Answer:
The dimensions of the yard are W=20ft and L=40ft.
Step-by-step explanation:
Let be:
W: width of the yard.
L:length.
Now, we can write the equation of that relates length and width:
[tex]L=5W-60[/tex] (Equation #1)
The area of the yard can be expressed as (using equation #1 into #2):
[tex]Area=W*L=W*(5W-60)[/tex] (Equation #2)
Since the Area of the yard is [tex]800 ft^2[/tex], then equation #2 turns into:
[tex]800=W*(5W-60)[/tex]
Now, we rearrange this equation:
[tex]800=W*(5W-60)//800=5W^2-60W//5W^2-60W-800=0[/tex]
We can divide the equation by 5 :
[tex]W^2-12W-160=0[/tex]
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 [tex](-20)*8=160[/tex] and [tex]-20+8=-12[/tex]. The equation factorised looks like this:
[tex](W-20)(W+8)=0[/tex]
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:
[tex]5*20-60=40[/tex]
Therefore, the dimensions of the yard are W=20ft and L=40ft.
