the area of a rectangle is 52 yd, and the length of the rectangle is 5yd less than twice the width. Find the dimensions of the rectangle

Question

I need to find the length and width. thank you so much 

Answer 1

Let A = the area = 52

Let W = the width

Let L = the length

 

The problem says the length is 5 less than twice the width, which means the following:

 

L = (W*2) - 5

 

We know that Area = Length * Width, so we have:

A = L * W

L * W = 52

 

We now have two equations and two variables, so we can solve. Substitute the value of L into the second equation:

 

(W*2 - 5)*W = 52

2W² - 5W - 52 = 0

 

We can factor this equation as follows:

 

(2W - 13)(W + 4) = 0

 

Negative lengths make no sense, so we can eliminate W+4.

 

Therefore we have 2W - 13 = 0

2W = 13

W = 6.5

 

We know that L = (W*2) - 5, so we can plug in W to solve for L:

 

L = (6.5*2) - 5

L = 13 - 5 = 8

 

Therefore the width is 6.5 and the length is 8.