Solve the following system of equations: $\;$ $x \left(x + y\right) = 9, \;\; y \left(x + y\right) = 16$
Given system of equations:
$x \left(x + y\right) = 9$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $y \left(x + y\right) = 16$ $\;\;\; \cdots \; (2)$
Dividing equations $(1)$ and $(2)$ gives
$\dfrac{x}{y} = \dfrac{9}{16}$
i.e. $\;$ $x = \dfrac{9y}{16}$ $\;\;\; \cdots \; (3)$
In view of equation $(3)$, equation $(1)$ becomes
$\dfrac{9y}{16} \left(\dfrac{9y}{16} + y\right) = 9$
i.e. $\;$ $\dfrac{81 y^2}{256} + \dfrac{9 y^2}{16} = 9$
i.e. $\;$ $\dfrac{225 y^2}{256} = 9$
i.e. $\;$ $y^2 = \dfrac{9 \times 256}{225}$ $\implies$ $y = \pm \dfrac{3 \times 16}{15} = \pm \dfrac{16}{5}$
Substituting the value of $y$ in equation $(3)$ gives
$x = \pm \dfrac{9}{16} \times \dfrac{16}{5} = \pm \dfrac{9}{5}$
$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(\dfrac{9}{5}, \dfrac{16}{5}\right), \; \left(\dfrac{-9}{5}, \dfrac{-16}{5}\right) \right\}$