Solve for x and y :

$\dfrac{y + 7}{5} = \dfrac{2y - x}{4} + 3x - 5$

$\dfrac{7 - 5x}{2} + \dfrac{3 - 4y}{6} = 5y - 18$

Simplifying first equation :

$\Rightarrow \dfrac{y + 7}{5} = \dfrac{2y - x}{4} + 3x - 5 \\[1em] \Rightarrow \dfrac{y + 7}{5} = \dfrac{2y - x + 12x - 20}{4} \\[1em] \Rightarrow 4(y + 7) = 5(2y + 11x - 20) \\[1em] \Rightarrow 4y + 28 = 10y + 55x - 100 \\[1em] \Rightarrow 55x + 10y - 4y = 100 + 28 \\[1em] \Rightarrow 55x + 6y = 128 \\[1em] \Rightarrow 55x = 128 - 6y \\[1em] \Rightarrow x = \dfrac{128 - 6y}{55}\text{ …….(1)}$

Simplifying second equation :

$\Rightarrow \dfrac{7 - 5x}{2} + \dfrac{3 - 4y}{6} = 5y - 18 \\[1em] \Rightarrow \dfrac{3(7 - 5x) + 3 - 4y}{6} = 5y - 18 \\[1em] \Rightarrow \dfrac{21 - 15x + 3 - 4y}{6} = 5y - 18 \\[1em] \Rightarrow 24 - 15x - 4y = 6(5y - 18) \\[1em] \Rightarrow 24 - 15x - 4y = 30y - 108 \\[1em] \Rightarrow 15x + 30y + 4y = 108 + 24 \\[1em] \Rightarrow 15x + 34y = 132 \text{ ……(2)}.$

Substituting value of x from equation (1) in (2), we get :

$\Rightarrow 15 \times \dfrac{128 - 6y}{55} + 34y = 132 \\[1em] \Rightarrow \dfrac{3}{11} \times (128 - 6y) + 34y = 132 \\[1em] \Rightarrow \dfrac{384 - 18y + 374y}{11} = 132 \\[1em] \Rightarrow 384 - 18y + 374y = 1452 \\[1em] \Rightarrow 384 + 356y = 1452 \\[1em] \Rightarrow 356y = 1452 - 384 \\[1em] \Rightarrow 356y = 1068 \\[1em] \Rightarrow y = \dfrac{1068}{356} = 3.$

Substituting value of y in equation (1), we get :

$\Rightarrow x = \dfrac{128 - 6 \times 3}{55} \\[1em] = \dfrac{128 - 18}{55}\\[1em] = \dfrac{110}{55} \\[1em] = 2.$

Hence, x = 2 and y = 3.