Make a $2$ unit bottle from $4$ and $3$ unit bottles Assume that we have only two bottles. The first one's volume is $3$ units and the second one's volume is $4$ units.  We can perform $3$ kinds of operations :  

  
*
  
*Fill an empty bottle with water.   
  
*Empty a full bottle.  
  
*Pouring the contents of one bottle into the other one.
  

How can we reach a bottle with $2$ units of volume watter inside it?

The way to solve this kind of problem is generally similar.
Given $a$ and $b$ with $gcd(a,b)=1$ we can apply Bezout to find $(u,v)$ such that $au+bv=1$.
In fact here $1$ (1 liter) is not interesting, but similarly there exists $(u,v)$ such that $au+bv=2$.
Here $4*(+2)+3*(-2) = 8-6=2$, the signs are also of interest:


*

*$+$ indicates the recipient which is filled 

*$-$ the recipient which is emptied (this is also the one we pour into).


\begin{array}{c|ccc}
action & \text{4 lit.} & \text{3 lit.} \\
\hline
 & 0 & 0  \\
fill & 4 & 0 & + \\
pour & 1 & 3 \\
empty & 1 & 0 & - \\
pour & 0 & 1 \\
fill & 4 & 1 & +\\
pour & 2 & 3 \\
empty & 2 & 0 & - 
\end{array}
So you see we filled twice the 4 liters bottle and emptied twice the 3 liters bottle to finally get $2l$ in the 4 liters bottle. 
Let's examine another situation: bottle of 7 liters and 11 liters and I want 6 liters.
Here $7*(+4)+11*(-2)=28-22=6$, the signs indicate we are filling the 7 liters bottle 4 times and we pour in the 11 liters bottle and empty it 2 times. Let's go...
\begin{array}{c|ccc}
action & \text{7 lit.} & \text{11 lit.} \\
\hline
 & 0 & 0  \\
fill & 7 & 0 & + \\
pour & 0 & 7 \\
fill & 7 & 7 & + \\
pour & 3 & 11 \\
empty & 3 & 0 & - \\
pour & 0 & 3 \\ 
fill & 7 & 3 & + \\ 
pour & 0 & 10 \\ 
fill & 7 & 10 & + \\ 
pour & 6 & 11  \\
empty & 6 & 0 & -
\end{array}
So next time you have a problem with two bottles of contenance $a,b$ and you want $c$, try to find first the equation $au+bv=c$. In general it can be found by hand quite quickly.
