Each of x, y, and z is a real number that satisfies this system of equations:

• 1/x+y+z=3
• x+1/y+z=3
• x+y+1/z=3

Determine all possible solutions to the above system of equations.

__Note__: Adapted from a problem that appeared on 2017 El Salvador Math Olympiad.

I will start by taking the difference of the first two equations.

Then x-1/x = y-1/y. There are two easy to spot solutions by inspection: x=y and x=-1/y.

These will be the only solutions: Notice that by multiplying through by x the equation becomes quadratic in x, so will have two solutions. And we already have two solutions.

The same process applies to the second and third equations: we have either y=z or y=-1/z.

So fixing x, we can use the relationships formed earlier to create (x,y,z) = (x,x,x) or (x,x,-1/x) or (x,-1/x,-1/x), or (x,-1/x,x).

Now all we need to do is plug each of these into one of the original equations. I will use the first equation.

Case 1 (x,y,z) = (x,x,x)

1/x+x+x=3 simplifies to 2x^2-3x+1=0 which has roots x=1 and x=1/2. Then (x,y,z)=(1,1,1) or (1/2,1/2,1/2)

Case 2 (x,y,z) = (x,x,-1/x)

1/x+x-1/x=3 simplifies to x=3. Then (x,y,z)=(3,3,-1/3)

Case 3 (x,y,z) = (x,-1/x,-1/x)

1/x-1/x-1/x=3 simplifies to x=-1/3. Then (x,y,z)=(-1/3,3,3)

Case 4 (x,y,z) = (x,-1/x,x)

1/x-1/x+x=3 simplifies to x=3. Then (x,y,z)=(3,-1/3,3)

The set of all solutions to the given system of equations is (x,y,z)={(1,1,1), (1/2,1/2,1/2), (3,3,-1/3), (3,-1/3,3), (-1/3,3,3)}.