# 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.

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

• do you have a shorter way? please – Phạm Mạnh Hùng Dec 28 '14 at 15:57
• Why shorter? You can solve for y very easily and then only a bit of algebra remains. This is a good hint. – 123 Dec 28 '14 at 16:02