time and work problem explanation I'm trying to understand the following question: 
If A and B work together they accomplish a piece of work in 10 days. If A would work at half of its capacity and B would work 5 times its capacity they would finish the work in 5 days. How many days will take for B to finish the work alone? 
My first try was to create the following system:
$$\left\{a+b=10\atop\frac12a+5b=5\right.$$
But apparently this leads to an incorrect result. I've seen that the solution actually starts with the following equation:
$$a+b=\frac1{10}$$.
Can you, please, explain why this equation is used and which are the 2 equations which I should build in order to correctly solve the problem? 
Thanks!
 A: If $A$ finishes the work in $a$ days then in one day makes $1/a$ of his work. So, if $A$ works at half of his capacity he will make $1/2a$ of his work in one day.
Now, assume $B$ takes $b$ days to finish the work. So, in a day he makes $1/b.$ If he works at $5$ times his capacity he will make $5/b$ of the work in one day.
If they work together at normal pace they will make $$\frac{1}{a}+\frac{1}{b}=\frac{1}{10}$$ of the work in one day, because it takes them $10$ days to finish.
If $A$ works at half of its capacity and $B$ at five times its capacity then
$$\frac{1}{2a}+\frac{5}{b}=\frac{1}{5},$$ since they finish in $5$ days. So, you have to solve the system
$$\left\{\begin{align}\frac{1}{a}+\frac{1}{b}&=\frac{1}{10} \\ \frac{1}{2a}+\frac{5}{b}&=\frac{1}{5}\end{align}\right.$$ Note that $a$ and $b$ are the number of days they need to finish work. If you call $x=1/a,y=1/b$ (the fraction of work they make in one day) you can write the system as 
$$\left\{\begin{align}x+y&=\frac{1}{10} \\ \frac{x}{2}+5y&=\frac{1}{5}\end{align}\right.$$
