SOLUTION: 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 areaof the second rectangle is 270 sq. centimet

Algebra ->  Inverses -> SOLUTION: 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 areaof the second rectangle is 270 sq. centimet      Log On


   

Question 6104: 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 areaof the second rectangle is 270 sq. centimeters greater than the first. What are the dimentions of the original rectangle?
Answer by Abbey(339) About Me  (Show Source):
You can put this solution on YOUR website!
let x = width of rectangle 1
let 4*x=length (it is 4 times as long as the width) of rectangle 1
area=length*width
area of rectangle 1 = x%2A4x=4x%5E2
let x+2 = width of rectangle 2
let 4x+5 = length of rectangle 2
area=length*width
area of rectangle 2=(4x+5)(x+2)=4x%5E2%2B13x%2B10
area of rec.2 is 270 cm greater than rec.1
4x%5E2%2B13x%2B10=270%2B4x%5E2
subtract 4x%5E2 from both sides:
13x%2B10=270
subtract 10 from both sides:
13x=260
divide both sides by 13:
x=20
now plug this back into your statements:
let x = width of rectangle 1 = 20
let 4*x=length = 80
let x+2 = width of rectangle 2 = 22
let 4x+5 = length of rectangle 2 = 85
to check, get the area of both and find the difference
area 1 = 20*80=1600
area 2 = 22*85= 1870
1870-1600=270, so the answer is correct