Solve the following equation and check your answer:

$\dfrac{3}{4}(2x - 5) - \dfrac{5}{6}(7 - 5x) = \dfrac{7x}{3}$

We have:

$\phantom{=} \dfrac{3}{4}(2x - 5) - \dfrac{5}{6}(7 - 5x) = \dfrac{7x}{3} \\[1em] \Rightarrow \dfrac{9(2x - 5) - 10(7 - 5x)}{12} = \dfrac{7x}{3} \\[1em] \Rightarrow \dfrac{18x - 45 - 70 + 50x}{12} = \dfrac{7x}{3} \\[1em] \Rightarrow \dfrac{68x - 115}{12} = \dfrac{7x}{3} \\[1em] \Rightarrow 3(68x - 115) = 12(7x) \quad \text{[By cross multiplication]} \\[1em] \Rightarrow 68x - 115 = 4(7x) \quad \text{[Dividing by 3 on both sides]} \\[1em] \Rightarrow 68x - 115 = 28x \\[1em] \Rightarrow 68x - 28x = 115 \quad \text{[Transposing -115 to RHS and +28x to LHS]} \\[1em] \Rightarrow 40x = 115 \\[1em] \Rightarrow x = \dfrac{115}{40} \\[1em] \Rightarrow x = \dfrac{23}{8}$

∴ x = $2\dfrac{7}{8}$

Check:

$\text{LHS} = \dfrac{3}{4}(2x - 5) - \dfrac{5}{6}(7 - 5x) \\[1em] \phantom{\text{LHS}} = \dfrac{3}{4}\left(\dfrac{23}{4} - 5\right) - \dfrac{5}{6}\left(7 - \dfrac{115}{8}\right) \\[1em] \phantom{\text{LHS}} = \dfrac{3}{4}\left(\dfrac{3}{4}\right) - \dfrac{5}{6}\left(-\dfrac{59}{8}\right) \\[1em] \phantom{\text{LHS}} = \dfrac{9}{16} + \dfrac{295}{48} \\[1em] \phantom{\text{LHS}} = \dfrac{322}{48} \\[1em] \phantom{\text{LHS}} = \dfrac{161}{24} \\[2em] \text{RHS} = \dfrac{7x}{3} \\[1em] \phantom{\text{RHS}} = \dfrac{7}{3} \times \dfrac{23}{8} \\[1em] \phantom{\text{RHS}} = \dfrac{161}{24}$

Hence, LHS = RHS.