SOLUTION: The width of a rectangle is 4 centimeters. If the length of the rectangle were doubled, its perimeter would increase by 12 centimeters. What is the area of the original rectangle?

Question 847415: The width of a rectangle is 4 centimeters. If the length of the rectangle were doubled, its perimeter would increase by 12 centimeters. What is the area of the original rectangle?
I try:
P = 2(w+l)
P+12 = 2(4+2l)
P+12 = 8+4l

Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!
w for width, L for Length.
w=4 and p for perimeter, .
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If 2L for length, then perimeter is p+12.
.

Use the given value of w and simplify both equations.
Original: 2*4+2L=p
8+2L=p
2L-p=-8
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Doubled Length: 8+4L=p+12
4L-p+8=12
4L-p=4
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The two equations are a system of equations in the variables, L and p.
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2L-p=-8
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4L-p=4
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Suggestion is, use elimination method. Subtract first equation from the second equation.
4L-p-(2L-p)=4-(-8)
2L=12
L=6-------PART OF ANSWER.
Use either equation to find p.
-p=-2L-8
p=2L+8
p=2*6+8
p=20---------(Nice to know but not really PART OF ANSWER).

All the question really asks is, what is the AREA of the original rectangle.
Area of that will be
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