SOLUTION: The perimeter of a cross section of a "two-by-four" piece of lumber is 10 1/2 in. The length is twice the width. Find the actual dimension of the cross section of a two-by-four. I

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: The perimeter of a cross section of a "two-by-four" piece of lumber is 10 1/2 in. The length is twice the width. Find the actual dimension of the cross section of a two-by-four. I       Log On


   



Question 160048This question is from textbook Elementary and Intermediate Algebra
: The perimeter of a cross section of a "two-by-four" piece of lumber is 10 1/2 in. The length is twice the width. Find the actual dimension of the cross section of a two-by-four. I have tried 2n+n=10 1/2 and that formula didn't work. I have tried 2(2n+n)=10 1/2 and thats wrong. Please help me? This question is from textbook Elementary and Intermediate Algebra

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
The perimeter of a cross section of a "two-by-four" piece of lumber is 10 1/2 in. The length is twice the width. Find the actual dimension of the cross section of a two-by-four.
:
You're right on the last equation 2(2n+n) = 10.5.
Let n = the width
"length is twice the width
2n = length
:
Perimeter is:
2L + 2W = 10.5
Replace L with 2n and W with n
2(2n) + 2(n) = 10.5
4n + 2n = 10.5
6n = 10.5
n = 10.5%2F6
n = 1.75" is the width, and 2(1.75) = 3.5" is the length
or
13%2F4 by 31%2F2 if you prefer