The length of a rectangle exceeds its width by 4 inches. How do you find the dimensions of the rectangle if its area is 96 square inches?

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Answer 1 · 2017-09-10T19:49:55

The rectangle is 8x12".

To solve this, we will use a system of equations.

First, we define the dimensions of the rectangle.

Letting the width be #x# and the length be #y#, we can say the following, since the width is four less than the length:

#y=x+4#

Next, we will define the rectangle's area. The area of a rectangle is equal to the product of its side lengths, so we can say:

#x*y=96#

Our next step will be to substitute. Using the first equation, we know that #y=x+4#, so we will replace #y# with #x+4# in the second equation.

#xcolor(red)((x+4))=96#

To solve, we will distribute the #x#.

#x^2+4x=96#

Next, we will subtract #96#.

#x^2+4x-96=0#

Now, we factor.

#(x+12)(x-8)=0#

Letting both factors equal zero, we get:

#x=-12# and #x=8#

Since length cannot be negative, we know the width must be #8# inches.

Our last step is to check the outputted #y# value in the first equation when #x# equals #8#.

#y=color(red)8+4#
#y=12#

Therefore, the length of the rectangle is #8# inches and the width is #12# inches.