SOLUTION: I am having issues with this one problem in my home work. I don't know why I cant get it...We have to set it up in the 5 step and I guess I am just confused. ?. The perimeter o

Algebra ->  Customizable Word Problem Solvers  -> Geometry -> SOLUTION: I am having issues with this one problem in my home work. I don't know why I cant get it...We have to set it up in the 5 step and I guess I am just confused. ?. The perimeter o      Log On

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Question 570089: I am having issues with this one problem in my home work. I don't know why I cant get it...We have to set it up in the 5 step and I guess I am just confused.
?. The perimeter of a rectangle is 36m. The length is 2m more than three thimes the width. Find the length and with of the rectangle.
Ok, so here is what I have done.
Data: I drew a rectangle with the length labeled as 3w + 2, and the width is just labeled as w.
Varaibles: P=2(3w+2)x 2w
Plan: P=2Lx2W
Equation: 36=2(3w+2)x 2w
Ok...so here is where I get stuck.
36=2(3w+2) x 2w
36=6w+4 x 2w
so I think now I subtract 4 from each side, but then I am stuck. Am I even close?



Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
I think you have the right idea, let's do it this way
Write an equation for each statement:
:
"The perimeter of a rectangle is 36m."
2L + 2W = 36
:
"The length is 2m more than three times the width."
L = 3W + 2
:
Replace L in the 2st equation with (3W+2); resulting in
2(3W+2) + 2W = 36
6W + 4 + 2W = 36: this is essentially what you did, just combine like terms, find W
8W = 36 - 4
8W = 32
W = 32/8
W = 4 m is the width
the
L = 3(4) + 2
L = 14 m is the length
;
:
Check our solutions by finding the perimeter with these values
2(14) + 2(4) =
28 + 8 = 36