+980 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 $xyz$?
+1009 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.
