Solve the following system of equations: $\;$ $x^2 - xy + y^2 = 21, \;\; y^2 - 2xy + 15 = 0$
Given system of equations:
$x^2 - xy + y^2 = 21$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $y^2 - 2xy + 15 = 0$ $\;\;\; \cdots \; (2)$
Least Common Multiple (LCM) of $15$ and $21$ is $105$.
Multiplying equation $(1)$ with $5$ gives $\;\;$ $5x^2 - 5xy + 5y^2 = 105$ $\;\;\; \cdots \; (3)$
Multiplying equation $(2)$ with $7$ gives $\;\;$ $-14xy + 7y^2 = -105$ $\;\;\; \cdots \; (4)$
Adding equations $(3)$ and $(4)$ gives
$5x^2 - 19xy + 12y^2 = 0$
i.e. $\;$ $5x^2 - 15xy - 4xy + 12y^2 = 0$
i.e. $\;$ $5x \left(x - 3y\right) - 4y \left(x - 3y\right) = 0$
i.e. $\;$ $\left(5x - 4y\right) \left(x - 3y\right) = 0$
i.e. $\;$ $x = \dfrac{4 y}{5}$ $\;\;$ or $\;\;$ $x = 3y$
Case 1:
When $\;$ $x = \dfrac{4y}{5}$ $\;\;\; \cdots \; (5)$, $\;$ equation $(1)$ becomes
$\left(\dfrac{4y}{5}\right)^2 - \left(\dfrac{4y}{5}\right) y + y^2 = 21$
i.e. $\;$ $\dfrac{16 \; y^2}{25} - \dfrac{4 \; y^2}{5} + y^2 = 21$
i.e. $\;$ $\left(16 - 20 + 25\right) y^2 = 21 \times 25$
i.e. $\;$ $21 \; y^2 = 21 \times 25$
i.e. $\;$ $y^2 = 25$ $\implies$ $y = \pm 5$
Substituting $\;$ $y = \pm 5$ $\;$ in equation $(5)$ gives
$x = \pm \dfrac{4}{5} \times 5 = \pm 4$
Case 2:
When $\;$ $x = 3y$ $\;\;\; \cdots \; (6)$, $\;$ equation $(1)$ becomes
$\left(3y\right)^2 - \left(3y\right) y + y^2 = 21$
i.e. $\;$ $9y^2 - 3y^2 + y^2 = 21$
i.e. $\;$ $7y^2 = 21$
i.e. $\;$ $y^2 = 3$ $\implies$ $y = \pm \sqrt{3}$
Substituting $\;$ $y = \pm \sqrt{3}$ $\;$ in equation $(6)$ gives
$x = \pm 3 \sqrt{3}$
$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(4, 5\right), \; \left(-4, -5\right), \; \left(3 \sqrt{3}, \sqrt{3}\right), \; \left(- 3\sqrt{3}, - \sqrt{3}\right) \right\}$