confusing algebra word problem?

Question

I'm stuck on this one last problem: A rectangle is 4 times as long as it is wide. A second rectangle is 5 centimeters longer and 2 centimeters wider than the first. The area of the second rectangle is 270 square centimeters greater than the first. What are the dimensions of the original rectangle? I'm supposed to solve and write down how I'm CHECKING it using algebra and the 5 step method. Ignore the 5 step method if you don't know it, I guess. =/

Answer 1

First rectangle:
w = width
4w = length

Second rectangle:
w+2 = width
4w + 5 = length

area of first plus 270 = area of second rectangle
w(4w) + 270 = (w+2)(4w + 5)
4w^2 + 270 = 4w^2 + 13w + 10
270 - 10 = 13 w
260 = 13 w
20 = w

First rectangle is 20 cm wide and 80 cm long.

check
20*80=1600 square cm = first rectangle's area
1600+270=1870 square cm= second rectangle's area
22*85= 1870 square cm

Answer 2

I don't do homework, but I think you're making it harder than it is. The first rectangle has length L and wiidth W with L=4W. The second has length L+5 and width W+2 or 1/4L+2, and (W+2) x (1/4L + 2) - 270 = W x L. You should be able to take it from there.

Answer 3

Area Rectangle One:
(4w) X (w) = 4w^2

Area Rectangle Two:
Area two = (w+2) X (4w+5) = 4w^2 + 5w + 8 W +10
4w^2 + 5w + 8 W +10= 4w^2 +13w +10

Area two = Area one + 270
4w^2 +13w +10 = 4w^2 + 270
-4w^2.....................-4w^2
...........13w + 10 = 270
..................-10......-10
....................13w=260
...................... w = 20

Orig Rect is 20 by 80

Answer 4

Did the teacher or book give you any examples as guides for solving them? That is a good way to develop a feel for working them. Maybe give some specific ones which are causing you trouble?