0
$\begingroup$

My question is:

Solve simultaneously:

$\left\{\begin{align*} &\frac{xy}{x+y}=1\\ &\frac{xz}{x+z}=2\\ &\frac{yz}{y+z}=3 \end{align*}\right.$

I am unable to solve this simultaneous equation. Any hints to solve this question would be of great help.

2 Answers 2

1

Clearly $x,y$, and $z$ are all non-zero: otherwise at least two of the fractions would be $0$. Thus, we may invert everything and rewrite the system as

$\left\{\begin{align*} &\frac{x+y}{xy}=1\\\\ &\frac{x+z}{xz}=\frac12\\\\ &\frac{y+z}{yz}=\frac13\;. \end{align*}\right.$

Now divide out on the lefthand side:

$\left\{\begin{align*} &\frac1x+\frac1y=1\\\\ &\frac1x+\frac1z=\frac12\\\\ &\frac1y+\frac1z=\frac13\;. \end{align*}\right.$

Now substitute $u=\dfrac1x,v=\dfrac1y$, and $w=\dfrac1z$ to get

$\left\{\begin{align*} &u+v=1\\\\ &u+w=\frac12\\\\ &v+w=\frac13\;. \end{align*}\right.$

This is a very easy system to solve, and once you’ve solved it, you can easily get $x,y$, and $z$.

  • 0
    @WimC: Yes, I caught that as soon as I hit *Post*; I figured that that’s what the waiting comment would be. :-) Thanks.2012-06-02
0

A more complicated approach:

As with Brian's solution, note that all $x,y,z \neq 0$.

Note that $z\neq2$, since if it was then the 2nd equation would give an immediate contradiction. Similarly, $z\neq 3$ using the 3rd equation.

Then the 2nd equation gives $x(z-2) = 2z$, or $x = \frac{2z}{z-2}$. The 3rd equation gives $y(z-3) = 3z$, or $y = \frac{3z}{z-3}$.

Substitution into the 1st equation gives: $\frac{6z^2}{(z-2)(z-3)} = \frac{2z}{z-2} + \frac{3z}{z-3},$ or equivalently, $z(z+12) = 0$. From this we get $z=-12$, $x = \frac{12}{7}$, $x = \frac{12}{5}$.