SOLUTION: The length of a rectangle is 15cm longer than two times its width.the perimeter of the rectangle is 252m determine the length and the width of the rectangle using algebraic equatio

Question 920183: The length of a rectangle is 15cm longer than two times its width.the perimeter of the rectangle is 252m determine the length and the width of the rectangle using algebraic equations/expressions
I am a parent trying to help my son on his assignment thank you so much
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
The perimeter formula for a rectangle is

P = 2*(L+W)

where L is the length and W is the width.

Since "The length of a rectangle is 15cm longer than two times its width" and "the perimeter of the rectangle is 252m", we know

W = x
L = 2x+15
P = 252

where x is some unknown number for now. Plug those values into P = 2*(L+W) and solve for x.

P = 2*(L+W)

252 = 2*(2x+15+x)

252 = 2*(3x+15)

252 = 6x+30

252-30 = 6x+30-30

222 = 6x

6x = 222

x = 222/6

x = 37

Now that we know x = 37, we can use this to find L and W. Recall that above we made

W = x
L = 2x+15

Since W = x and x = 37, we know W = 37.

Use this x value to find L

L = 2x+15

L = 2(37)+15

L = 74+15

L = 89

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Final Answer:

So the length is 89 and the width is 37.

Let me know if you need more help or if you need me to explain a step in more detail.
Feel free to email me at jim_thompson5910@hotmail.com
or you can visit my website here: http://www.freewebs.com/jimthompson5910/home.html

Thanks,

Jim
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