Set up an algebraic equation and to solve to find the dimensions of the fence

Question

A person wishes to build a rectangular fence that will enclose an area of 317 square feet. The width of the fence is to be 3 feet less than twice it's length. Set up and solve an algebraic equation to find the dimensions of the fence to the nearest tenth of a foot

Answer 1

Hi, Brittany!

 

I'm going to set this up as a system of equations, because that's what makes the most sense to me. 

 

We know that the area of a rectangle is given by the formula: area = length * width, and they've told us that the area is 317 ft², so let's let L be length, W be width, and write our first equation:

Equation 1: 317 ft² = L*W

 

Next, based on the phase "the width of the fence is to be 3 feet less than twice it's length," we can set up a second equation:

Equation 2: W = 2*L - 3 ft

 

Now we have a system of equations that we can solve by substitution:

1: 317 ft² = L*W

2: W = 2*L - 3 ft

 

317 ft² = L*(2*L - 3 ft) (plugging eq 2 into eq 1)

317 ft² = 2L² - 3L ft

2L² - 3L ft - 317 ft² = 0

L = 13.362 ft (from the quadratic equation)

 

317 ft² = 13.362 ft * W (plugging back into eq 1)

W = 23.724 ft

 

Remembering to account for proper significant figures, our solution is:

Length = 13.4 ft

Width = 23.7 ft

 

Hope that helps!