions for Class 8 Maths Chapter 2 - Linear Equations in One Variable

Solve the following linear equations.
Ex 2.2 Question 1.

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

Answer.

$
\begin{aligned}
& \text {} \frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4} \\
& \Rightarrow \frac{x}{2}-\frac{x}{3}=\frac{1}{4}+\frac{1}{5} \\
& \Rightarrow \frac{3 x-2 x}{6}=\frac{5+4}{20} \\
& \Rightarrow \frac{x}{6}=\frac{9}{20} \\
& \Rightarrow x=\frac{9 \times 6}{20}=\frac{27}{10}
\end{aligned}
$

To check:
$
\begin{aligned}
& \frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4} \\
& \Rightarrow \frac{27}{10 \times 2}-\frac{1}{5}=\frac{27}{10 \times 3}+\frac{1}{4}
\end{aligned}
$

$\begin{aligned}
& \Rightarrow \frac{27}{20}-\frac{1}{5}=\frac{9}{10}+\frac{1}{4} \\
& \Rightarrow \frac{27-4}{20}=\frac{18+5}{20}
\end{aligned}$

$
\begin{aligned}
& \Rightarrow \frac{23}{20}=\frac{23}{20} \\
& \Rightarrow \text { L.H.S. }=\text { R. H. S. }
\end{aligned}
$

Therefore, it is correct.

Ex 2.2 Question 2.

$
\begin{aligned}
& \text {} \frac{n}{2}-\frac{3 n}{4}+\frac{5 n}{6}=21 \\
& \text { Answer. } \frac{n}{2}-\frac{3 n}{4}+\frac{5 n}{6}=21 \\
& \Rightarrow \frac{6 n-9 n+10 n}{12}=21 \\
& \Rightarrow \frac{7 n}{12}=21 \\
& \Rightarrow n=\frac{21 \times 12}{7} \\
& \Rightarrow n=36
\end{aligned}
$

To check:

$
\begin{aligned}
& \frac{n}{2}-\frac{3 n}{4}+\frac{5 n}{6}=21 \\
& \Rightarrow \frac{36}{2}-\frac{3 \times 36}{4}+\frac{5 \times 36}{6}=21 \\
& \Rightarrow 18-27+30=21 \\
& \Rightarrow 21=21 \\
& \Rightarrow \text { L.H.S. }=\text { R.H.S. }
\end{aligned}
$

Therefore, it is correct.

Ex 2.2 Question 3.

$x+7-\frac{8 x}{3}=\frac{17}{6}-\frac{5 x}{2}$

Answer.

$x+7-\frac{8 x}{3}=\frac{17}{6}-\frac{5 x}{2}$
$\Rightarrow \frac{x}{1}-\frac{8 x}{3}+\frac{5 x}{2}=\frac{17}{6}-\frac{7}{1}$
$\Rightarrow \frac{6 x-16 x+15 x}{6}=\frac{17-42}{6}$
$\Rightarrow \frac{5 x}{6}=\frac{-25}{6}$
$\Rightarrow x=\frac{-25 \times 6}{6 \times 5}$
$\Rightarrow x=-5$
To check:
$
\begin{aligned}
& x+7-\frac{8 x}{3}=\frac{17}{6}-\frac{5 x}{2} \\
& \Rightarrow-5+7-\frac{8 \times(-5)}{3}=\frac{17}{6}-\frac{5 \times(-5)}{2}
\end{aligned}
$

$\begin{aligned}
& \Rightarrow 2+\frac{40}{3}=\frac{17}{6}+\frac{25}{2} \\
& \Rightarrow \frac{6+40}{3}=\frac{17+75}{6} \\
& \Rightarrow \frac{46}{3}=\frac{92}{6} \\
& \Rightarrow \frac{46}{3}=\frac{46}{3}
\end{aligned}$

$\Rightarrow$ L.H.S. $=$ R. H. S.
Therefore, it is correct.
Ex 2.2 Question 4.

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

Answer.

$\frac{x-5}{3}=\frac{x-3}{5}$
$
\begin{aligned}
& \Rightarrow 5 \times(x-5)=3(x-3) \\
& \Rightarrow 5 x-25=3 x-9 \\
& \Rightarrow 5 x-3 x=-9+25 \\
& \Rightarrow 2 x=16 \\
& \Rightarrow x=\frac{16}{2}=8
\end{aligned}
$

To check:
$
\frac{x-5}{3}=\frac{x-3}{5}
$

$\Rightarrow$ L.H.S. $=$ R. H. S.
Therefore, it is correct.

