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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: 7,675 Thanks: 2655 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: 7,675 Thanks: 2655 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. Tags system Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post SGH Number Theory 6 August 30th, 2013 09:02 AM SGH Number Theory 3 August 27th, 2013 02:09 PM consigliere- Algebra 2 April 13th, 2013 09:47 AM ARTjoMS Calculus 2 November 20th, 2011 04:28 AM zain Elementary Math 1 April 7th, 2008 02:23 PM

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