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

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.
