Let x, y, z be real numbers that satisfy this system of equations:
• xy+4z=60
• yz+4x=60
• zx+4y=60
Let S be the set of possible values of x.
Determine the sum of the squares of the elements of S.
Take the difference of the first and second equations: xy-yz+4z-4x=0
Then factor: (x-z)*(y-4) = 0
Then either x=z or y=4.
Do the same thing with the second and third equations and get either x=y or z=4.
So we have four cases:
x=y and x=z
x=z and z=4
x=y and y=4
y=4 and z=4
Case 1 substituted into any of the original equations yields x^2+4x=60. This quadratic has roots x=-10 and x=6. Then (x,y,z) = (-10,-10,-10) or (6,6,6).
Case 2 substituted into any of the original equations yields 4y+16=60, which makes y=11. Then (x,y,z)=(4,11,4)
Case 3 and 4 are essentially the same as case 2, but cyclically rotated. Then (x,y,z)=(11,4,4) or (4,4,11).
x takes on four possible values, so then set S is {-10,4,6,11} and the sum of the squares of it's elements is (-10)^2+4^2+6^2+11^2 = 273.
Solution
