Math Help - Find x,y,z

1. Find x,y,z

Find natural numbers $x,y,z$ such that

$\frac1{x}+\frac1{y}+\frac1{z}=\frac1{5}$

where $x

2. x = 10, y = 15, z = 30

\begin{aligned}
\frac{1}{10} + \frac{1}{15} + \frac{1}{30} &= \frac{1}{5} \\
\frac{3}{30} + \frac{2}{30} + \frac{1}{30} &= \frac{1}{5} \\
\frac{6}{30} &= \frac{1}{5}
\end{aligned}

I used guess-and-check, though...

Here's another solution:
x = 8, y = 16, z = 80

\begin{aligned}
\frac{1}{8} + \frac{1}{16} + \frac{1}{80} &= \frac{1}{5} \\
\frac{10}{80} + \frac{5}{80} + \frac{1}{80} &= \frac{1}{5} \\
\frac{16}{80} &= \frac{1}{5}
\end{aligned}

Here's a third solution (again, using guess-and-check):
x = 10, y = 12, z = 60

\begin{aligned}
\frac{1}{10} + \frac{1}{12} + \frac{1}{60} &= \frac{1}{5} \\
\frac{6}{60} + \frac{5}{60} + \frac{1}{60} &= \frac{1}{5} \\
\frac{12}{60} &= \frac{1}{5}
\end{aligned}

...and a fourth (this is fun ):
x = 12, y = 15, z = 20

\begin{aligned}
\frac{1}{12} + \frac{1}{15} + \frac{1}{20} &= \frac{1}{5} \\
\frac{5}{60} + \frac{4}{60} + \frac{3}{60} &= \frac{1}{5} \\
\frac{12}{60} &= \frac{1}{5}
\end{aligned}

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3. Solution #5 (I'm on a roll ):
x = 8, y = 20, z = 40

\begin{aligned}
\frac{1}{8} + \frac{1}{20} + \frac{1}{40} &= \frac{1}{5} \\
\frac{10}{80} + \frac{4}{80} + \frac{2}{80} &= \frac{1}{5} \\
\frac{16}{80} &= \frac{1}{5}
\end{aligned}

Solution #6:
x = 9, y = 15, z = 45

\begin{aligned}
\frac{1}{9} + \frac{1}{15} + \frac{1}{45} &= \frac{1}{5} \\
\frac{5}{45} + \frac{3}{45} + \frac{1}{45} &= \frac{1}{5} \\
\frac{9}{45} &= \frac{1}{5}
\end{aligned}

Solution #7:
x = 10, y = 14, z = 35

\begin{aligned}
\frac{1}{10} + \frac{1}{14} + \frac{1}{35} &= \frac{1}{5} \\
\frac{7}{70} + \frac{5}{70} + \frac{2}{70} &= \frac{1}{5} \\
\frac{14}{70} &= \frac{1}{5}
\end{aligned}

Solution #8 (this is getting ridiculous):
x = 6, y = 40, z = 120

\begin{aligned}
\frac{1}{6} + \frac{1}{40} + \frac{1}{120} &= \frac{1}{5} \\
\frac{20}{120} + \frac{3}{120} + \frac{1}{120} &= \frac{1}{5} \\
\frac{24}{120} &= \frac{1}{5}
\end{aligned}

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