How to solve this system of equations

Question

I just can't seem to get this one.. I know the solutions are $(x, y) = (0, 8)$ (I can find this one) and $(x, y) = (-1, 4)$ but I can't work out how to find $(-1, 4)$.

The system is: $$y(4x-y+8)=0$$ and $$x(4x-3y+16)=0$$

Any help would be appreciated.

Cheers

Answer 1

First remove the cases $x=0,y=0$ by considering them separately.

Then solve the system

$$4x-y+8=0$$ $$4x-3y+16=0$$

Answer 2

$$If (x=0)$$

$$y*(-y+8)=0$$ $$0*(-3y+16)=0$$ $$ y=0 $$ $$ y=8 $$

$$Else If(y=0)$$

$$0*(4x+8)=0$$ $$x*(4x+16)=0$$ $$x=-4$$

$$Else$$

$$4x+8-y=4x+16-3y$$ $$2y=8$$ $$y=4$$ $$4x+8-4=0$$ $$x=-1$$