SOLUTION: A rectangle is 30 feet longer than it is wide. If its length were increased by 50 feet and its width were diminished by 8 feet, its area would be increased by 200 square feet. Find

Question 591120: A rectangle is 30 feet longer than it is wide. If its length were increased by 50 feet and its width were diminished by 8 feet, its area would be increased by 200 square feet. Find its dimensions. Also, is there a good way to learn to solve word problems? Having a hard time with them. Thank you
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
width =x
length = x+30
The area = x(x+30)
x^2+30x sq. feet
New dimensions
width = x-8
Length = x+30+50= x+80
New area = (x-8)(x+80)
x(x+80)-8(x+80)
x^2+80x-8x-640
New Area - old area = 200
x^2+72x-640-x^2-30x=200

42x=200+640
42x=840
/42
x= 20
The width is 20 feet
Length = x+30 = 20+30 = 50 feet

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