Solve the following system of equations: $\;$ $\dfrac{3x + y}{x - 1} - \dfrac{x - y}{2y} = 2, \;\; x - y = 4$
Given system of equations:
$\dfrac{3x + y}{x - 1} - \dfrac{x - y}{2y} = 2$ $\;\;\; \cdots \; (1)$
[Equation $(1)$ is not valid when $\;$ $x - 1 = 0$ $\;$ i.e. $\;$ $x = 1$ $\;$ or when $\;$ $y = 0$] $\;\;\; \cdots \; (1a)$
and $\;$ $x - y = 4$ $\;$ i.e. $\;$ $x = y + 4$ $\;\;\; \cdots \; (2)$
In view of equation $(2)$, equation $(1)$ becomes
$\dfrac{3 \left(y + 4\right) + y}{y + 4 - 1} - \dfrac{y + 4 - y}{2y} = 2$
i.e. $\;$ $\dfrac{4y + 12}{y + 3} - \dfrac{4}{2y} = 2$
i.e. $\;$ $\dfrac{2y + 6}{ y + 3} - \dfrac{1}{y} = 1$
i.e. $\;$ $2y^2 + 6y - y - 3 = y^2 + 3y$
i.e. $\;$ $y^2 + 2y - 3 = 0$
i.e. $\;$ $\left(y + 3\right) \left(y - 1\right) = 0$
i.e. $\;$ $y = -3$ $\;$ or $\;$ $y = 1$
When $\;$ $y = -3$, $\;$ equation $(2)$ gives $\;$ $x = -3 + 4 = 1$
and when $\;$ $y = 1$, $\;$ equation $(2)$ gives $\;$ $x = 1 + 4 = 5$
But according to $(1a)$, $x \ne 1$
$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(5, 1\right) \right\}$