Question

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

Answer 1

$\begin{array}{l}\\ \mathrm{We}\mathrm{have}:\\ \frac{x+2}{6}-\left(\frac{11-x}{3}-\frac{1}{4}\right)=\frac{3x-4}{12}\\ \Rightarrow \frac{x+2}{6}-\left(\frac{11-x}{3}\right)+\frac{1}{4}=\frac{3x-4}{12}\\ \Rightarrow \frac{x+2}{6}-\left(\frac{11-x}{3}\right)-\frac{3x-4}{12}=-\frac{1}{4}\text{(By transposition)}\\ \Rightarrow \frac{2(x+2)-4(11-x)-1(3x-4)}{12}=-\frac{1}{4}\\ \Rightarrow \frac{2x+4-44+4x-3x+4}{12}=-\frac{1}{4}\\ \Rightarrow 3x-36=-\frac{1}{4}\times \text{12}\\ \Rightarrow 3x=-3+36\\ \Rightarrow x=\frac{33}{3}\\ \Rightarrow x=11\\ \text{CHECK: Substituting}x\text{= 11 in the given equation, we get:}\\ \\ \text{LHS=}\frac{x+2}{6}-(\frac{11-x}{3}-\frac{1}{4})\\ \text{=}\frac{11+2}{6}-(\frac{11-11}{3}-\frac{1}{4})\\ \text{=}\frac{13}{6}-(-\frac{1}{4})\\ \text{=}\frac{13}{6}+\frac{1}{4}\\ \text{=}\frac{13\times 2+3}{12}\\ =\frac{29}{12}\\ \text{RHS=}\frac{3x-4}{12}\\ =\frac{3\times 11-4}{12}\\ =\frac{33-4}{12}\\ =\frac{29}{2}\\ \text{∴ LHS=RHS}\\ \text{Hence,}x\text{= 11 is a solution of the given equation}\text{.}\\ \text{Verified.}\end{array}$