smc

If we solve the second equation for $x$ in terms of $y$, we find $x=-\frac{y+3}{2}$ which we can substitute to find: $$2(-\frac{y+3}{2}{)}^2+6(-\frac{y+3}{2})+5y+1=0$$ Multiplying by two and simplifying, we find: $$\begin{array}{rl}2\cdot [2(-\frac{y+3}{2}{)}^2+6(-\frac{y+3}{2})+5y+1]& =2\cdot 0\\ (y+3{)}^2-6y-18+10y+2& =0\\ y^2+6y+9-6y-18+10y+2& =0\\ y^2+10y-7& =0\end{array}$$ Therefore the answer is $\boxed{y^2+10y-7=0}$