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February 26th, 2013, 09:42 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, 10:42 AM  #2 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,396 Thanks: 829  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, 11:41 AM  #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, 11:43 AM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,805 Thanks: 1045 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, 01:06 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, 01:10 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 18,954 Thanks: 1601 
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, 01:21 PM  #7 
Newbie Joined: Feb 2013 Posts: 16 Thanks: 0  Re: Simultaneous Equations
Thank you!!


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