SOLUTION: A rectangle is 4 times as long as it is wide. A second rectangle is 5 cm longer and 2 cm wider than the first. The area of the second rectangle is 270 square cm greater than the fi

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Question 91208: A rectangle is 4 times as long as it is wide. A second rectangle is 5 cm longer and 2 cm wider than the first. The area of the second rectangle is 270 square cm greater than the first. What are the dimensions of the original rectangle?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Translate each sentence into an equation/expression
:
A rectangle is 4 times as long as it is wide.
x = width of the 1st rectangle
then
4x = length of the 1st rectangle
:
:
A second rectangle is 5 cm longer and 2 cm wider than the first.
(4x + 5) = length of the 2nd rectangle
(x+2) = width of the 2nd rectangle
:
:
The area of the second rectangle is 270 square cm greater than the first.
:
Area of 2nd rectangle - area of 2st rectangle = 270
(4x+5)*(x+2) - x(4x) = 270
(4x^2 + 8x + 5x + 10) - 4x^2 = 270; FOILed (4x+5)(x+2)
4x^2 + 13x + 10 - 4x^2 = 270;
13x + 10 = 270; 4x^2's cancel each other
13x = 270 - 10
13x = 260
x = 260/13
x = 20
:
What are the dimensions of the original rectangle?
1st rectangle: 4*20 = 80: 80 by 20
2nd rectangle: 80+5; 20+2; 85 by 22
:
:
Check solution
85*22 = 1870
80*20 = 1600
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diffence 270
:
:
How about this? Did I make it understandable to you?