Question

Question 62

Solve the following:

$\frac{x}{2}-\frac{1}{4}(x-\frac{1}{3})=\frac{1}{6}(x+1)+\frac{1}{12}$

Answer 1

Given, $\frac{x}{2}-\frac{1}{4}(x-\frac{1}{3})=\frac{1}{6}(x+1)+\frac{1}{12}$
$\Rightarrow \frac{x}{2}-\frac{x}{4}+\frac{1}{12}=\frac{x}{6}+\frac{1}{6}+\frac{1}{12}\Rightarrow \frac{2x-x}{4}+\frac{1}{12}=\frac{x}{6}+\frac{2+1}{12}\Rightarrow \frac{x}{4}+\frac{1}{12}=\frac{x}{6}+\frac{3}{12}\Rightarrow \frac{x}{4}-\frac{x}{6}=\frac{3}{12}-\frac{1}{12}\Rightarrow \frac{3x-2x}{12}=\frac{3-1}{12}\Rightarrow \frac{x}{12}=\frac{2}{12}\Rightarrow 12x=2\times 12\Rightarrow 12x=24\Rightarrow \frac{12x}{12}=\frac{24}{12}\therefore x=2$