solve that

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

$\Rightarrow \quad \frac{x}{2}-\frac{x}{4}+\frac{1}{12}=\frac{x}{6}+\frac{1}{6}+\frac{1}{12}$

$\Rightarrow \quad \frac{2 x-x}{4}+\frac{1}{12}=\frac{x}{6}+\frac{2+1}{12}$

$\Rightarrow \quad \frac{2 x-x}{4}+\frac{1}{12}=\frac{x}{6}+\frac{2+1}{12}$

$\Rightarrow$ $\frac{x}{4}+\frac{1}{12}=\frac{x}{6}+\frac{3}{12}$

$\Rightarrow$ $\frac{x}{4}-\frac{x}{6}=\frac{3}{12}-\frac{1}{12} \quad$ transposing $\frac{x}{6}$ to LHS and $\frac{1}{12}$ to RHS $]$

$\Rightarrow$ $\frac{6 x-4 x}{24}=\frac{3-1}{12}$

$\Rightarrow$ $\frac{2 x}{24}=\frac{2}{12}$

$\Rightarrow \quad 2 \times 12 x=2 \times 24$ [by cross-multiplication]

$\Rightarrow$ $24 x=48$

$\Rightarrow$ $\frac{24 x}{24}=\frac{48}{24}$ [dividing both sides by 24 ]

$\therefore$ $x=2$