1. ## Finding ordered triples.

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

3. ## Re: Finding ordered triples.

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?
\begin{align*}X+Y&=120 \\Y+Z&=96 \\X+Z&=72\end{align*}

4. ## Re: Finding ordered triples.

D'oh! Just solve for x, y, and z. It's that easy and I didn't even see it?

Thanks!

5. ## Re: Finding ordered triples.

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 $A$ where $A^2=X+Y+Z$
The final answer looks $\pm x,~\pm y,~\pm z$

6. ## Re: Finding ordered triples.

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 $A$ where $A^2=x+y+z$
What does "divide A" mean?

Originally Posted by Plato
The final answer looks $\pm x,~\pm y,~\pm z$
It seems to me that there are only two triples that satisfy the original system.

7. ## Re: Finding ordered triples.

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 $x=X/A,~y=Y/A,~z=Z/A$ where $A^2=X+Y+Z$.

Take a look at this.