Determine all possible triplets (x,y,z) of real numbers that satisfy this system of equations:
• x2-xy-xz = 5
• y2-yz-xy = -4
• z2-xz-yz = -7
I'll start by multiplying the first equation by y and the second by x, then taking the sum of those.
That results in -2xyz = 5y-4x.
Similarly, multiply the first equation by z and the third by x, then taking the sum of those.
That results in -2xyz = 5z-7x.
And also, multiply the second equation by z and the third by y, then taking the sum of those.
That results in -2xyz = -4z-7y.
Now for each of the three derived equations, divide through by xyz. That leaves us with:
5/(xz) - 4/(yz) = -2
5/(xy) - 7/(yz) = -2
-4/(xy) - 7/(xz) = -2
Treat this as a system of linear equations in 1/(xy), 1/(xz) and 1/(yz). Then: 1/(yz)=4/7, 1/(xz)=2/35 and 1/(xy)=5/2
Taking reciprocals yields yz=7/4, xz=35/2, and xy=2/5.
Then multiplying two and dividing out the third gives z^2=49/4, or z=+/-7/2; then y=+/-1/2 and x=+/-5.
The signs are matched, so then (x,y,z) = (5, 1/2, 7/2) or (-5, -1/2, -7/2).
Solution
