SOLUTION: Hello, I need help on my Algebra 2 homework. I tried different ways to reach an answer, but the answers I got doesn't make sense. The problem is: The width of a rectangle is 12 un

Question 9498: Hello,
I need help on my Algebra 2 homework. I tried different ways to reach an answer, but the answers I got doesn't make sense. The problem is: The width of a rectangle is 12 units less than its length. If you add 30 units to both the length and width, you double the perimeter. Find the length and width of the original rectangle. Please reply as soon as possible. I greatly appreciate your help and time.
Julia
Answer by rapaljer(4671) (Show Source): You can put this solution on YOUR website!
Let x = length of the rectangle
x-12 = width of the rectangle
2x + 2(x-12) = original perimeter
2x + 2x - 24 or 4x - 24 = original perimeter

New rectangle
x + 30 = new length
x - 12 + 30 = x+ 18 = new width
2(x+30) + 2 (x+ 18) = new perimeter
2x + 60 + 2x + 36 or 4x + 96 = new perimeter

THE EQUATION:
New perimter = 2* original perimeter
4x + 96 = 2( 4x - 24)
4x + 96 = 8x - 48
-4x = -144
x= 36 = original length
x - 12 = 36- 12 = 24 = original width

ANSWER: Origianl rectangle is 24 by 36

Check: New perimeter = twice original perimeter:
Perimeter = 2(36) + 2(24) = 72 + 48 = 120

New rectangle length = 36 + 30 = 66
New rectangle width = 24 + 30 = 54
Perimeter = 2(66) + 2(54) = 132 + 108 = 140

R^2 at SCC
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