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
Add the three equations
\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 )

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About mvtrinh

Retired high school math teacher.
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