Question:
$\frac{1}{2}(x+1)+\frac{1}{3}(x-1)=\frac{5}{12}(x-2)$
Solution:
Given, $\frac{1}{2}(x+1)+\frac{1}{3}(x-1)=\frac{5}{12}(x-2)$
$\Rightarrow \quad \frac{x}{2}+\frac{1}{2}+\frac{x}{3}-\frac{1}{3}=\frac{5 x}{12}-\frac{5}{6}$
$\Rightarrow \quad \frac{x}{2}+\frac{x}{3}-\frac{5 x}{12}=\frac{1}{3}-\frac{1}{2}-\frac{5}{6} \quad\left[\right.$ transposing $\frac{1}{2}, \frac{1}{3}$ to RHS and $\frac{5 x}{12}$ to LHS
$\Rightarrow$ $\frac{6 x+4 x-5 x}{12}=\frac{2-3-5}{6}$
$\Rightarrow$ $\frac{5 x}{12}=\frac{-6}{6}$
$\Rightarrow$ $5 x \times 6=(-6) \times 12$ [by cross-multiplication]
$\Rightarrow$ $x=\frac{(-6) \times 12}{5 \times 6}$
$\therefore$ $x=\frac{-12}{5}$