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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.

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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.

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

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

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