The system is:
[tex]\frac{x}{y}=2x + 4[/tex]
[tex]\frac{x-3}{y-3}=\frac{2}{3}[/tex] (y musn't be 3).
Solving for y in the first equation gives you:
[tex]y=\frac{x}{2x+4}[/tex]
Simplify the second equation to:
[tex]2(y-3)=3(x-3)[/tex]
Insert y you solved in first equation to second equation to get:
[tex]\frac{x}{x+2}-6=3x-3[/tex]
Further simplification gets you to quadratic equation:
[tex]x=(3x+3)(x+2)\implies x=3x^2+9x+6[/tex] [tex]3x^2+8x+6=0[/tex].
Solving quadratic equation using quadratic formula gives you two solutions of x in complex plane:
[tex]x_1=\boxed{-\frac{8}{6}+\frac{\sqrt{8}}{6}i}[/tex]
[tex]x_2=\boxed{-\frac{8}{6}-\frac{\sqrt{8}}{6}i}[/tex]
Insert the found x-es into first equation to find the y-s:
[tex]y_1=\boxed{-\frac{1}{2}+\frac{\sqrt{2}}{2}i}[/tex]
[tex]y_2=\boxed{-\frac{1}{2}-\frac{\sqrt{2}}{2}i}[/tex]
So there are two solutions (points) [tex]P_1(x_1,y_1),P_2(x_2,y_2)[/tex] both in [tex]\mathbb{C}[/tex] plane.
And the two fractions are:
[tex]\frac{-\frac{8}{6}+\frac{\sqrt{8}}{6}i}{-\frac{1}{2}+\frac{\sqrt{2}}{2}i}[/tex].
[tex]\frac{-\frac{8}{6}-\frac{\sqrt{8}}{6}i}{-\frac{1}{2}-\frac{\sqrt{2}}{2}i}[/tex].
Hope this helps.
