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

Given:

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

To do:

We have to solve \( \frac{x}{2}-\frac{1}{4}\left(x-\frac{1}{3}\right)=\frac{1}{6}(x+1)+\frac{1}{12} \).

Solution:

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

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

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

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

$\Rightarrow 6x-3x+1=2x+3$

$\Rightarrow 3x-2x=3-1$

$\Rightarrow x=2$

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Updated on: 10-Oct-2022

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