## Triples of Real Numbers

What are all ordered triples of real numbers $(x,y,z)$ that satisfy
$\left (x+y\right )\left (x+y+z\right )=120$
$\left (y+z\right )\left (x+y+z\right )=96$
$\left (x+z\right )\left (x+y+z\right )=72$?
Source: NCTM Mathematics Teacher, October 2006

Solution
$\left (x+y+z\right )\left [\left (x+y\right )+\left (y+z\right )+\left (x+z\right )\right ]=288$
$\left (x+y+z\right )\left (2x+2y+2z\right )=288$
$\left (x+y+z\right )\left (x+y+z\right )=144$
$x+y+z=\pm 12$
Substitute the value of $x+y+z=+12$ into the equations
$x+y=10$
$y+z=8$
$x+z=6$
Solve for $x,y,z$ yields $\left (x,y,z\right )=\left (4,6,2\right )$.
Substitute the value of $x+y+z=-12$ into the equations
$x+y=-10$
$y+z=-8$
$x+z=-6$
Solve for $x,y,z$ yields $\left (x,y,z\right )=\left (-4,\!-6,\!-2\right )$.

Answer: $\left (4,6,2\right )$ and $\left (-4,\!-6,\!-2\right )$