SOLUTION: a photograph of a child has a width 2 inches longer than its length. the proud parents decide the photo is cute enough that they want a larger version. the picture is made bigger b
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Question 208045This question is from textbook algebra 1 an integrated approach
: a photograph of a child has a width 2 inches longer than its length. the proud parents decide the photo is cute enough that they want a larger version. the picture is made bigger by increasing both its length and its width by 2 inches, its area increases by 20 square inches. what are the dimensions of the original picture?use the 5 step method.
This question is from textbook algebra 1 an integrated approach
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
a photograph of a child has a width 2 inches longer than its length.
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we let L = length and W = width.
we get:
L = W + 2
that we hold for future reference.
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the proud parents decide the photo is cute enough that they want a larger version. the picture is made bigger by increasing both its length and its width by 2 inches, its area increases by 20 square inches.
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we get (L + 2) * (W + 2) = A + 20
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original Area is given by the equation:
L * W = A
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what are the dimensions of the original picture?
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we have 3 equations to work with:
equation 1: L = W + 2
equation 2: (L + 2) * (W + 2) = A + 20
equation 3: (L * W) = A
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we can use these relationships to reduce the number of unknowns down to 1.
that would be W.
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first we replace A with (L * W).
equation 2 becomes: (L + 2) * (W + 2) = (L * W) + 20
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next we replace L with W + 2.
equation 2 becomes: (W + 2 + 2) * (W + 2) = ( (W + 2) * W ) + 20
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we now solve for W.
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equation 2 start out as:
(W + 2 + 2) * (W + 2) = ( (W + 2) * W ) + 20
this becomes:
(W + 4) * (W + 2) = ( (W + 2) * W ) + 20
multiplying the factors out we get:
W^2 + 6W + 8 = W^2 + 2W + 20
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if we subtract W^2 + 2W from both sides of the equation, we get:
4W + 8 = 20
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if we subtract 8 = from both sides of the equation, we get:
4W = 12
divide both sides by 4 to get:
W = 3
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since L = W + 2, we get:
L = 5
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original dimensions of the photograph are
Length = 5 and width = 3
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new photo becomes length = 7 and width = 5
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area of the original photo is 5 * 3 = 15 square inches.
area of the larger photo is 7 * 5 = 35 square inches.
area has increased by 20 square inches.
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all requirements have been satisfied so the answer is:
length of the original photo is 5 inches.
width of the original photo is 3 inches.
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