Non-linear systems help! I have a non-linear system of equations, $$\left\{ \begin{array}{rcl} x^2 - xy + 8 = 0 \\ x^2 - 8x + y = 0 \\ \end{array} \right.$$
I have tried equating the expressions (because both equal 0), which tells me: $$x^2 - xy + 8 = x^2 - 8x + y$$
Moving all expressions to the right yields: $$0 = xy - 8x + y - 8$$
Factoring the equation: $$0 = x(y-8) + 1(y-8)$$
$$0 = (x+1)(y-8)$$
$$x=-1$$
$$y=8$$
Problem solved, right? No. When you plug in the values into the equations above, you get a false statement. Allow me to demonstrate:
$$x^2 - 8x + y = 0$$
$$(-1)^2 - 8(-1) + (8) = 0$$
$$1 - (-8) + 8 = 0$$
$$1 + 8 + 8 = 0$$
$$17 = 0$$
Can someone please help me solve this system of nonlinear equations? I am stuck.
Hint: You got $x=-1$  OR $y=8$. BTW, cool idea to equate the expressions.
Выражаешь из второго $y$. Потом подставляешь в первое вместо $y$, далее решаешь квадратное уравнение относительно $x$. Находишь два корня. Потом для каждого из корней находишь $y$. В итоге получится два корня $x$ и два корня $y$. Всё банально просто.  

Quoth google translate:


  Expresses the second $y$. Then you substitute the first instead of $y$, continue to solve quadratic equation with respect to $x$. There are two root. Then for each of the roots are $y$. That will have two roots and two root $x$ $y$. All corny easy.