Suppose that $x,$ $y,$ and $z$ are three positive numbers that satisfy the equations $xyz = 1,$ $x + \frac {1}{z} = 5,$ and $y + \frac {1}{x} = 29.$ Find $z + \frac {1}{y}.$
Well, we can start by solving for $x$. We can take the second equation and multiply both sides by $z$ to get $xz + 1 = 5z$. Yes, that gives us $x+\frac{1}{\frac{1}{xy}}=5$, so $x+xy=5$. Now, let's solve for $y$. We can substitute $x=\frac{5}{1+y}$ into $y + \frac {1}{x} = 29$. That gives us $y+\frac{1}{\frac{5}{1+y}}=29$, so $y+\frac{1+y}{5}=29$. We can multiply both sides by 5 to get $5y+1+y=145$, so $6y=144$ and $y=24$. Now let's solve for $x$. We have $x=5-xy$, so $x=5-24x$. Finally, we plug in $x=\frac{1}{5}$ to find $z$. We already know $xyz=1$, so $24*\frac{1}{5}*z=1$, so $z=\frac{5}{24}$. So $z+\frac{1}{y}=\frac{5}{24}+\frac{1}{24}=\frac{1}{4}$.

