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avatar+941 

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 $xyz$?

 May 4, 2020
 #1
avatar+986 
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.

 May 4, 2020
 #2
avatar+111566 
-1

Nicely answered Hugo,

It is not finished of course but qwerty can do that.

 

(Unfinished answers are better) 

 May 4, 2020

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