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 February 26th, 2013, 10:42 AM #1 Newbie   Joined: Feb 2013 Posts: 16 Thanks: 0 Simultaneous Equations Hi, I've been trying to solve these simultaneous equations, but I just can't. Any help would be greatly appreciated. x + y + z = 5 1/x + 1/y + 1/z = 1/5 xy + xz + yz = -9
 February 26th, 2013, 11:42 AM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,060 Thanks: 722 Re: Simultaneous Equations Solution a bit long...go enter x+y+z+1/x+1/y+1/z+x*y+x*z+y*z+19/5 = 0 here: http://www.numberempire.com/equationsolver.php Perhaps time to quit?
 February 26th, 2013, 12:41 PM #3 Newbie   Joined: Feb 2013 Posts: 16 Thanks: 0 Re: Simultaneous Equations Thanks for the reply, but that website just gives me the solutions. I'm actually just interested in the method used to solve.
 February 26th, 2013, 12:43 PM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,660 Thanks: 965 Math Focus: Elementary mathematics and beyond Re: Simultaneous Equations Are x, y and z integers? Or are they real numbers? Rational numbers, perhaps?
 February 26th, 2013, 02:06 PM #5 Newbie   Joined: Feb 2013 Posts: 16 Thanks: 0 Re: Simultaneous Equations The solutions are integers. By inspection, I can see that the solutions are -3, 3, and 5, but that doesn't tell me which value corresponds to each variable, which is why I'm looking for an algebraic method for solving.
 February 26th, 2013, 02:10 PM #6 Global Moderator   Joined: Dec 2006 Posts: 18,245 Thanks: 1439 Multiplying the second equation by xyz gives xy + xz + yz = xyz/5, so xyz = -45. From the third equation, xyz + xz² + yz² = -9z, so -9z = -45 + z²(5 - z). Hence z³ - 5z² - 9z + 45 = 0, which factorizes as (z + 3)(z - 3)(z - 5) = 0. It's easy to finish from there.
 February 26th, 2013, 02:21 PM #7 Newbie   Joined: Feb 2013 Posts: 16 Thanks: 0 Re: Simultaneous Equations Thank you!!

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