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