Solve the following equation and check your answer:

$\dfrac{x-4}{7}-\dfrac{x+4}{5} = \dfrac{x+3}{7}$

We have:

$\phantom{=} \dfrac{x-4}{7}-\dfrac{x+4}{5} = \dfrac{x+3}{7} \\[1em] \Rightarrow \dfrac{5(x - 4) - 7(x + 4)}{35} = \dfrac{x+3}{7} \\[1em] \Rightarrow 5x - 20 - 7x - 28 = \left(\dfrac{x+3}{7}\right) \times 35 \\[1em] \Rightarrow -2x - 48 = \left(\dfrac{x+3}{1}\right) \times 5 \\[1em] \Rightarrow -2x - 48 = 5(x+3) \\[1em] \Rightarrow -2x - 48 = 5x + 15 \\[1em] \Rightarrow -2x - 5x = 15 + 48 \quad \text{[Transposing -48 to RHS and +5x to LHS]} \\[1em] \Rightarrow -7x = 63 \\[1em] \Rightarrow -x = \dfrac{63}{7} \\[1em] \Rightarrow -x = 9$

∴ x = -9

Check:

$\text{LHS} = \dfrac{x-4}{7} - \dfrac{x+4}{5} \\[1em] \phantom{\text{LHS}} = \dfrac{-13}{7} - \dfrac{-5}{5} \\[1em] \phantom{\text{LHS}} = -\dfrac{13}{7} + 1 \\[1em] \phantom{\text{LHS}} = -\dfrac{6}{7} \\[2em] \text{RHS} = \dfrac{x+3}{7} \\[1em] \phantom{\text{RHS}} = \dfrac{-9+3}{7} \\[1em] \phantom{\text{RHS}} = -\dfrac{6}{7}$

Hence, LHS = RHS.