SOLUTION: Hi!
I have a word problem for college algebra course and I'm not quite sure how to begin it. The problem says:
A rectangle is 30 feet longer than it is wide. If its length were
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I have a word problem for college algebra course and I'm not quite sure how to begin it. The problem says:
A rectangle is 30 feet longer than it is wide. If its length were
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Question 1127028: Hi!
I have a word problem for college algebra course and I'm not quite sure how to begin it. The problem says:
A rectangle is 30 feet longer than it is wide. If its length were increased by 50 ft and its width were diminished by 8 feet, its area would be increased by 200 square feet. Find its dimensions.
Thank you so much!
-Maddie
You can put this solution on YOUR website! width=x
length=x+30
x+80 (length + 50)
x-8 (width -8)
area is increased by 200 ft^2
initial area is x(x+30) or x^2+30x
final area is (x+80)(x-8)=x^2+72x-640
add 200 to initial area and have final area
so x^2+30x+200=x^2+72x-640
-42x=-840
x=20 feet width
x+30=50 feet length, area is 1000 ft^2
length is increased to 100
width is decreased to 12
area is 1200 ft^2,