Solve the following system of equations: $\;$ $\dfrac{x}{y} - \dfrac{y}{x} = \dfrac{5}{6}, \;\; x^2 - y^2 = 5$
Given system of equations:
$\dfrac{x}{y} - \dfrac{y}{x} = \dfrac{5}{6}$ $\;$ i.e. $\;$ $x^2 - y^2 = \dfrac{5 xy}{6}$ $\;\;\; \cdots \; (1)$
$x^2 - y^2 = 5$ $\;\;\; \cdots \; (2)$
In view of equation $(2)$, equation $(1)$ becomes
$\dfrac{5xy}{6} = 5$
i.e. $\;$ $x \; y = 6$ $\implies$ $y = \dfrac{6}{x}$ $\;\;\; \cdots \; (3)$
Substituting the value of $y$ from equation $(3)$ in equation $(2)$ gives
$x^2 - \left(\dfrac{6}{x}\right)^2 = 5$
i.e. $\;$ $x^2 - \dfrac{36}{x^2} = 5$
i.e. $\;$ $x^4 - 5x^2 - 36 = 0$
i.e. $\;$ $\left(x^2 - 9\right) \left(x^2 + 4\right) = 0$
i.e. $\;$ $x^2 = 9$ $\;$ or $\;$ $x^2 = -4$
But $x^2 = -4$ is not a valid solution since square of any number cannot be negative.
Now, $\;$ $x^2 = 9$ $\implies$ $x = \pm 3$
Substituting the value of $x$ in equation $(3)$ gives
when $\;$ $x = 3$, $\;$ $y = \dfrac{6}{3} = 2$
and when $\;$ $x = -3$, $\;$ $y = \dfrac{6}{-3} = -2$
$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(3, 2\right), \left(-3, -2\right) \right\}$