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August 7th, 2014, 07:28 AM   #1
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system

solve xy+x^2=15
x^3+y^3+xy=41
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August 7th, 2014, 07:51 AM   #2
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By inspection, $x \ne 0$, $y \ne 0$ and one solution is $(x,y)=(3,2)$.

It seems a bit nasty to find other solutions.

Edit: graphically, that's the only solution (twice). The two curves are asymptotically equal to $x+y=0$.

Last edited by v8archie; August 7th, 2014 at 08:14 AM.
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August 7th, 2014, 09:21 AM   #3
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Actually there are two more roots, but they are unpleasant beasts.

Write $$xy = 15-x^2 \qquad \text{and} \qquad x+y= \tfrac{15}{x} \qquad \text{from the first equation}\\[12pt]
\text{the second equation can be written} \\
(x+y)^3 - 3xy(x+y) + xy = 41$$

substituting in gives us a quintic in $x$, the roots of which we put into the first equation to find $y$.
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August 7th, 2014, 10:23 AM   #4
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Quote:
Originally Posted by v8archie View Post
. . . graphically, that's the only solution (twice).
Twice? There are three distinct real solutions and two other complex solutions.

Quote:
Originally Posted by v8archie View Post
. . . gives us a quintic in $x$
The quintic has one integer root: x = 3.
That leaves one with the quartic equation $x^4 - 42x^3 - 100x^2 + 375x + 1125 = 0$,
which has two real roots (3.2206527019659... and 44.063175458193...).
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August 7th, 2014, 10:47 AM   #5
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Twice?
It wasn't clear from my graph that the two roots were distinct. I didn't make it clear that I'd revised my thoughts on that.
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