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

Answer

Given, $\frac{\text{x}}{2}-\frac{1}{4\Big(\text{x}-\frac{1}{3}\Big)}=\frac{1}{6}(\text{x}+1)+\frac{1}{12}$$\frac{\text{x}}{2}-\frac{\text{x}}{4}+\frac{1}{12}=\frac{\text{x}}{6}+\frac{1}{6}+\frac{1}{12}$
$\frac{2\text{x}-\text{x}}{4}+\frac{1}{12}=\frac{\text{x}}{6}+\frac{2+1}{12}$
$\frac{\text{x}}{4}+\frac{1}{12}=\frac{\text{x}}{6}+\frac{3}{12}$
$\frac{\text{x}}{4}-\frac{\text{x}}{6}=\frac{3}{12}-\frac{1}{12}$
$\frac{6\text{x}-4}{24}=\frac{3-1}{12}$
$\frac{2\text{x}}{24}=\frac{2}{12}$
$2 \times 12\text{x} = 2 \times 24$
$24\text{x}= 48$
$\frac{24\text{x}}{24}=\frac{48}{24}$
$\text{x} = 2$

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