```Question 77419
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:
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{{{A = L * W}}}
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The problem tells you that the area is {{{3x^2 + 5x}}} so you can put this into the equation
in place of A to get:
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{{{3x^2 + 5x = L * 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^2 + 5x = x * 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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{{{(3x^2 + 5x)/x = (x*W)/x}}}
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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^2}}} you can do so by recognizing
that {{{x^2 = x*x}}}.  So you are really dividing {{{3*x*x/x}}} 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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{{{(3*x*x + 5*x)/x = (x*W)/x}}}
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and canceling like x terms in the numerators and denominator results in:
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{{{(3*x*cross(x) + 5*cross(x))/cross(x) = (cross(x)*W)/cross(x)}}}
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