Answer
$y=3+i3\sqrt{2}$, $x=1-i\sqrt{2}$
$y=3-i3\sqrt{2}$, $x=1+i\sqrt{2}$
Work Step by Step
Working from the first equation, $3x+y=6$, we make $x$ the subject of the formula, $x=\frac{6-y}{3}$
We then sub this modified version of the first equation into the second equation , $xy=9$ to get $(\frac{6-y}{3})y=9$.
Expanding this equation and moving all thew terms to the left hand side, we get: $-\frac{y^{2}}{3}+2y-9=0$
The method used in the introduction then uses the quadratic method to solve for the equation. Therefore, we get:
$y=\frac{-2\pm\sqrt{2^{2}-4(-\frac{1}{3})(-9)}}{2(-\frac{1}{3})}\\=3\pm\frac{3\sqrt{4-12}}{2}\\=3\pm\frac{3\sqrt{-8}}{2}\\=3\pm\frac{6\sqrt{(2)(-1)}}{2}\\=3\pm 3\sqrt{2}\sqrt{-1}\\=3\pm i3\sqrt{2}$
From this, we can deduce the corresponding values of $x$.
When $y=3+i3\sqrt{2}$, $x=\frac{6-(3+i3\sqrt{2})}{3}\\=1-i\sqrt{2}$
when $y=3-i3\sqrt{2}$, $x=\frac{6-(3-i3\sqrt{2})}{3}\\=1+i\sqrt{2}$
