Solve the following system of equations: $\;$ $xy + x + y = 11, \;\; x^2 y + x y^2 = 30$
Given system of equations:
$xy + x + y = 11$ $\;\;\; \cdots \; (1)$
$x^2 y + x y^2 = 30$ $\;$ i.e. $\;$ $xy \left(x + y\right) = 30$ $\;\;\; \cdots \; (2)$
Let $\;$ $xy = u$ $\;\;\; \cdots \; (3)$ $\;$ and $\;$ $x + y = v$ $\;\;\; \cdots \; (4)$
In view of equations $(3)$ and $(4)$, equations $(1)$ and $(2)$ become
$u + v = 11$ $\implies$ $u = 11 - v$ $\;\;\; \cdots \; (5)$
and $\;$ $uv = 30$ $\;\;\; \cdots \; (6)$
Substituting the value of $u$ from equation $(5)$ in equation $(6)$ gives
$\left(11 - v\right) v = 30$
i.e. $\;$ $v^2 - 11v + 30 = 0$
i.e. $\;$ $\left(v - 6\right) \left(v - 5\right) = 0$
i.e. $\;$ $v = 6$ $\;$ or $\;$ $v = 5$
Substituting the value of $v$ in equation $(5)$ gives
when $\;$ $v = 6$, $\;$ $u = 11 - 6 = 5$;
when $\;$ $v = 5$, $\;$ $u = 11 - 5 = 6$
Substituting $u = 5$ and $v = 6$ in equations $(3)$ and $(4)$ gives
$xy = 5$ $\;\;\; \cdots \; (7)$
and $\;$ $x + y = 6$ $\implies$ $x = 6 - y$ $\;\;\; \cdots \; (8)$
In view of equation $(8)$ equation $(7)$ becomes
$\left(6 - y\right) y = 5$
i.e. $\;$ $y^2 - 6y + 5 = 0$
i.e. $\;$ $\left(y - 5\right) \left(y - 1\right) = 0$
i.e. $\;$ $y = 5$ $\;$ or $\;$ $y = 1$
$\therefore \;$ We have from equation $(8)$
when $\;$ $y = 5$, $\;$ $x = 6 - 5 = 1$
and when $\;$ $y = 1$, $\;$ $x = 6 - 1 = 5$
Next, substituting $u = 6$ and $v = 5$ in equations $(3)$ and $(4)$ gives
$xy = 6$ $\;\;\; \cdots \; (9)$
and $\;$ $x + y = 5$ $\implies$ $x = 5 - y$ $\;\;\; \cdots \; (10)$
In view of equation $(10)$ equation $(9)$ becomes
$\left(5 - y\right) y = 6$
i.e. $\;$ $y^2 - 5y + 6 = 0$
i.e. $\;$ $\left(y - 2\right) \left(y - 3\right) = 0$
i.e. $\;$ $y = 2$ $\;$ or $\;$ $y = 3$
$\therefore \;$ We have from equation $(10)$
when $\;$ $y = 2$, $\;$ $x = 5 - 2 = 3$
and when $\;$ $y = 3$, $\;$ $x = 5 - 3 = 2$
$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(1, 5\right), \; \left(5, 1\right), \; \left(3, 2\right), \; \left(2, 3\right) \right\}$