Solve the following system of equations: $\;$ $x^2 = \left(x - a\right)y, \;\; y^2 - ay = 9ax$
Given system of equations:
$x^2 = \left(x - a\right)y$ $\;\;\; \cdots \; (1)$, and $\;$ $\;$ $y^2 - xy = 9ax$ $\;\;\; \cdots \; (2)$
Case 1: When $a \neq 0$
We have from equation $(1)$,
$y = \dfrac{x^2}{x - a}$ $\;\;\; \cdots \; (3)$
Substituting the value of $y$ from equation $(3)$ in equation $(2)$ gives
$\left(\dfrac{x^2}{x - a}\right)^2 - x \left(\dfrac{x^2}{x - a}\right) = 9ax$
i.e. $\;$ $\dfrac{x^3}{\left(x - a\right)^2} - \dfrac{x^2}{x - a} = 9a$
i.e. $\;$ $x^3 - x^2 \left(x - a\right) = 9a \left(x - a\right)^2$
i.e. $\;$ $x^3 - x^3 + ax^2 = 9ax^2 - 18a^2 x + 9a^3$
i.e. $\;$ $8ax^2 - 18 a^2 x + 9a^3 = 0$
i.e. $\;$ $8ax^2 - 12 a^2 x - 6 a^2 x + 9a^3 = 0$
i.e. $\;$ $4ax \left(2x - 3a\right) - 3a^2 \left(2x - 3a\right) = 0$
i.e. $\;$ $\left(2x - 3a\right) \left(4ax - 3a^2\right) = 0$
i.e. $\;$ $x = \dfrac{3a}{2}$ $\;\;$ or $\;\;$ $x = \dfrac{3a}{4}$
When $\;$ $x = \dfrac{3a}{2}$, $\;$ we have from equation $(3)$
$y = \dfrac{\left(\dfrac{3a}{2}\right)^2}{\dfrac{3a}{2} - a}$ $\implies$ $y = \dfrac{9a^2}{4} \times \dfrac{2}{a} = \dfrac{9a}{2}$
When $\;$ $x = \dfrac{3a}{4}$, $\;$ we have from equation $(3)$
$y = \dfrac{\left(\dfrac{3a}{4}\right)^2}{\dfrac{3a}{4} - a}$ $\implies$ $y = \dfrac{9a^2}{16} \times \dfrac{2}{-a} = \dfrac{-9a}{4}$
$\therefore \;$ The solution to the given system of equations is $\;\;$ $\left(x, y\right) = \left\{\left(\dfrac{3a}{2}, \dfrac{9a}{2}\right), \left(\dfrac{3a}{4}, \dfrac{-9a}{4}\right) \right\}$
Case 2: When $a = 0$
Then equation $(1)$ becomes $\;\;\;$ $x^2 = xy$ $\;\;\; \cdots \; (4)$
and equation $(2)$ becomes $\;\;\;$ $y^2 - xy = 0$ $\;\;$ i.e. $\;$ $y^2 = xy$ $\;\;\; \cdots \; (5)$
$\therefore \;$ We have from equations $(4)$ and $(5)$,
$x^2 = y^2$ $\;\;\;$ i.e. $\;$ $x = y$
$\therefore \;$ The solution to the given system of equations is $\;\;$ $\left(x, y\right) = \left\{\left(k, k\right) \mid k \in R \right\}$