Question

The length of a rectangle is 4 feet longer than twice the width of the same rectangle. If the perimeter of the rectangle is 68 feet, what are the dimensions (the length and width) of this rectangle? Draw a picture to help, and use algebra to solve.

Answer 1

@snowflake @darkknight @az @carmelle @ihelpuuhelpme

Answer 2

So.



Width = W

Length = 2W + 4

Perimeter = 68

Now you plug the values you have within the formula: 2L + 2w = perimeter becomes
2(2w+4) + 2w = 68

Use distribution: 4w + 8 + 2w = 68

combine like terms and you get: 6W + 8 = 68

subtract 8 from both sides to get 6w alone and you get: 6w = 60

Divide by 6 to get the w alone, don't forget what you do to one side you have to do to the other so: 6w/6 = w and 60 /6 = 10
and we now know that W = 10

Now you can check your work by plugging 10 into the equation anywhere there is a w.

So for length 2w + 4 become 2(10) + 4 = 20 + 4
L = 24
and W = 10
so 2L + 2W = 68 becomes
2(24) +2(10) = 68

Does that make sense? xd Correct me if I'm wrong.

My source: https://www.algebra.com/algebra/homework/Rectangles/Rectangles.faq.question.341706.html

Answer 3

omg so smort that im confuzzled :0

Answer 4

Shall i "dum it down" xd

Answer 5

i understand it tyy

Answer 6

wait so whats the l and w?

Answer 7

Length = 24

Width = 10

Equation: 2(24) +2(10) = 68

Answer 8

ohhh ok THANK YOUU WAAAAAAHHH

Answer 9

MAY GOD BLESS YOU 😭😭😭😭🙏

Answer 10

may God bless you the most <3

Answer 11

👍🥺🥺🤧🤧