Solve the following equation and check your answer:

$\dfrac{3y - 2}{7} - \dfrac{5y - 8}{4} = \dfrac{1}{14}$

We have:

$\phantom{=} \dfrac{3y - 2}{7} - \dfrac{5y - 8}{4} = \dfrac{1}{14} \\[1em] \Rightarrow \dfrac{4(3y - 2) - 7(5y - 8)}{28} = \dfrac{1}{14} \\[1em] \Rightarrow 12y - 8 - 35y + 56 = \dfrac{1}{14} \times 28 \\[1em] \Rightarrow -23y + 48 = \dfrac{1}{1} \times 2 \\[1em] \Rightarrow -23y + 48 = 2 \\[1em] \Rightarrow -23y = 2 - 48 \quad \text{[Transposing +48 to RHS]} \\[1em] \Rightarrow -23y = - 46 \\[1em] y = \dfrac{-46}{-23} \\[1em]$

∴ y = 2

Check:

$\text{LHS} = \dfrac{3y - 2}{7} - \dfrac{5y - 8}{4} \\[1em] \phantom{\text{LHS}} = \dfrac{6 - 2}{7} - \dfrac{10 - 8}{4} \\[1em] \phantom{\text{LHS}} = \dfrac{4}{7} - \dfrac{1}{2} \\[1em] \phantom{\text{LHS}} = \dfrac{8 - 7}{14} \\[1em] \phantom{\text{LHS}} = \dfrac{1}{14} \\[2em] \text{RHS} = \dfrac{1}{14}$

Hence, LHS = RHS.