<s>[INST] 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?  
 [/INST] 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,$$x=z-a=5-1=4$$ $$y=z+b=5+1=6$$
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