Solve the following equation and check your answer:

$\dfrac{3}{4}(7x - 1)-\Big(2x -\dfrac{1-x}{2}\Big) = x + \dfrac{3}{2}$

We have:

$\phantom{=} \dfrac{3}{4}(7x - 1)-\Big(2x -\dfrac{1-x}{2}\Big) = x + \dfrac{3}{2} \\[1em] \Rightarrow \dfrac{21x - 3}{4}-\left(\dfrac{4x -(1 - x)}{2}\right) = \dfrac{2x + 3}{2} \\[1em] \Rightarrow \dfrac{21x - 3}{4}-\left(\dfrac{4x - 1 + x}{2}\right) = \dfrac{2x + 3}{2} \\[1em] \Rightarrow \dfrac{21x - 3}{4}-\dfrac{5x - 1}{2} = \dfrac{2x + 3}{2} \\[1em] \Rightarrow \dfrac{21x - 3 - 10x + 2}{4} = \dfrac{2x + 3}{2} \\[1em] \Rightarrow \dfrac{11x - 1}{4} = \dfrac{2x + 3}{2} \\[1em] \Rightarrow 2(11x - 1) = 4(2x + 3) \quad \text{[By cross multiplication]} \\[1em] \Rightarrow 22x - 2 = 8x + 12 \\[1em] \Rightarrow 22x - 8x = 12 + 2 \quad \text{[Transposing -2 to RHS and +8x to LHS]} \\[1em] \Rightarrow 14x = 14 \\[1em] \Rightarrow x = \dfrac{14}{14}$

∴ x = 1

Check:

$\text{LHS} = \dfrac{3}{4}(7x - 1) - \left(2x - \dfrac{1-x}{2}\right) \\[1em] \phantom{\text{LHS}} = \dfrac{3}{4}(6) - \left(2 - 0\right) \\[1em] \phantom{\text{LHS}} = \dfrac{9}{2} - 2 \\[1em] \phantom{\text{LHS}} = \dfrac{5}{2} \\[2em] \text{RHS} = x + \dfrac{3}{2} \\[1em] \phantom{\text{RHS}} = 1 + \dfrac{3}{2} \\[1em] \phantom{\text{RHS}} = \dfrac{5}{2}$

Hence, LHS = RHS.