Answer
$\{(0,3), (-\frac{36}{17}, -\frac{3}{17})\}$
Work Step by Step
1. Isolate $y$ from the second equation to get $y=\frac{3}{2}x+3$
2. Use the above relation in the first equation to get $2x^2+(\frac{3}{2}x+3)^2=9$ or $2x^2+\frac{9}{4}(x+2)^2=9$ or $8x^2+9(x^2+4x+4)=36$ which gives $17x^2+36x=0$
3. The above equation gives two solutions $x=0, -\frac{36}{17}$
4. The corresponding $y$ can be found as $y=3, -\frac{3}{17}$
5. Thus the solution set is $\{(0,3), (-\frac{36}{17}, -\frac{3}{17})\}$
