Question

$\frac{x-2}{4}+\frac{1}{3}=x-\frac{2x-1}{3}$

Answer 1

$\begin{array}{l}\\ \text{We have:}\\ \frac{x-2}{4}+\frac{1}{3}=x-\frac{2x-1}{3}\\ \Rightarrow \frac{x-2}{4}+\frac{2x-1}{3}-x=-\frac{1}{3}(\mathrm{T}\text{ransposing -}\frac{2x-1}{3}\text{to LHS and}\frac{1}{3}\text{to RHS)}\\ \Rightarrow \left(\frac{3(x-2)+4(2x-1)-12x}{12}\right)=-\frac{1}{3}\\ \Rightarrow \left(\frac{3x-6+8x-4-12x}{12}\right)=-\frac{1}{3}\\ \Rightarrow 11x-12x-10=-\frac{1}{{\sout{3}}^1}\times \sout{1}{\sout{2}}^4\\ \Rightarrow -x=-4+10\\ \Rightarrow -x=6\\ \Rightarrow x=-6\\ \text{CHECK: Substituting}x\text{=}-\text{6 in the given equation, we get:}\\ \\ \text{LHS=}\frac{x-2}{4}+\frac{1}{3}\\ \text{=}\frac{-6-2}{4}+\frac{1}{3}\\ \text{=}-2+\frac{1}{3}\\ \text{=}\frac{-5}{3}\\ \text{RHS=}x-\frac{2x-1}{3}\\ \text{=}-6-\frac{2\times (-6)-1}{3}\\ \text{=}-6-\frac{(-13)}{3}\\ \text{=}-6+\frac{13}{3}\\ \text{=}\frac{-5}{3}\\ \text{∴ LHS=RHS}\\ \text{Hence, y=10 is a solution of the given equation}\text{.}\end{array}$