+0

0
95
1

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 $$xyz$$?

Jan 4, 2020

#1
+42
+2

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.

Jan 4, 2020