Question

Kathy is fencing in her yard that has an area of 800 square feet. Her yard is in the shape of a rectangle where the length is 60 less than 5 times the width. Find the length and width

Answer 1

The dimensions of the yard are W=20ft and L=40ft.

Explanation:

Let be:

W: width of the yard.

L:length.

Now, we can write the equation of that relates length and width:

`L=5W-60` (Equation #1)

The area of the yard can be expressed as (using equation #1 into #2):

`Area=W*L=W*(5W-60)` (Equation #2)

Since the Area of the yard is
`800 ft^2`, then equation #2 turns into:

`800=W*(5W-60)`

Now, we rearrange this equation:

`800=W*(5W-60)//800=5W^2-60W//5W^2-60W-800=0`

We can divide the equation by 5 :

`W^2-12W-160=0`

We need to find the solution for this quadratic. Let's find the factors of 160 that multiplied yields -160 and added yields -12. Let's choose -20 and 8, since
`(-20)*8=160` and
`-20+8=-12`. The equation factorised looks like this:

`(W-20)(W+8)=0`

Therefore the possible solutions are W=20 and W=-8. We discard W=-8 since width must be a positive number. To find the length, we substitute the value of W in equation #1:

`5*20-60=40`

Therefore, the dimensions of the yard are W=20ft and L=40ft.