Answer
$(-\frac{1}{2}, -\frac{3}{2})$ and $(\frac{1}{2},\frac{3}{2})$
Work Step by Step
Add the equations:
$x^2+2xy+y^2=4$
Factor the equation:
$(x+y)^2=4$
We have
$x+y=\pm 2$
If $x+y=-2$ or $y=-2-x$, substituting to the first equation of the system then
$x^2+x(-2-x)=1$
$x^2-2x-x^2=1$
$-2x=1$
$x=-\frac{1}{2}$
and so $y=-2-(-\frac{1}{2})=-\frac{3}{2}$.
Now, if $x+y=2$ or $y=2-x$, substituting to the first equation of the system then
$x^2+x(2-x)=1$
$x^2+2x-x^2=1$
$2x=1$
$x=\frac{1}{2}$
and so $y=2-\frac{1}{2}=\frac{3}{2}$.
Therefore, the solutions are $(-\frac{1}{2}, -\frac{3}{2})$ and $(\frac{1}{2},\frac{3}{2})$.
