3. Solve the system of equations:

$$
\begin{array}{r}
x-y+x y=17 \\
x^{2}+y^{2}=34
\end{array}
$$
3. By multiplying the first equation by -2 and adding it to the second, we get

$$
0=x^{2}-2 x y+y^{2}-2(x-y)=(x-y)^{2}-2(x-y)=(x-y)(x-y-2)
$$

from which it follows that $x=y$ or $x=y+2$. In the first case, the second equation reduces to $2 x^{2}=34$, so we have two solutions: $(x, y)=(\sqrt{17}, \sqrt{17})$ and $(x, y)=(-\sqrt{17},-\sqrt{17})$. In the second case, the second equation reduces to $2 y^{2}+4 y-30=0$, i.e., $y^{2}+2 y-15=0$, whose solution yields $y=\frac{-2 \pm \sqrt{64}}{2}=\frac{-2 \pm 8}{2}$, i.e., $y=-5$ or $y=3$, giving us two more solutions: $(x, y)=(-3,-5)$ and $(x, y)=(5,3)$, so the given system has a total of 4 solutions.

![](https://cdn.mathpix.com/cropped/2024_06_04_f254ba4f25ce7f8d9df1g-33.jpg?height=450&width=582&top_left_y=246&top_left_x=192)

Op 2017 3B 4