Paul got a box of w erasers and x crayons for his birthday. At the end of the school year he only had y crayons left while not having lost a single eraser. How many more crayons than erasers did he have left?
Let's say that Paul had c crayons and e erasers at the beginning of the school year. Then:

c + e = w + x

c - y = e

y = c - e

Now we can solve for c and e:

c = (w + x + y)/2

e = (w + x - y)/2

Plugging in y = c - e yields:

c = (w + x + c - e)/2

2c = w + x + c

c = (w + x)/2

e = w/2 - x/2 + y/2

Now we can solve for y:

y = c - e

y = (w + x)/2 - w/2 + x/2 - y/2

y = 0

Therefore, y = 0, c = (w + x)/2, and e = (w + x)/2. So Paul had the same number of crayons and erasers left at the end of the school year.