Solve the following simultaneous equations:

$\dfrac{x}{2} + y = \dfrac{4}{5}, x + \dfrac{y}{2} = \dfrac{7}{10}$

Simplifying, equation : $\dfrac{x}{2} + y = \dfrac{4}{5}$

$\Rightarrow \dfrac{x}{2} + y = \dfrac{4}{5} \\[1em] \Rightarrow \dfrac{x + 2y}{2} = \dfrac{4}{5} \\[1em] \Rightarrow 5(x + 2y) = 4 \times 2 \\[1em] \Rightarrow 5x + 10y = 8 \\[1em] \Rightarrow 10y = 8 - 5x \\[1em] \Rightarrow y = \dfrac{8 - 5x}{10} \text{ ….(1)}$

Substituting value of y from equation (1) in $x + \dfrac{y}{2} = \dfrac{7}{10}$, we get :

$\Rightarrow x + \dfrac{\dfrac{8 - 5x}{10}}{2} = \dfrac{7}{10} \\[1em] \Rightarrow x + \dfrac{8 - 5x}{20} = \dfrac{7}{10} \\[1em] \Rightarrow \dfrac{20x + 8 - 5x}{20} = \dfrac{7}{10} \\[1em] \Rightarrow 15x + 8 = \dfrac{7}{10} \times 20 \\[1em] \Rightarrow 15x + 8 = 7 \times 2 \\[1em] \Rightarrow 15x = 14 - 8 \\[1em] \Rightarrow 15x = 6 \\[1em] \Rightarrow x = \dfrac{6}{15} = \dfrac{2}{5}.$

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

$\Rightarrow y = \dfrac{8 - 5x}{10} \\[1em] \Rightarrow y = \dfrac{8 - 5 \Big(\dfrac{2}{5}\Big)}{10} \\[1em] \Rightarrow y = \dfrac{8 - 2}{10} \\[1em] \Rightarrow y = \dfrac{6}{10} = \dfrac{3}{5}.$

Hence, $x = \dfrac{2}{5}, y = \dfrac{3}{5}$.