Question

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?

Answer 1

The rectangle is 8x12".

Explanation:

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.