Finding $1998x + 5y + 3z$ given $\frac{1}{x} + \frac{1}{y} = \frac{3}{z}$ Suppose $x,y,z$ are positive integers so that $y$ and $z$ are prime and $x = yz.$ If they satisfy $$\frac{1}{x} + \frac{1}{y} = \frac{3}{z},$$ find $1998x + 5y + 3z.$

I first moved $\frac{1}{y}$ over to the other side and expanded out, but from there I got stuck. Can someone help me?

You are on the right track. After that you should have
$$\frac{1}x=\frac{3}z-\frac{1}y\Longrightarrow \frac{1}{yz}=\frac{3y-z}{yz}\Longrightarrow 1=3y-z\Longrightarrow 3y=z+1$$
Now note that if $y$ is odd, then $z$ has to be even. And since $z$ is a prime, $z$ has to be $2$, but in this case we will have $y=1$ which is not a prime.
Therefore $y$ has to be an even prime. That is, $\color{red}{y=2}$.
Can you end it now?
