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Problem of the week #64 - June 17th, 2013

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Jameson

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Jan 26, 2012
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Jameson

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Jan 26, 2012
4,043
Congratulations to the following members for their correct solutions:

1) MarkFL
2) BAdhi
3) kaliprasad
4) soroban
5) anemone
6) Sudharaka

Solution (from BAdhi):
$\begin{align}
xy(x+y)&=30\qquad(1)\\
x^3+y^3&=35\qquad(2)\\
3(1)+(2)\implies \qquad x^3+3x^2y+3xy^2+y^3&=90+35\\
(x+y)^3&=125\\
(x+y)&=5\qquad (a)\\
\text{from (1)} \implies \qquad xy&=6\qquad (b)\\
\text{from (a)} \implies \qquad x(5-x)&=6\\
x^2-5x+6&=0\\
(x-2)(x-3)&=0\\
x=2,& x=3
\newline
\text{from (b)} \implies \qquad y=3,& y=2
\end{align}$

Therefore final answer is,
$$\text{$x=2$ and $y=3$}\\
\text{or}\\
\text{$x=3$ and $y=2$}$$
 
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