The equation xyz=1500 can be thought of as a system of equations. This particular system has 3 unknowns and 1 equation. As a result of this, there is not enough information to determine the exact values of x,y and z.
Consider this (incorrect) argument. Let's suppose that x=1, y=1 and z=1500. Then xyz=(1)(1)(1500)=1500. This satisfies our given equation, so these must be the values of our variables.
But this is incorrect, because if x=10, y=10, and z=15, then xyz=(10)(10)(15)=1500. This still satisfies our equation, but none of the variables have the same value as previously concluded.
In general, if you have n equations and k unknowns with k>n, then there is not a unique solution to the system.
