document.write( "Question 160048This question is from textbook Elementary and Intermediate Algebra
\n" ); document.write( ": 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? \n" ); document.write( "

Algebra.Com's Answer #118050 by ankor@dixie-net.com(22740) $\"\"$ $\"About$

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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.
\n" ); document.write( ":
\n" ); document.write( "You're right on the last equation 2(2n+n) = 10.5.
\n" ); document.write( "Let n = the width
\n" ); document.write( "\"length is twice the width
\n" ); document.write( "2n = length
\n" ); document.write( ":
\n" ); document.write( "Perimeter is:
\n" ); document.write( "2L + 2W = 10.5
\n" ); document.write( "Replace L with 2n and W with n
\n" ); document.write( "2(2n) + 2(n) = 10.5
\n" ); document.write( "4n + 2n = 10.5
\n" ); document.write( "6n = 10.5
\n" ); document.write( "n = $\"10.5%2F6\"$
\n" ); document.write( "n = 1.75\" is the width, and 2(1.75) = 3.5\" is the length
\n" ); document.write( "or
\n" ); document.write( "1 $\"3%2F4\"$ by 3 $\"1%2F2\"$ if you prefer \n" ); document.write( "

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