# Solving these three independent equations

Could anyone help step me through solving these three equations, I get lost in massive fractions and squares etc:

\begin{align}\frac{\frac{yz}{y+z}}{x + \frac{yz}{y+z}} &= \frac{1}{2} \tag{1} \\ \frac{\frac{xz}{x+z}}{y + \frac{xz}{x+z}} &= \frac{1}{6} \tag{2} \\z &= 10 \tag{3} \end{align}

• First simplify fractions like $\frac{yz}{y}=z$ with a reminder that $y\ne 0$. Then substitute $z=10$ in the other two equations. Then, if you still get stuck along the way, edit your question and post the progress, then point at where in particular you got stuck. – dxiv May 9 '17 at 6:45
• Hi @dxiv, thanks, someone else edited my post and it changed the question. I've edited it back so you'll see yz/y = z situations don't exist. – edd91 May 9 '17 at 6:50
• That's one more good reason for you to use MathJax when posting. That said, the second part of my previous comment still applies. – dxiv May 9 '17 at 6:54

The first equation can be written as

$$\frac{yz}{xy+xz+yz}=\frac{1}{2}$$

The second equation can be written as

$$\frac{xz}{xy+xz+yz}=\frac{1}{6}$$

So we have

$$\frac{xy}{xy+xz+yz}=1-\frac{yz}{xy+xz+yz}-\frac{xz}{xy+xz+yz}=1-\frac{1}{2}-\frac{1}{6}$$

$$\frac{xy}{xy+xz+yz}=\frac{1}{3}$$

Therefore,

$$\frac{xz}{xy}=\frac{1/6}{1/3}=\frac{1}{2}$$

So $y=20$.

$$\frac{yz}{xz}=\frac{1/2}{1/6}=3$$

$\displaystyle x=\frac{20}{3}$.