MATH SOLVE

3 months ago

Q:
# 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

Accepted Solution

A:

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: Β [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.