Prove that solving equation is possible The equation:
$${2 \over x} + {3 \over y} = {5 \over z} ,\,\,x < z$$
$x,y,$ and $z$ must be natural numbers.
How would I go about proving this equation possible to solve? 
I can't seem to figure out how to prove it without doing some "random" calculations to see how numbers fit in the equation.
Example:
The variables for the possibly lowest answer is: $$x=1,y=6,z=2$$
Since:
$$
{2 \over 1} + {3 \over 6} = {5 \over 2} ,\,\,1 < 2
$$
How would it be possible to find this answer, or any other correct answer for that matter, without using "random" placeholder variables to see how to equation evolves?
Simpler put: How to is it possible to solve this equation faster and more effectively? Is it possible to determine whether or not the equation is solvable before trying to solve it?  

Firstly if $x<z$ then $y>z$.Because if $y<z$,$$\frac{2}{x}+\frac{3}{y}>\frac{2}{z}+\frac{3}{z}=\frac{5}{z}$$Contradiction!
So let $x=z-a$ and $y=z+b$.Plug these values in the equation and you will get,$$z=\frac{5ab}{3b-2a}=\frac{5}{\frac{3}{a}-\frac{2}{b}}$$
Hence minimum value of z can be 5.
Therefore,
$$\frac{3}{a}-\frac{2}{b}=1\implies b=\frac{2a}{3-a}=\frac{2}{\frac{3}{a}-1}$$
The minimum value of $\frac{3}{a}-1$ is 2 (because if $\frac{3}{a}-1=1$ a = 1.5(a fraction))
Hence minimum value of b=2/2=1.$$\frac{3}{a}-1=2\implies a=1$$
So,$$x=z-a=5-1=4$$ $$y=z+b=5+1=6$$
