## Trigonometry 7th Edition

When $y=4+2i$, $x=4-2i$ when $y=4-2i$, $x=4+2i$
Working from the first equation, $x+y=8$, we make $x$ the subject of the formula, $x=8-y$ We then sub this modified version of the first equation into the second equation , $xy=20$ to get $(8-y)y=20$. Expanding this equation and moving all terms to the left hand side, we get: $-y^{2}+8y-20=0$ The method used in the introduction then uses the quadratic method to solve for the equation. Therefore, we get: $y=\frac{-8\pm\sqrt{8^{2}-4(-1)(-20)}}{2(-1)}\\=4\pm\frac{\sqrt{64-80}}{-2}\\=4\pm\frac{\sqrt{-16}}{-2}\\=4\pm\frac{\sqrt{(16)(-1)}}{2}\\=4\pm2\sqrt{(-1)}\\=4\pm2i$ From this, we can deduce the corresponding values of $x$. When $y=4+2i$, $x=8-(4+2i)\\=4-2i$ when $y=4-2i$, $x=8-(4-2i)\\=4+2i$