Squares change in area part2

[Dillinger4]

The length of a rectangle is 8 less than 5 time it's width. The width is increased by three while the length is unchanged. Using a as the variable, express the change in area in terms of a

Im not quite sure what is being asked. Are we simply supposed to calculate the area? If that is the case i end up getting 2(5w - 8) + 2(w + 3) my books says 15a - 24

[twanvl]

You must calculate the difference in area between the 'before' and the 'after' situations.

Before the change:
l = 5w - 8
A = w*l

After:
l' = l+3
w' = w
A' = w'*l'

The exercise asks for A'-A.

Using a as the variable, express the change in area in terms of a

This part doesn't make sense, there is no a mentioned anywhere. Unless your book uses a as a standard notation for length, width or area.

[Dillinger4]

Yeah i mean the wording is crap. No wonder people get confused when taking math tests.

The length of a rectangle is 8 less than 5 time it's width. a = l * w
(5w-8) * w

The width is increased by three while the length is unchanged.
(5w - 8) * (w + 3)

So we should end up with. (5w-8)(w+3) - (5w-8)w
6w + 15 w -8w -24 -(6w - 8w)
6w + 15w -8w -24 -6w + 8w
15w - 24

I think that's it. Thanks for the help.