Linear Equations in One Variable — Maths STD 7 — Question

As, $\frac{\text{x}-2}{3}=\frac{2\text{x}-1}{3}-1$
$\Rightarrow\frac{\text{x}-2}{3}-\frac{2\text{x}-1}{3}=-1 ($By transposing $\frac{2\text{x}-1}{3}$ to $\text{L.H.S.)}$
$\Rightarrow\frac{(\text{x}-2)-(\text{2x}-1)}{3}=-1$
$\Rightarrow\frac{\text{x}-2-2\text{x}+1}{3}=-1$
$\Rightarrow\frac{-\text{x}-1}{3}=-1$
$\Rightarrow-\text{x}-1=-1\times3 ($By transposing $3$ to $\text{R.H.S.)}$
$\Rightarrow-\text{x}-1=-3$
$\Rightarrow-\text{x}=-3-1 ($By transposing $-1$ to $\text{R.H.S.)}$
$\Rightarrow-\text{x}=-2$
$\therefore\text{x}=2$
Hence, the correct alternative is option $(a).$