Solve the following system of equations: $\;$ $2y^2 - 4xy + 3x^2 = 17, \;\; y^2 - x^2 = 16$
Given system of equations:
$2y^2 - 4xy + 3x^2 = 17$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $y^2 - x^2 = 16$ $\;\;\; \cdots \; (2)$
Subtracting equations $(1)$ and $(2)$ gives
$y^2 - 4xy + 4x^2 = 1$
i.e. $\;$ $\left(y - 2x\right)^2 = 1$
i.e. $\;$ $y - 2x = \pm 1$ $\;\;$ i.e. $\;$ $y = 2x \pm 1$ $\;\;\; \cdots \; (3)$
Substituting the value of $y$ from equation $(3)$ in equation $(2)$
Case 1:
When $\;$ $y = 2x + 1$ $\;\;\; \cdots \; (3a)$ $\;$ we have $\;$ $\left(2x + 1\right)^2 - x^2 = 16$
i.e. $\;$ $4x^2 + 4x + 1 - x^2 = 16$
i.e. $\;$ $3x^2 + 4x - 15 = 0$
i.e. $\;$ $\left(3x - 5\right) \left(x + 3\right) = 0$
i.e. $\;$ $x = \dfrac{5}{3}$, $\;\;$ or $\;\;$ $x = -3$
When $\;$ $x = \dfrac{5}{3}$, $\;$ we have from equation $(3a)$, $\;$ $y = \left[2 \times \dfrac{5}{3}\right] + 1 = \dfrac{13}{3}$
and when $\;$ $x = -3$, $\;$ we have from equation $(3a)$, $\;$ $y = \left[2 \times \left(-3\right)\right] + 1 = -5$
Case 2:
When $\;$ $y = 2x - 1$ $\;\;\; \cdots \; (3b)$ $\;$ we have $\;$ $\left(2x - 1\right)^2 - x^2 = 16$
i.e. $\;$ $4x^2 - 4x + 1 - x^2 = 16$
i.e. $\;$ $3x^2 - 4x - 15 = 0$
i.e. $\;$ $\left(3x + 5\right) \left(x - 3\right) = 0$
i.e. $\;$ $x = \dfrac{-5}{3}$, $\;\;$ or $\;\;$ $x = 3$
When $\;$ $x = \dfrac{-5}{3}$, $\;$ we have from equation $(3b)$, $\;$ $y = \left[2 \times \left(\dfrac{-5}{3}\right)\right] - 1 = \dfrac{-13}{3}$
and when $\;$ $x = 3$, $\;$ we have from equation $(3b)$, $\;$ $y = \left[2 \times 3\right] - 1 = 5$
$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(-3, -5\right), \; \left(\dfrac{-5}{3}, \dfrac{-13}{3}\right), \; \left(\dfrac{5}{3}, \dfrac{13}{3}\right), \; \left(3, 5\right) \right\}$