SOLUTION: There is a rectangle whose perimeter is 18 centimeters. If its length is decreased by 5 centimeters and it's width is increased by 12 centimeters, it's area is doubled. Find its le

Question 102689: There is a rectangle whose perimeter is 18 centimeters. If its length is decreased by 5 centimeters and it's width is increased by 12 centimeters, it's area is doubled. Find its length and width.
This Question is so hard to understand! Please help!
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Let's take it step-by-step
:
"There is a rectangle whose perimeter is 18 centimeters."
Let the length and width be x & y
Perimeter:
2x + 2y = 18
Simplify, divide equation by 2
x + y = 9
y = (9-x); we substitute (9-x) for y
Find the area:
x(9-x) = -x^2 + 9x
:
"If its length is decreased by 5 centimeters and it's width is increased by 12 centimeters, it's area is doubled. Find its length and width."
:
New length: (x-5)
:
New width:
(9-x) + 12
(-x + 21)
:
New area (double the old area): 2(-x^2 + 9x) = -2x^2 + 18x
:
New width * new length = twice the area
(-x + 21)*(x - 5) = -2x^2 + 18x
FOIL the left side
-x^2 + 5x + 21x -105 = -2x^2 + 18
-x^2 + 26x - 105 = -2x^2 + 18x
:
Combine to form a quadratic equation:
-x^2 + 2x^2 + 26x - 18x - 105 = 0
x^2 + 8x - 105 = 0
Factor this to
(x + 15)*(x - 7) = 0
x = +7 is the original length (only the positive solution is used)
:
Remember y = 9 - x:
y = 9 - 7
y = 2 is the original width
Original area = 7 * 2 = 14 sq/cm
:
New length and width:
L = 7 - 5 = 2 cm
W = 21 - 7 = 14
2 * 14 = 28 sq/cm, double the original
:
Could you follow this OK?

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