Solve :

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

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

Given,

1st equation :

$\Rightarrow \dfrac{7 + x}{5} - \dfrac{2x - y}{4} = 3y - 5 \\[1em] \Rightarrow \dfrac{4(7 + x) - 5(2x - y)}{20} = 3y - 5 \\[1em] \Rightarrow 28 + 4x - 10x + 5y = 20(3y - 5) \\[1em] \Rightarrow 28 - 6x + 5y = 60y - 100 \\[1em] \Rightarrow 6x = 28 + 100 + 5y - 60y \\[1em] \Rightarrow 6x = 128 - 55y \\[1em] \Rightarrow x = \dfrac{128 - 55y}{6} ……(1)$

2nd equation :

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

Substituting value of x from equation (1) in above equation :

$\Rightarrow 34\Big(\dfrac{128 - 55y}{6}\Big) + 15y = 132 \\[1em] \Rightarrow \dfrac{4352 - 1870y}{6} + 15y = 132 \\[1em] \Rightarrow \dfrac{4352 - 1870y + 90y}{6} = 132 \\[1em] \Rightarrow 4352 - 1780y = 792 \\[1em] \Rightarrow 1780y = 4352 - 792 \\[1em] \Rightarrow 1780y = 3560 \\[1em] \Rightarrow y = \dfrac{3560}{1780} = 2.$

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

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

Hence, x = 3 and y = 2.