Ex 2.2 Question 5.

$\frac{3 t-2}{4}-\frac{2 t+3}{3}=\frac{2}{3}-t$

Answer.

$\frac{3 t-2}{4}-\frac{2 t+3}{3}=\frac{2}{3}-t$
$
\begin{aligned}
& \Rightarrow \frac{3 t-2}{4}-\frac{2 t+3}{3}+t=\frac{2}{3} \\
& \Rightarrow \frac{3(3 t-2)-4(2 t+3)+12 t}{12}=\frac{2}{3} \\
& \Rightarrow \frac{9 t-6-8 t-12+12 t}{12}=\frac{2}{3} \\
& \Rightarrow \frac{13 t-18}{12}=\frac{2}{3} \\
& \Rightarrow 3 \times(13 t-18)=2 \times 12 \\
& \Rightarrow 39 t-54=24 \\
& \Rightarrow 39 t=24+54 \\
& \Rightarrow 39 t=78
\end{aligned}
$

$
\Rightarrow t=\frac{78}{39}=2
$

To check:
$
\begin{aligned}
& \frac{3 t-2}{4}-\frac{2 t+3}{3}=\frac{2}{3}-t \\
\Rightarrow & \frac{3 \times 2-2}{4}-\frac{2 \times 2+3}{3}=\frac{2}{3}-2 \\
\Rightarrow & \frac{6-2}{4}-\frac{4+3}{3}=\frac{2-6}{3}
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow \frac{4}{4}-\frac{7}{3}=\frac{-4}{3} \\
& \Rightarrow \frac{1}{1}-\frac{7}{3}=\frac{-4}{3} \\
& \Rightarrow \frac{3-7}{3}=\frac{-4}{3} \\
& \Rightarrow \frac{-4}{3}=\frac{-4}{3} \\
& \Rightarrow \text { L.H.S. }=\text { R. H.S. }
\end{aligned}
$

Therefore, it is correct.
Ex 2.2 Question 6.

$m-\frac{m-1}{2}=1-\frac{m-2}{3}$

Answer.

$m-\frac{m-1}{2}=1-\frac{m-2}{3}$
$
\Rightarrow \frac{m}{1}-\frac{m-1}{2}+\frac{m-2}{3}=1
$

$\begin{aligned}
& \Rightarrow \frac{6 m-3(m-1)+2(m-2)}{6}=1 \\
& \Rightarrow \frac{6 m-3 m+3+2 m-4}{6}=1 \\
& \Rightarrow \frac{5 m-1}{6}=1 \\
& \Rightarrow 5 m-1=6 \\
& \Rightarrow 5 m=6+1 \\
& \Rightarrow 5 m=7
\end{aligned}$

$
\Rightarrow m=\frac{7}{5}
$

To check:

$
\begin{aligned}
& m-\frac{m-1}{2}=1-\frac{m-2}{3} \\
\Rightarrow & \frac{7}{5}-\frac{\frac{7}{5}-1}{2}=1-\frac{\frac{7}{5}-2}{3} \\
\Rightarrow & \frac{7}{5}-\frac{\frac{7-5}{5}}{2}=1-\frac{\frac{7-10}{5}}{3} \\
\Rightarrow & \frac{7}{5}-\frac{2}{5 \times 2}=1-\frac{-3}{5 \times 3} \\
\Rightarrow & \frac{7}{5}-\frac{1}{5}=1+\frac{1}{5} \\
\Rightarrow & \frac{7-1}{5}=\frac{5+1}{5} \\
\Rightarrow & \frac{6}{5}=\frac{6}{5} \\
\Rightarrow & \text { L.H.S. }=\text { R. H.S. }
\end{aligned}
$

Therefore, it is correct.

Simplify and solve the following linear equation.
Ex 2.2 Question 7.

$3(t-3)=5(2 t+1)$

Answer.

