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

You can Download the Examples (Revised) - Chapter 2 - Linear Equations In One Variable - Ncert Solutions class 8 - Maths with expert answers for all chapters. Perfect for Tamil & English Medium students to revise the syllabus and score more marks in board exams. Download and share it with your friends

Share this to Friend on WhatsApp

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

Example 1:

Solve $2 x-3=x+2$
Solution:

We have
$
\begin{aligned}
2 x & =x+2+3 \\
2 x & =x+5 \\
2 x-x & =x+5-x \quad \text { (subtracting } x \text { from both sides) } \\
x & =5
\end{aligned}
$
or
(subtracting $x$ from both sides)
or
Here we subtracted from both sides of the equation, not a number (constant), buta term involving the variable. We can do this as variables are also numbers. Also, note that subtracting $x$ from both sides amounts to transposing $x$ to LHS.

Example 2:

Solve $5 x+\frac{7}{2}=\frac{3}{2} x-14$
Solution:

Multiply both sides of the equation by 2 . We get
$
2 \times\left(5 x+\frac{7}{2}\right)=2 \times\left(\frac{3}{2} x-14\right)
$

$
(2 \times 5 x)+\left(2 \times \frac{7}{2}\right)=\left(2 \times \frac{3}{2} x\right)-(2 \times 14)
$
or
$
10 x+7=3 x-28
$
or
$
10 x-3 x+7=-28
$
(transposing $3 x$ to LHS)
or
$
\begin{aligned}
7 x+7 & =-28 \\
7 x & =-28-7 \\
7 x & =-35
\end{aligned}
$
or $\quad x=\frac{-35}{7}$
or
$x=-5$
(solution)

Example 3:

Solve $\frac{6 x+1}{3}+1=\frac{x-3}{6}$
Solution:

Multiplying both sides of the equation by 6 ,
$
\frac{6(6 x+1)}{3}+6 \times 1=\frac{6(x-3)}{6}
$
or
$
2(6 x+1)+6=x-3
$
or
$
12 x+2+6=x-3
$
(opening the brackets )
or
$
12 x+8=x-3
$
or
$
\begin{array}{r}
12 x-x+8=-3 \\
11 x+8=-3
\end{array}
$
or
$
11 x=-3-8
$
or
$
11 x=-11
$
or
$
x=-1
$
(required solution)

$
\begin{gathered}
\text { Check }: \text { LHS }=\frac{6(-1)+1}{3}+1=\frac{-6+1}{3}+1=\frac{-5}{3}+\frac{3}{3}=\frac{-5+3}{3}=\frac{-2}{3} \\
\text { RHS }=\frac{(-1)-3}{6}=\frac{-4}{6}=\frac{-2}{3} \\
\text { LHS }=\text { RHS. } \quad \text { (as required) }
\end{gathered}
$
(as required)

Example 4:

Solve $5 x-2(2 x-7)=2(3 x-1)+\frac{7}{2}$
Solution:

Let us open the brackets,
$
\begin{aligned}
\text { LHS }=5 x-4 x+14 & =x+14 \\
\text { RHS } & =6 x-2+\frac{7}{2}=6 x-\frac{4}{2}+\frac{7}{2}=6 x+\frac{3}{2}
\end{aligned}
$

The equation is $x+14=6 x+\frac{3}{2}$
or
$
14=6 x-x+\frac{3}{2}
$
or
$
14=5 x+\frac{3}{2}
$
or
$
14-\frac{3}{2}=5 x
$
(transposing $\frac{3}{2}$ )

$
\begin{aligned}
\text { or } & \frac{28-3}{2} & =5 x \\
\text { or } & \frac{25}{2} & =5 x \\
\text { or } & x & =\frac{25}{2} \times \frac{1}{5}=\frac{5 \times 5}{2 \times 5}=\frac{5}{2}
\end{aligned}
$

Therefore, required solution is $x=\frac{5}{2}$.

$\begin{aligned}
& \text { Check }: \text { LHS }=5 \times \frac{5}{2}-2\left(\frac{5}{2} \times 2-7\right) \\
& =\frac{25}{2}-2(5-7)=\frac{25}{2}-2(-2)=\frac{25}{2}+4=\frac{25+8}{2}=\frac{33}{2} \\
& \text { RHS }=2\left(\frac{5}{2} \times 3-1\right)+\frac{7}{2}=2\left(\frac{15}{2}-\frac{2}{2}\right)+\frac{7}{2}=\frac{2 \times 13}{2}+\frac{7}{2} \\
& =\frac{26+7}{2}=\frac{33}{2}=\text { LHS. (as required) } \\
&
\end{aligned}$