Find the number of ordered triples (x,y,z) of real numbers that satisfy

\(\begin{align*} x + y - z &= 0, \\ xz - xy + yz &= 27, \\ xyz &= 54. \end{align*}\)

Guest Jun 27, 2019

#1**0 **

I don't really have a good method, but this is what I did:

notice x*y*z = 54

So prime factorize 54

3 * 3 * 3 * 2

Ok so with the prime factorization, it isn't tedious at all to guess and check with those numbers into the other equations.

x + y - z = 0 (the first equation) is also:

x + y = z

and

xyz = 54

It is much easier now to guess and check with only the prime factorization.

The second equation is very complicated we can use it to check our three x, y, and z values later.

So with our prime factorization 3 * 3 * 3 * 2

I guessed that (3, 3, 6) is a solution to the equations we have above. Then we check with the second equation to make sure it is true, which it is.

Ok, notice how our answer also could be negative. However, that can't be true because xyz = 54 has to have two negative numbers but the only solution to x + y = z has three.

So the answer is 1 I believe...

This problem was more of a logic puzzle with guessing in my opinion.

CalculatorUser Jun 27, 2019