Let x,y,z be nonzero real numbers, such that no two are equal, and

\(x + \frac{1}{y} = y + \frac{1}{z} = z + \frac{1}{x}. \)

Find all possible numeric values of xyz

I got 1 as one of the answers.

MeldHunter Jul 23, 2024

#1**+1 **

Ok, first off, let's split the one equation into a system of equations so that each term is equal to another. We get

\(x + 1/y = y + 1/z\\ x + 1/y = z + 1/x\\ y + 1/z = z + 1/x\)

I won't really show the steps, although I will if needed. Let me know if you want to.

\((x, y, z) = (-\frac{1}{z-1}, \frac{z-1}{z}, z)\\ (x, y,z) = (-\frac{1}{z+1}, -\frac{z+1}{z}, z)\)

We want to find the value of xyz since we need the product.

Thus, multiplying these together, we get

\( (-\frac{1}{z-1})(\frac{z-1}{z})z=-\frac{1}{z-1}\left(z-1\right)=-1\)

\( (-\frac{1}{z+1} )(-\frac{z+1}{z}) z=\frac{1}{z}z=1\)

So good job! You were correct about 1. The other value is -1.

I didn't go into too much detail, but let me know if you need more assistance.

Thanks! :)

NotThatSmart Jul 23, 2024