Question

Solve the linear equation $\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}$

Answer 1

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

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

$\Rightarrow \frac{3x-2x}{6}=\frac{5+4}{20}$

$\Rightarrow \frac{x}{6}=\frac{9}{20}$

$\Rightarrow x=\frac{9\times 6}{20}=\frac{27}{10}$

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

$L.H.S=\frac{27}{10\times 2}-\frac{1}{5}$

$=\frac{27}{20}-\frac{1}{5}$

$=\frac{27}{20}-\frac{1\times 4}{5\times 4}$

$=\frac{27}{20}-\frac{4}{20}$

$=\frac{23}{20}$

$R.H.S=\frac{27}{10\times 3}+\frac{1}{4}$

$=\frac{9}{10}+\frac{1}{4}$

$=\frac{9\times 2}{10\times 2}+\frac{1\times 5}{4\times 5}$

$=\frac{18+5}{20}$

$=\frac{23}{20}$

$\therefore L.H.S=R.H.S$
Therefore, $x=\frac{27}{10}$ is the solution of the given equation.