Solve the simultaneous equations $(x+y)^2+3y^{2}=7$ and $x+2y(x+1)=5$

Question

Solve this pair of simultaneous equations: $$\begin{cases} (x+y)^2+3y^{2}&\!\!\!\!\!=7, \\[2pt] x+2y\,(x+1)&\!\!\!\!\!=5. \end{cases} $$ I tried expanding the equations and differencing them, which gives $$x^2-x+4y^2-2y=2,$$ but I don't know what to do next.

Answer 1

Write the first equation in the form $$ x^2+4xy+4y^2-2xy=7 $$ or $$ (x+2y)^2-2xy=7 $$ Then rewrite the second equation as $$ (x+2y)+2xy=5 $$ Now set $u=x+2y$ and $v=xy$, to get the system $$ \begin{cases} u^2-2v=7\\ u+2v=5 \end{cases} $$ Long, but safe.

Answer 2

Hint: Solve for $y$ in the second equation and then substitute in the first one.