SOLUTION: The length of a rectangle is 3 cm is less than twice its width. A second rectangle is such that each dimension is the reciprocal of the corresponding dimension of the first rectang

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Question 137206This question is from textbook algebra: structure and method: book1
: The length of a rectangle is 3 cm is less than twice its width. A second rectangle is such that each dimension is the reciprocal of the corresponding dimension of the first rectangle. The perimeter of the of the second rectangle is 1/5 the perimeter of the first. Find the perimeter of the first rectangle. This question is from textbook algebra: structure and method: book1

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Let x = width of the 1st rectangle
:
I says,"The length of a rectangle is 3 cm less than twice its width." therefore:
L = (2x-3)
:
Then it says, "A second rectangle is such that each dimension is the reciprocal of the corresponding dimension of the first rectangle." Therefore 2nd rectangle:
1%2F%28%282x-3%29%29 by 1%2Fx
:
The perimeter of the of the second rectangle is 1/5 the perimeter of the first. Find the perimeter of the first rectangle.
:
old rectangle perimeter = 5 times new rectangle perimeter
2(2x-3) + 2x = 5(2(1%2F%28%282x-3%29%29%29) + 2(1%2Fx))
:
4x - 6 + 2x = 5(2%2F%28%282x-3%29%29%29) + 2%2Fx))
:
6x - 6 = 10%2F%28%282x-3%29%29%29 + 10%2Fx
Put the two fractions over a single common denominator
6x - 6 = %2810x+%2B+10%282x-3%29%29%2Fx%282x-3%29%29
:
6x - 6 = %2810x+%2B+20x+-+30%29%2F%282x%5E2-3x%29
:
6x - 6 = %2830x+-+30%29%2F%282x%5E2-3x%29
Multiply both sides by the denominator:
(2x^2 - 3x)(6x - 6) = 30x - 30
Foil the left side
12x^3 - 18x^2 - 12x^2 + 18x = 30x - 30
:
Combine like terms on the left
12x^3 - 30x^2 + 18x - 30x + 30 = 0
:
12x^3 - 30x^2 - 12x + 30 = 0
:
Factor this equation by grouping
(12x^3 - 12x) - (30x^2 - 30) = 0
:
12x(x^2 - 1) - 30(x^2 - 1) = 0
:
Factor out (x^2 - 1) and you have:
(x^2 - 1)(12x - 30) = 0
Three solutions:
x^2 = 1
x = +/-Sqrt(1)
x = +1
x = -1
and
12x = +30
x = 30%2F12
x = 2.5
:
We find that neither + or -1 will work, therefore
:
x = +2.5 is our solution
:
Check solution by find the perimeters
L = 2(2.5) - 3 = 2, W = 2.5
2(2) + 2(2.5) = 5(2%2F2 + 2%2F2.5)
go to decimals
4 + 5 = 5(1 + .8)
9 = 5(1.8); confirms this solution