Solve the following equation and check your answer:

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

We have:

$\phantom{=} \dfrac{x-2}{3}+\dfrac{x-3}{4} = \dfrac{x-1}{2} \\[1em] \Rightarrow \dfrac{4(x - 2) + 3(x - 3)}{12} = \dfrac{x-1}{2} \\[1em] \Rightarrow 4x - 8 + 3x - 9 = \left(\dfrac{x-1}{2}\right) \times 12 \\[1em] \Rightarrow 7x - 17 = \left(\dfrac{x-1}{1}\right) \times 6 \\[1em] \Rightarrow 7x - 17 = 6(x-1) \\[1em] \Rightarrow 7x - 17 = 6x - 6 \\[1em] \Rightarrow 7x - 6x = -6 + 17 \quad \text{[Transposing -17 to RHS and +6x to LHS]} \\[1em]$

∴ x = 11

Check:

$\text{LHS} = \dfrac{x-2}{3} + \dfrac{x-3}{4} \\[1em] \phantom{\text{LHS}} = \dfrac{11-2}{3} + \dfrac{11-3}{4} \\[1em] \phantom{\text{LHS}} = 3 + 2 \\[1em] \phantom{\text{LHS}} = 5 \\[2em] \text{RHS} = \dfrac{x-1}{2} \\[1em] \phantom{\text{RHS}} = \dfrac{11-1}{2} \\[1em] \phantom{\text{RHS}} = 5$

Hence, LHS = RHS.