Answer
$\{(\frac{14}{5},-\frac{2}{5}),(-2,-2)\}$
Work Step by Step
1. Use substitution, the first equation gives $x=3y+4$ and the second equation becomes $(3y+4)^2+y^2=8$, $10y^2+24y+8=0$, or $5y^2+12y+4=0$
2. Factor the above to get $(5y+2)(y+2)=0$, thus $y=-\frac{2}{5}, -2$
3. Back substitute the above to the first equation to get $x=\frac{14}{5},-2$
4. Thus the solution set is $\{(\frac{14}{5},-\frac{2}{5}),(-2,-2)\}$
