Question 137206
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)
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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/((2x-3))}}} by {{{1/x}}}
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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/((2x-3)))}}}) + 2({{{1/x}}}))
:
4x - 6 + 2x = 5({{{2/((2x-3)))}}}) + {{{2/x}}}))
:
6x - 6 = {{{10/((2x-3)))}}} + {{{10/x}}}
Put the two fractions over a single common denominator
6x - 6 = {{{(10x + 10(2x-3))/x(2x-3))}}}
:
6x - 6 = {{{(10x + 20x - 30)/(2x^2-3x)}}}
:
6x - 6 = {{{(30x - 30)/(2x^2-3x)}}}
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
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Combine like terms on the left
12x^3 - 30x^2 + 18x - 30x + 30 = 0
:
12x^3 - 30x^2 - 12x + 30 = 0
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Factor this equation by grouping
(12x^3 - 12x) - (30x^2 - 30) = 0
:
12x(x^2 - 1) - 30(x^2 - 1) = 0
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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 = {{{30/12}}}
x = 2.5
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We find that neither + or -1 will work, therefore
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x = +2.5 is our solution 
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Check solution by find the perimeters
L = 2(2.5) - 3 = 2, W = 2.5
2(2) + 2(2.5) = 5({{{2/2}}} + {{{2/2.5}}})
go to decimals
4 + 5 = 5(1 + .8)
9 = 5(1.8); confirms this solution