Work Problem: Painting a Home A and B working together can finish painting a home in six days. A working alone can finish it in 5 days less than B. How long will it take each of them to finish the work alone?
The answer is $B = 15$, $A = 10$. I tried:
$${ { \frac{1}{A} }+{ \frac{1}{B} }={ \frac{1}{6} } }$$
$${ { \frac{1}{A} }={ \frac{1}{B}-{ \frac{1}{5} } } }$$
I get $B = 5.454545$. What am I doing wrong? Any hints?
 A: Your second equation should be
$$
A = B-5
$$
since in the first equation, you are using $A$ and $B$ to represent the number of days each person would require to complete the job alone.
Alternatively, you could rewrite the equations as
$$
A+B = \frac{1}{6}
$$
$$
\frac{1}{A} = \frac{1}{B}-5
$$
where $A$ and $B$ now represent the fraction of the job each person completes in a single day.
A: If $A$ and $B$ are meant to represent the time required by each painter to finish the house solo, then I agree with your first equation.  Thinking of time in this way I also agree with @Brian that the second equation should be as he has stated it: how much less time the first painter requires relative to the second painter.  Equivalently, $B=A+5$.  
For me, a convenient way to think of the problem is in terms of rates which leads to a different meaning of $A$ and $B$. Let $T$ be the amount of time it takes for the first painter to complete one house.  Then define the rate of work for the first painter to be
$$ A = \frac{1 \ house}{T \ days} ,$$
the rate of work for the second painter to be
$$ B = \frac{1 \ house}{(T+5) \ days},$$
and the combined rate of work of both painters to be 
$$ A + B = \frac{1 \ house}{6 \ days}.$$
Then by substitution in the third equation we have that
$$\frac{1}{T} + \frac{1}{(T+5)} = \frac{1}{6}.$$
Solve for feasible values of $T$ to get the time required for the first painter to do the job solo.  Add $5$ to get the time for the second painter. 
