Answer
Either $x=13$ and $y=7$ or $x=7$ and $y=13$
Work Step by Step
equation 1: $x + y =20$
equation 2: $x^{2} + y^{2} = 218$
From equation 1: $x = 20-y$ substituting this to equation 2: $$(20-y)^{2} + y^{2} = 218$$ $$400 - 40y + y^{2} + y^{2} = 218$$
Simplify:
$$2y^{2} - 40y + 182 = 0$$
Use the quadratic formula to get the roots: $a=2$, $b=-40$, $c=182$ $$y = \frac{-b±\sqrt{b^2-4ac}}{2a}$$ $$y = \frac{-(-40)±\sqrt{(-40)^2-4(2)(182)}}{2(2)}$$
$$y = 13, y=7$$
Substituting to equation 1:
$$x + y = 20$$
If $y =13$,
$$x +13 = 20$$ $$x = 7$$
If $y=7$,
$$x + 7 = 20$$ $$x = 13$$
