Question

$\frac{x+2}{6}-(\frac{11-x}{3}-\frac{1}{4})=\frac{3x-4}{12}$

Answer 1

$\frac{x+2}{6}-(\frac{11-x}{3}-\frac{1}{4})=\frac{3x-4}{12}$

$\Rightarrow$ $\frac{x+2}{6}$ - $\frac{(11-x)}{3}$ + $\frac{1}{4}$ = $\frac{(3x-4)}{12}$

$\Rightarrow$ $\frac{(x+2)\times 2}{6\times 2}$ - $\frac{(11-x)\times 4}{3\times 4}$ + $\frac{1\times 3}{4\times 3}$ = $\frac{(3x-4)}{12}$ [L.C.M of 6, 3, 4 is 12, so we will change all denominator to 12]

$\Rightarrow$ $\frac{(2x+4)}{12}$ - $\frac{(44-4x)}{12}$+$\frac{3}{12}$=$\frac{(3x-4)}{12}$

$\Rightarrow$ $\frac{(2x+4)-(44-4x)+3}{12}$ = $\frac{(3x-4)}{12}$

$\Rightarrow$ $(2x+4)-(44-4x)+3=3x-4$ [cancelling 12 from both sides]

$\Rightarrow$ $2x+4-44+4x+3=3x-4$

$\Rightarrow$ $6x-37=3x-4$

$\Rightarrow$ $3x=33$

$\therefore$ $x=11$