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 $$xy = 15x^2 \qquad \text{and} \qquad x+y= \tfrac{15}{x} \qquad \text{from the first equation}\\[12pt] \text{the second equation can be written} \\ (x+y)^3  3xy(x+y) + xy = 41$$ substituting in gives us a quintic in $x$, the roots of which we put into the first equation to find $y$. 
August 7th, 2014, 10:23 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,469 Thanks: 2038  Twice? There are three distinct real solutions and two other complex solutions. 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 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,635 Thanks: 2620 Math Focus: Mainly analysis and algebra  

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