# Given x, y, and zinRR^+, what is the solution to the system of equations xy=a, yz=b, and xz=c?

##### 1 Answer
Dec 28, 2015

$\left\{\begin{matrix}x = \sqrt{\frac{a c}{b}} \\ y = \sqrt{\frac{a b}{c}} \\ z = \sqrt{\frac{b c}{a}}\end{matrix}\right.$

#### Explanation:

Before continuing, let's note that as $x , y , z > 0$ we have $a , b , c > 0$ and so we may take square roots of any product or quotient thereof, and do so without considering negative solutions.

$\left\{\begin{matrix}x y = a \\ y z = b \\ x z = c\end{matrix}\right.$

From the first equation, we have

$y = \frac{a}{x}$

Substituting this into the second equation:

$b = y z = \frac{a}{x} z$

$\implies z = \frac{b}{a} x$

Substituting this into the third equation:

$c = x z = \frac{b}{a} {x}^{2}$

$\implies {x}^{2} = \frac{a c}{b}$

$\implies x = \sqrt{\frac{a c}{b}}$

Substituting our solution for $x$ into our intermediate result from working on the first equation:

$y = \frac{a}{x} = \frac{a}{\sqrt{\frac{a c}{b}}} = \sqrt{\frac{a b}{c}}$

Substituting our solution for $x$ into our intermediate result from working on the second equation:

$z = \frac{b}{a} x = \frac{b}{a} \sqrt{\frac{a c}{b}} = \sqrt{\frac{b c}{a}}$

$\therefore \left\{\begin{matrix}x = \sqrt{\frac{a c}{b}} \\ y = \sqrt{\frac{a b}{c}} \\ z = \sqrt{\frac{b c}{a}}\end{matrix}\right.$