previous question. the distance around the room is 64 yds, it is 8 longer than it is wide. what is the L&W?

Question

please explain in 3rd grade terms.

Answer 1

P = 2L + 2B
64 = 2(B + 8) + 2B
64 = 4B + 16
4B = 48
B = 12 yds
L = 20 yds

Answer 2

<>While this is slightly above a 3rd grade problem (and como has got the math about as simple as it gets), here goes:
Distance around the room P (perimeter) = 64 yds
The perimeter is made up of 2 L (lengths) and 2 W (widths)
So P = 2L + 2W
The length is 8 yds more than the width
So L = W + 8
Substitute and you get 64 = 2 (W+8) + 2W (your P (which is 64) is made of 2 times the width + 8 plus 2 more widths)
Multiply your brackets:
64 = 2 (W+8) + 2W
64 = 2W + 16 +2W and add your like terms
64 = 4W +16 subtract 16 from both sides to reduce
48 = 4W divide both sides by 4 to reduce
12 = W or W = 12
Your Width is 12 yds.
Now, you know that the Length is 6 yds more than the Width, so
L = W +8 or 12 +8 = 20
Your Length is 20 yds

Answer 3

Your first respondent gave you the answer, but not how to do it.

first, the formula

We assume the room is rectangular, right?


2L + 2W =perimeter.
Let w = width
let (w+8) = length
then the length = w + 8, because you said it was 8 feet longer than the width. So, pretend you are walking around the perimeter

w+w+w+8+w+8 = 64, or
2(w) +2(w+8) = 64
Solve for w (the width) then solve for length which is w+8.

Always in a story problem read what the question is that they want FIRST (the last line) before you even read the first line.
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