Assuming \(x \), \(y\), and \(z\) are positive real numbers satisfying: \(\begin{align*} xy-z&=15, \\ xz-y&=0, \text{ and} \\ yz-x&=0, \end{align*} \)
then, what is the value of \(xyz\)?
+41 Hello!
Notice our equations 2 and 3 are equivalent.
xy-y=yz-x
y(z+1)=x(z+1)
x=y
Now we can rewrite our 2nd equation as x(z-1)=0
First we note that x cannot =0, as x,y, and z are all positive. Thus, z=1.
We can plug this in, giving us x^2-1=15. Hence, x=4 (x cannot be -4)
Therefore, x=4, y=4, and z=1. Thus, our answer is 4*4*1= 16.