$3(t-3)=5(2 t+1)$
$
\Rightarrow 3 t-9=10 t+5
$

$
\begin{aligned}
& \Rightarrow 3 t-10 t=5+9 \\
& \Rightarrow-7 t=14 \\
& \Rightarrow t=\frac{14}{-7} \\
& \Rightarrow t=-2
\end{aligned}
$

To check:
$
\begin{aligned}
& 3(t-3)=5(2 t+1) \\
& \Rightarrow 3(-2-3)=5\{2 \times(-2)+1\} \\
& \Rightarrow 3 \times-5=5(-4+1) \\
& \Rightarrow-15=5 \times(-3) \\
& \Rightarrow-15=-15 \\
& \Rightarrow \text { L.H.S. }=\text { R. H.S. }
\end{aligned}
$

Therefore, it is correct.

Ex 2.2 Question 8.

$\text {} 15(y-4)-2(y-9)+5(y+6)=0$

$\begin{aligned}
& \text { Answer. } 15(y-4)-2(y-9)+5(y+6)=0 \\
& \Rightarrow 15 y-60-2 y+18+5 y+30=0 \\
& \Rightarrow 18 y-12=0 \\
& \Rightarrow 18 y=12 \\
& \Rightarrow y=\frac{12}{18}
\end{aligned}$

$
\Rightarrow y=\frac{2}{3}
$

To check:
$
\begin{aligned}
& 15(y-4)-2(y-9)+5(y+6)=0 \\
& \Rightarrow 15\left(\frac{2}{3}-4\right)-2\left(\frac{2}{3}-9\right)+5\left(\frac{2}{3}+6\right)=0 \\
& \Rightarrow 15\left(\frac{2-12}{3}\right)-2\left(\frac{2-27}{3}\right)+5\left(\frac{2+18}{3}\right)=0 \\
& \Rightarrow 15 \times \frac{-10}{3}-2 \times \frac{-25}{3}+5 \times \frac{20}{3}=0 \\
& \Rightarrow-50+\frac{50}{3}+\frac{100}{3}=0 \\
& \Rightarrow-50+\frac{50+100}{3}=0 \\
& \Rightarrow-50+\frac{150}{3}=0
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow-50+50=0 \\
& \Rightarrow 0=0 \\
& \Rightarrow \text { L.H.S. = R. H. S. }
\end{aligned}
$

Therefore, it is correct.
Ex 2.2 Question 9.

$3(5 z-7)-2(9 z-11)=4(8 z-13)-17$

Answer.

$3(5 z-7)-2(9 z-11)=4(8 z-13)-17$
$
\Rightarrow 15 z-21-18 z+22=32 z-52-17
$

$
\begin{aligned}
& \Rightarrow-3 z+1=32 z-69 \\
& \Rightarrow-3 z-32 z=-69-1 \\
& \Rightarrow-35 z=-70 \\
& \Rightarrow z=\frac{-70}{-35}=2
\end{aligned}
$

To check:
$
\begin{aligned}
& 3(5 z-7)-2(9 z-11)=4(8 z-13)-17 \\
\Rightarrow & 3(5 \times 2-7)-2(9 \times 2-11)=4(8 \times 2-13)-17 \\
\Rightarrow & 3(10-7)-2(18-11)=4(16-13)-17 \\
\Rightarrow & 3 \times 3-2 \times 7=4 \times 3-17 \\
\Rightarrow & 9-14=12-17 \\
\Rightarrow & -5=-5 \\
\Rightarrow & \text { L.H.S. }=\text { R. H.S. }
\end{aligned}
$

Therefore, it is correct.

Ex 2.2 Question 10.

$\text {} 0.25(4 f-3)=0.05(10 f-9)$

Answer.

$\begin{aligned}
& \text {} 0.25(4 f-3)=0.05(10 f-9) \\
& \Rightarrow 1.00 f-0.75=0.50 f-0.45 \\
& \Rightarrow 1.00 f-0.50 f=-0.45+0.75 \\
& \Rightarrow 0.50 f=0.3 \\
& \Rightarrow f=\frac{0.3}{0.50}
\end{aligned}$

$
\Rightarrow f=0.6
$

To check:
$
\begin{aligned}
& 0.25(4 f-3)=0.05(10 f-9) \\
& \Rightarrow 0.25(4 \times 0.6-3)=0.05(10 \times 0.6-9) \\
& \Rightarrow 0.25(2.4-3)=0.05(6.0-9) \\
& \Rightarrow 0.25 \times(-0.6)=0.05 \times(-3) \\
& \Rightarrow-0.150=-0.150 \\
& \Rightarrow \text { L.H.S. }=\text { R. H.S. }
\end{aligned}
$

Therefore, it is correct.