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 $xyz$?

qwertyzz May 4, 2020

#1**0 **

Haven't done these in some time!

So.

Bullets 2 and 3 tell us that xz = y, and yz = x.

xz = y

yz = x

Add the two equations gives

xz + yz = x+y

= z(x + y)

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

Ayyyy!

xy - 1 = 15

xy = 16

16*1 = 16

If you don't understand anything feel free to ask.

Of course idk if this is the right answer but i'm reasonably sure that this is.

hugomimihu May 4, 2020