Solve the following equation and check your answer:

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

We have:

$\phantom{=} \dfrac{2x + 3}{3}-\dfrac{3x - 2}{4} = 1 \\[1em] \Rightarrow \dfrac{4(2x + 3) - 3(3x - 2)}{12} = 1 \\[1em] \Rightarrow 8x + 12 - 9x + 6 = 1 \times 12 \\[1em] \Rightarrow -x + 18 = 12 \\[1em] \Rightarrow -x = 12 - 18 \quad \text{[Transposing +18 to RHS]} \\[1em] \Rightarrow -x = -6 \\[1em] \Rightarrow x = 6 \\[1em]$

∴ x = 6

Check:

$\text{LHS} = \dfrac{2x + 3}{3} - \dfrac{3x - 2}{4} \\[1em] \phantom{\text{LHS}} = \dfrac{15}{3} - \dfrac{16}{4} \\[1em] \phantom{\text{LHS}} = 5 - 4 \\[1em] \phantom{\text{LHS}} = 1 \\[2em] \text{RHS} = 1$

Hence, LHS = RHS.