Solve the following system of equations: $\;$ $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{3}, \;\; x^2 + y^2 = 160$
Given system of equations:
$\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{3}$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $x^2 + y^2 = 160$ $\;\;\; \cdots \; (2)$
Equation $(1)$ can be written as $\;\;$ $\dfrac{x + y}{xy} = \dfrac{1}{3}$ $\;\;\; \cdots \; (1a)$
Equation $(2)$ can be written as $\;\;$ $\left(x + y\right)^2 - 2xy = 160$ $\;\;\; \cdots \; (2a)$
Let $\;\;$ $x + y = u$ $\;\;\; \cdots \; (3)$; $\;\;$ $x \; y = v$ $\;\;\; \cdots \; (4)$
In view of equations $(3)$ and $(4)$. equations $(1a)$ and $(2a)$ respectively become
$\dfrac{u}{v} = \dfrac{1}{3}$ $\;\;$ i.e. $\;$ $v = 3u$ $\;\;\; \cdots \; (1b)$
and $\;\;$ $u^2 - 2v = 160$ $\;\;\; \cdots \; (2b)$
Substituting for $v$ from equation $(1b)$ in equation $(2b)$ gives
$u^2 - 2\times \left(3u\right) = 160$
i.e. $\;$ $u^2 - 6u - 160 = 0$
i.e. $\;$ $\left(u - 16\right) \left(u + 10\right) = 0$
i.e. $\;$ $u = 16$ $\;\;$ or $\;\;$ $u = -10$
$\therefore \;$ We have from equation $(1b)$
when $\;\;$ $u = 16$, $\;$ $v = 3 \times 16 = 48$
and when $\;\;$ $u = -10$, $\;$ $v = 3 \times \left(-10\right) = -30$
Case 1:
Substituting $\;\;$ $u = 16, \; v = 48$ $\;\;$ in equations $(3)$ and $(4)$ gives
$x + y = 16$ $\;\;\;$ i.e. $\;$ $x = 16 - y$ $\;\;\; \cdots \; (3a)$
and $\;\;$ $x \; y = 48$
i.e. $\;$ $\left(16 - y\right) y = 48$ $\;\;\;$ [in view of equation $(3a)$]
i.e. $\;$ $y^2 - 16 y + 48 = 0$
i.e. $\;$ $\left(y - 12\right) \left(y - 4\right) = 0$
i.e. $\;$ $y = 12$ $\;\;$ or $\;\;$ $y = 4$
When $\;\;$ $y = 12$, $\;\;$ we have from equation $(3a)$, $\;\;$ $x = 16 - 12 = 4$
and when $\;\;$ $y = 4$, $\;\;$ we have from equation $(3a)$, $\;\;$ $x = 16 - 4 = 12$
Case 2:
Substituting $\;\;$ $u = -10, \; v = -30$ $\;\;$ in equations $(3)$ and $(4)$ gives
$x + y = -10$ $\;\;\;$ i.e. $\;$ $x = -y - 10$ $\;\;\; \cdots \; (3b)$
and $\;\;$ $x \; y = -30$
i.e. $\;$ $-\left(y + 10\right) y = -30$ $\;\;\;$ [in view of equation $(3b)$]
i.e. $\;$ $y^2 + 10y - 30 = 0$
i.e. $\;$ $y = \dfrac{-10 \pm \sqrt{100 + 120}}{2} = \dfrac{-10 \pm 2 \sqrt{55}}{2}$
i.e. $\;$ $y = - 5 \pm \sqrt{55}$
When $\;\;$ $y = -5 + \sqrt{55}$, $\;\;$ we have from equation $(3b)$,
$x = - \left(-5 + \sqrt{55}\right) - 10 = -5 - \sqrt{55}$
and when $\;\;$ $y = -5 - \sqrt{55}$, $\;\;$ we have from equation $(3b)$,
$x = - \left(-5 - \sqrt{55}\right) - 10 = -5 + \sqrt{55}$
$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(4, 12\right), \; \left(12, 4\right), \; \left(-5 - \sqrt{55}, -5 + \sqrt{55}\right), \; \left(-5 + \sqrt{55}, -5 - \sqrt{55}\right) \right\}$