Solve the following system of equations: $\;$ $5 \left(x + y\right) + 2xy = -19, \;\; 15xy + 5 \left(x + y\right) = -175$
Given system of equations:
$5 \left(x + y\right) + 2xy = -19$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $15xy + 5 \left(x + y\right) = -175$ $\;\;\; \cdots \; (2)$
Let $\;$ $x + y = u$ $\;\;\; \cdots \; (3)$ $\;$ and $\;$ $x \; y = v$ $\;\;\; \cdots \; (4)$
Then equations $(1)$ and $(2)$ become
$5u + 2v = -19$ $\;\;\; \cdots \; (5)$; $\;\;$ $5u + 15v = -175$ $\;\;\; \cdots \; (6)$
Solving equations $(5)$ and $(6)$ simultaneously gives
$13v = -156$ $\implies$ $v = -12$
Substituting $\;$ $v = -12$ $\;$ in equation $(5)$ gives
$5u = -19 + 24 = 5$ $\;\;$ i.e. $\;$ $u = 1$
Substituting the values of $u$ and $v$ in equations $(3)$ and $(4)$ gives
$x + y = 1$ $\;\;$ i.e. $\;$ $x = 1 - y$ $\;\;\; \cdots \; (7)$ $\;\;$ and $\;\;$ $x \; y = -12$ $\;\;\; \cdots \; (8)$
In view of equation $(7)$ equation $(8)$ becomes
$\left(1 - y\right) y = - 12$
i.e. $\;$ $y^2 - y - 12 = 0$
i.e. $\;$ $\left(y - 4\right) \left(y + 3\right) = 0$
i.e. $\;$ $y = 4$ $\;\;$ or $\;\;$ $y = -3$
When $\;$ $y = 4$, $\;$ equation $(7)$ gives $\;\;$ $x = 1 - 4 = -3$
and when $\;$ $y = -3$, $\;$ equation $(7)$ gives $\;\;$ $x = 1 - \left(-3\right) = 4$
$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(4, -3\right), \; \left(-3, 4\right) \right\}$