the length of a rectangle is 7 yd more than twice the width x. the area is 697yd^2

Question

the length of a rectangle is 7 yd more than twice the width x. the area is 697yd^2

 

write and equation

Answer 1

Let w = the width

Let 2w + 7 = the length

A = lw

w(2w + 7) = 697

2w² + 7w - 697 = 0

(2w + 41)(w - 17) = 0

2w + 41 = 0           and   w - 17 = 0

        2w = -41        and         w = 17 yards

          w = -41/2, not a logical answer

length 2w + 7 = 2(17) + 7

length = 41 yards

 

Answer 2

Hi Chris, let x represent the width. Let y represent the length, which is two times the width plus seven.

 

The area of a rectangle is: A = l * w.

 

Substituting:

 

697 = (2x + 7) * x

 

697 = 2x² + 7x

 

2x² + 7x - 697 = 0

 

Now use the quadratic formula to solve for x.

 

Remember, only positive answers are valid.

 

Questions.