Solve the following system of equations: $\;$ $\dfrac{1}{2x - y} + y = -5, \;\; \dfrac{y}{2x - y} = 6$
Given system of equations:
$\dfrac{1}{2x - y} + y = -5$ $\;\;\; \cdots \; (1)$
and $\;$ $\dfrac{y}{2x - y} = 6$ $\;\;\; \cdots \; (2)$
Let $\;\;$ $\dfrac{1}{2x - y} = p$ $\;\;\; \cdots \; (3)$
Then, in view of equation $(3)$, equations $(1)$ and $(2)$ respectively become
$p + y = -5$ $\;\;\; \cdots \; (4)$
and $\;$ $p \; y = 6$ $\;\;$ i.e. $\;$ $p = \dfrac{6}{y}$ $\;\;\; \cdots \; (5)$
In view of equation $(5)$, equation $(4)$ can be written as
$\dfrac{6}{y} + y = -5$
i.e. $\;$ $y^2 + 5y + 6 = 0$
i.e. $\;$ $\left(y + 3\right) \left(y + 2\right) = 0$
i.e. $\;$ $y = -3$ $\;$ or $\;$ $y = -2$
Substitute $\;$ $y = -3$ $\;$ in equation $(1)$ to get
$\dfrac{1}{2x + 3} - 3 = -5$
i.e. $\;$ $\dfrac{1}{2x + 3} = -2$
i.e. $\;$ $2x + 3 = \dfrac{-1}{2}$
i.e. $\;$ $2x = \dfrac{-7}{2}$ $\implies$ $x = \dfrac{-7}{4}$
Substitute $\;$ $y = -2$ $\;$ in equation $(1)$ to get
$\dfrac{1}{2x + 2} - 2 = -5$
i.e. $\;$ $\dfrac{1}{2x + 2} = -3$
i.e. $\;$ $2x + 2 = \dfrac{-1}{3}$
i.e. $\;$ $2x = \dfrac{-7}{3}$ $\implies$ $x = \dfrac{-7}{6}$
$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(\dfrac{-7}{4}, -3\right), \; \left(\dfrac{-7}{6}, -2\right) \right\}$