SOLUTION: I'm totally lost. My problem: A rectangle is 4 times as long as it is wide. A second rectangle is 5cm longer and 2cm wider than the first. The area of the second rectangle is 530 s
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-> SOLUTION: I'm totally lost. My problem: A rectangle is 4 times as long as it is wide. A second rectangle is 5cm longer and 2cm wider than the first. The area of the second rectangle is 530 s
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Question 417356: I'm totally lost. My problem: A rectangle is 4 times as long as it is wide. A second rectangle is 5cm longer and 2cm 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? P.S. I really stink at these kind of problems.Any help would be really appreciated. Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! I rectangle
width = x
length = 4x
..
II rectangle
width = x+2
length = 4x+5
...
Area II = Area I +530
(x+2)(4x+5)= 4x^2+530
4x^2+5x+8x+10=4x^2+530
13x+10=530
13x=520
/13
x=40 cm width of I
length of I = 160 cm
