Question 95253: If the length of a rectangle is decreased by 2 inches and its width is increased by 2 inches, the area of the rectangle increases by 16 square inches. If the length is increased by 5 inches and the width is decreased by 3 inches, the area of the rectangle increases by 15 square inches. What is the area of the original rectangle?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! If the length of a rectangle is decreased by 2 inches and its width is increased by 2 inches, the area of the rectangle increases by 16 square inches. If the length is increased by 5 inches and the width is decreased by 3 inches, the area of the rectangle increases by 15 square inches. What is the area of the original rectangle?
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Let x = length, and y = width of the original rectangle
x*y = area of original rectangle
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1st rectangle equation from the statement:
"If the length of a rectangle is decreased by 2 inches and its width is increased by 2 inches, the area of the rectangle increases by 16 square inches.
(x-2)*(y+2) = xy + 16
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xy + 2x - 2y - 4 = xy + 16; FOILed (x-2)*(y+2)
xy - xy + 2x - 2y = 16 + 4; arrange the unknowns on the left, numbers on right
2x - 2y = 20
x - y = 10; simplified, divided by 2
x = (y + 10)
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1nd rectangle equation from the statement:
" If the length is increased by 5 inches and the width is decreased by 3 inches, the area of the rectangle increases by 15 square inches."
(x+5)*(y-3) = xy + 15
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xy - 3x + 5y - 15 = xy + 15
xy - xy - 3x + 5y = 15 + 15
-3x + 5y = 30
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Substitute (y+10) for x from the 1st equation:
-3(y+10) + 5y = 30
-3y - 30 + 5y = 30
-3y + 5y = 30 + 30
2y = 60
y = 30 inches, width of the original square
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Find x:
x = y + 10
x = 40 inches, length of the original square
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What is the area of the original rectangle?
40 * 30 = 1200 sq inches
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Check solutions using the changed rectangles
1st rectangle: 38 * 32 = 1216 sq inches
2nt rectangle: 45 * 27 = 1215 sq inches
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Did this make sense to you? Any questions?
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