SOLUTION: Geometry. The area of a rectangle of length x is given by 3x^2 + 5x. Find the width of the rectangle.

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Question 77419: Geometry. The area of a rectangle of length x is given by 3x^2 + 5x. Find the width of the rectangle.
Answer by bucky(2189) About Me  (Show Source):
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The area of a rectangle of length x is given by 3x^2 + 5x. Find the width of the rectangle.
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The area of a rectangle (A) is given by the equation L * W where L represents the length and
W represents the width. In equation form this is:
.
A+=+L+%2A+W
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The problem tells you that the area is 3x%5E2+%2B+5x so you can put this into the equation
in place of A to get:
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3x%5E2+%2B+5x+=+L+%2A+W
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and the problem also tells you that the length of the rectangle is x. So you can put x
into the equation in place of L. The result is:
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3x%5E2+%2B+5x+=+x+%2A+W
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You can now solve for W by dividing both sides of the equation by the multiplier of W which
is x. Dividing both sides by x results in:
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%283x%5E2+%2B+5x%29%2Fx+=+%28x%2AW%29%2Fx
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On the left side, the x in the denominator must be divided into each of the two terms in
the numerator. When you divide the x into 3x%5E2 you can do so by recognizing
that x%5E2+=+x%2Ax. So you are really dividing 3%2Ax%2Ax%2Fx and the effect is that
the x in the denominator cancels with one of the x terms in the numerator. So the problem
can be written as:
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%283%2Ax%2Ax+%2B+5%2Ax%29%2Fx+=+%28x%2AW%29%2Fx
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and canceling like x terms in the numerators and denominator results in:
.

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and the answer becomes:
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3%2Ax+%2B+5+=+W
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and that's the answer ... W+=+3x+%2B+5
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Hope this helps you to understand the problem and the steps you can take to solve it.