General way to solve motion problems Is there a good way to solve the following type of problem ?.

  
*
  
*A tank can be filled using pipes A, B, or both.
  
*It takes pipe A, running alone, $x$ hours to fill the tank.
  
*It takes both pipes, running together, $y$ hours to fill the tank.
  

How long does it take pipe B, running alone, to fill the tank ?.
I am thinking something along the lines of
$$
{A \over x} + {B \over z} = {1 \over y}
$$
 A: I would approach the problem in the following way.  We will use several times the equation
$$P = rt,$$
where $P$ is the percentage of the tank filled over time $t$ by a pipe filling with rate $r$. The first step is to fill in the known information:
Denote the rate at which pipe A fills by $r_A$. Then: $1 = r_A x$. Note that we use $P=1$ since the tank is completely filled (i.e., 100% is filled) after time $x$.
Similarly, for both pipes we know: $1 = (r_A+r_B) y$.  
Now we use a little bit of algebra to solve for $r_A$ in the first equation: $r_A = 1/x$, and plug this into the second equation:
$$1 = (1/x+r_B)y.$$
Solving for $r_B$:
$$r_B = \frac{1}{y} - \frac{1}{x} = \frac{x-y}{xy}.$$
Then, we deduce that the necessary time for pipe B to fill the tank by itself is $z$ solving:
$$1 = r_B z.$$
Since we already know $r_B$ we can solve for $z$:
$$z = \frac{1}{r_B} = \frac{xy}{x-y}.$$
A: Start by specifying what $A$ and $B$ are.  If they are the volume of water to be filled, your equation is correct, but if they are each the volume of the tank, you would have them equal to each other.  Not helpful.
So, how about $A$ = the rate (say, gallons per hour) with which pipe A delivers water to the tank, similarly for $B$.  Then the volume of the tank is $Ax$, but this same volume is also $Bz$.  Having the pipes deliver water simultaneously will fill the tank in $y$ hours: the volume of the tank is also $(A+B)y$.
