Find all ordered triples (x,y,z) of real numbers which satisfy: (x+y)(x+y+z)=120 (y+z)(x+y+z)=96 (x+z)(x+y+z)=72
Last edited by wannabebrainiac; Jun 4th 2012 at 12:42 PM.
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Hint: Add all equations.
Originally Posted by wannabebrainiac Find all ordered triples (x,y,z) of real numbers which satisfy: (x+y)(x+y+z)=120 (y+z)(x+y+z)=96 (x+z)(x+y+z)=72 Can you solve this system?
Last edited by Plato; Jun 4th 2012 at 02:36 PM.
D'oh! Just solve for x, y, and z. It's that easy and I didn't even see it? Thanks!
Originally Posted by wannabebrainiac D'oh! Just solve for x, y, and z. It's that easy and I didn't even see it? Thanks! It is not that easy You now need to divide where The final answer looks
Last edited by Plato; Jun 4th 2012 at 02:37 PM.
I am a bit lost. What does the system in post #3 have to do with the original system? Originally Posted by Plato You now need to divide where What does "divide A" mean? Originally Posted by Plato The final answer looks It seems to me that there are only two triples that satisfy the original system.
Originally Posted by emakarov I am a bit lost. What does the system in post #3 have to do with the original system? What does "divide A" mean? It seems to me that there are only two triples that satisfy the original system. You are correct there are only two triples. I realize that was misleading, sorry. When I types up the original the caps got changed. I will edit it. Then where . Take a look at this.
Last edited by Plato; Jun 4th 2012 at 02:50 PM.
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