Can someone help me with some Algebra

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 530 square centimeters greater than the first. What are the dimensions of the original rectangle?

original rectangle: x cm wide by 4x cm long...

area of 1st = 4x^2 cm^2

second rectangle:

length = (4x + 5) cm
width = (x + 2) cm

area = (4x + 5)(x + 2) = 4x^2 + 13x + 10 cm^2

this is 530 cm^2 greater than 4x^2

so,

4x^2 + 13x + 10 = 4x^2 + 530

solve for x

13x = 520
x = 40

the first rectangle is:
40 cm wide
160 cm long
area = 6400 cm^2

the second rectangle is:

165 cm long
42 cm wide
area = 6930 cm^2

practice...read your textbook.

The first rectangle is x W and 4x L. A1 = x * 4x = 4x^2
The second rectangle is (x + 2) W and (4x + 5) L. A2 = (x + 2)(4x + 5) = 4x^2 + 13x + 10

4x^2 + 13x + 10 = 4x^2 + 530
13x + 10 = 530
13x = 520
x = 40

So the original rectangle is 40 * 160, and it's area is 6400

The second rectangle is 42 * 165, and it's area is 6930, which is 530 greater than the first one.

suppose the original width is x and length 4x

for the 1st rect, area = x * 4x = 4x^2
for the 2nd rect, area = (4x + 5)(x + 2)
now,
(4x + 5)(x + 2) = 530 + 4x^2
=> 4x^2 + 13x + 10 = 530 + 4x^2
=> 13x = 520
=> x = 40

hence the dimensions of the original rect are 40 and 160