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 August 7th, 2014, 07:28 AM #1 Senior Member   Joined: Nov 2013 Posts: 434 Thanks: 8 system solve xy+x^2=15 x^3+y^3+xy=41
 August 7th, 2014, 07:51 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,635 Thanks: 2620 Math Focus: Mainly analysis and algebra 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.
 August 7th, 2014, 09:21 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,635 Thanks: 2620 Math Focus: Mainly analysis and algebra 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 $y$. Thanks from topsquark
August 7th, 2014, 10:23 AM   #4
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Quote:
 Originally Posted by v8archie . . . graphically, that's the only solution (twice).
Twice? There are three distinct real solutions and two other complex solutions.

Quote:
 Originally Posted by v8archie . . . 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...).

August 7th, 2014, 10:47 AM   #5
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Quote:
 Originally Posted by skipjack 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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