Solve the following simultaneous equations:

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

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

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

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

$\Rightarrow \dfrac{4x - 3}{6} + \dfrac{5y - 7}{2} = 18 - 5x \\[1em] \Rightarrow \dfrac{2(4x - 3) + 6(5y - 7)}{12} = 18 - 5x \\[1em] \Rightarrow \dfrac{8x - 6 + 30y - 42}{12} = 18 - 5x \\[1em] \Rightarrow 8x - 6 + 30y - 42 = 12(18 - 5x) \\[1em] \Rightarrow 8x + 30y - 48 = 216 - 60x \\[1em] \Rightarrow 8x + 30y + 60x = 216 + 48 \\[1em] \Rightarrow 68x + 30y = 264 \text{ ….(2) }$

Substituting value of x from equation (1) in 68x + 30y = 264, we get :

$\Rightarrow 68\Big(\dfrac{128 - 55y}{6}\Big) + 30y = 264 \\[1em] \Rightarrow 34\Big(\dfrac{128 - 55y}{3}\Big) + 30y = 264 \\[1em] \Rightarrow \Big(\dfrac{4352 - 1870y}{3}\Big) + 30y = 264 \\[1em] \Rightarrow \Big(\dfrac{4352 - 1870y + 90y}{3}\Big) = 264 \\[1em] \Rightarrow (4352 - 1780y) = 264 \times 3 \\[1em] \Rightarrow (4352 - 1780y) = 792 \\[1em] \Rightarrow 1780y = 4352 - 792 \\[1em] \Rightarrow 1780y = 3560 \\[1em] \Rightarrow y = \dfrac{3560}{1780} = 2.$

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

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

Hence, x = 3, y = 2.