Answer
$\left(\frac{7}{3},\frac{23}{3}\right)$ and $\left(\frac{23}{5},-\frac{7}{5}\right)$
Work Step by Step
$x^2-y^2-32x+128=0.....(1)$
$\underline{y^2-x^2-8y+8=0.....(2)}\ +$ (Add both equations to eliminate $x^2$ and $y^2$)
$-32x-8y+136=0$ (Divide by $-8$)
$4x+y-17=0$ (Write as $y=$)
$y=-4x+17.........(3)$
Substituting (3) into (1),
$x^2-(-4x+17)^2-32x+128=0$
$x^2-(16x^2-136x+289)-32x+128=0$
$x^2-16x^2+136x-289-32x+128=0$
$-15x^2+104x-161=0$ (Divide by $-1$)
$15x^2-104x+161=0$ (Use the factoring method)
$(3x-7)(5x-23)=0$
$x=\frac{7}{3}\vee x=\frac{23}{5}$
For $x=\frac{7}{3}$, $y=-4\cdot \frac{7}{3}+17=-\frac{28}{3}+\frac{51}{3}=\frac{23}{3}$
For $x=\frac{23}{5}$, $y=-4\cdot \frac{23}{5}+17=-\frac{92}{5}+\frac{85}{5}=-\frac{7}{5}$
Thus, the solutions are $\left(\frac{7}{3},\frac{23}{3}\right)$ and $\left(\frac{23}{5},-\frac{7}{5}\right)$.
