Find all pairs of positive whole numbers Find all pairs of positive whole numbers x and y which are a solution for $ \dfrac{2}{x} + \dfrac {3}{y} = 1 $.
I don't really understand how to tackle this question. I rewrote $ \dfrac{2}{x} + \dfrac {3}{y} = 1 $ as $2y+ 3x =xy$  but that's it..
 A: You can write the equation as $xy-2y-3x=0$. If you factor this you will get
\begin{equation*}
(x-2)(y-3)=6.
\end{equation*}
Do you see how to proceed?
A: If we multiply both sides the original equation by $xy$, we ge $2y+3x=xy$. We can rewrite this as $2y+3x-xy=0$. We now perform a little trick.
Note that what we have is very much like $(3-y)(x-2)=3x+2y-xy-6$. If we subtract $6$ from both sides, we get $2y+3x-xy-6=-6$ and $(3-y)(x-2)=-6$. Multiplying through by $-1$ gives $(y-3)(x-2)=6$. You can now search for pairs of small $x$ and $y$ that satisfy the condition. (Hint: Use the prime factorization of $6$ to help you.)
A: The "trick" of rewriting $2y+3x=xy$ as $(x-2)(y-3)=6$ is nice. Note its resemblance to its far more important cousin, completing the square. 
But that's not what this answer is about.  We claim that even someone totally innocent of algebra can find all the solutions of $\dfrac{2}{x}+\dfrac{3}{y}=1$, and prove that all of them have been found. All one needs to do is to engage with the numbers. 
The key observation is that $x$ and $y$ cannot both be big. We will use this observation fairly efficiently. But even an inefficient approach leads quickly to a full answer.
If $x\ge 5$ and $y\ge 5$, we are in trouble unless $x=y=5$. This satisfies the equation. Except in this case, one of $x$ or $y$ must be $\le 4$. A short list of candidates! And many of them are obviously no good. 
Can $x=1$? Of course not, the result would be too big. Similarly, $x$ cannot be $2$. 
If $x=3$, then $y=9$.  If $x=4$, then $y=6$.
Can $y=1$, $2$, or $3$? Of course not. Put $y=4$. We get $x=8$.
