Answer
$y=2+i2\sqrt{3}$, $x=1-i\sqrt{3}$
$y=2-i2\sqrt{3}$, $x=1+i\sqrt{3}$
Work Step by Step
Working from the first equation, $2x+y=4$, we make $x$ the subject of the formula, $x=\frac{4-y}{2}$
We then sub this modified version of the first equation into the second equation , $xy=8$ to get $(\frac{4-y}{2})y=8$.
Expanding this equation and moving all thew terms to the left hand side, we get: $-\frac{y^{2}}{2}+2y-8=0$
The method used in the introduction then uses the quadratic method to solve for the equation. Therefore, we get:
$y=\frac{-2\pm\sqrt{2^{2}-4(-\frac{1}{2})(-8)}}{2(-\frac{1}{2})}\\=2\pm\frac{\sqrt{4-16}}{-1}\\=2\pm\sqrt{-12}\\=2\pm\sqrt{(12)(-1)}\\=2\pm\sqrt{12}\sqrt{-1}\\=2\pm i2\sqrt{3}$
From this, we can deduce the corresponding values of $x$.
When $y=2+i2\sqrt{3}$, $x=\frac{4-(2+i2\sqrt{3})}{2}\\=1-i\sqrt{3}$
when $y=2-i2\sqrt{3}$, $x=\frac{4-(2-i2\sqrt{3})}{2}\\=1+i\sqrt{3}$
