Question 981149
What did you want with these?  Solution for the two as a system?   If that, you can solve first equation for x in terms of y, and then substitute this for x in the second  (quadratic) equation.  


{{{(2y+2)^2+5(2y+2)+3y=3}}}


{{{4y^2+8y+4+10y+10+3y=3}}}


{{{4y^2+21y+14=3}}}


{{{4y^2+21y+11=0}}}
This is not easily factored.
{{{y=(-21+- sqrt(567))/8}}}-------after steps from formula of general solution to quadratic equation
.
{{{y=(-21+- 9sqrt(7))/8}}}-----------and you can find the x values corresponding to each of these.