Solve the following system of equations: $\;$ $xy + 4 = 0, \;\; x + y = 3$
Given system of equations:
$xy + 4 = 0$ $\;\;\; \cdots \; (1)$
$x + y = 3$ $\;$ i.e. $\;$ $x = 3 - y$ $\;\;\; \cdots \; (2)$
In view of equation $(2)$, equation $(1)$ becomes
$\left(3 - y\right) y + 4 = 0$
i.e. $\;$ $y^2 - 3y - 4 = 0$
i.e. $\;$ $\left(y - 4\right) \left(y + 1\right) = 0$
i.e. $\;$ $y = 4$ $\;$ or $\;$ $y = -1$
Substituting the value of $y$ in equation $(2)$ gives
when $\;$ $y = 4$, $\;$ $x = 3 - 4 = -1$
and when $\;$ $y = -1$, $\;$ $x = 3 - \left(-1\right) = 4$
$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(-1, 4\right), \; \left(4, -1\right) \right\}$