document.write( "Question 120731: The length of a rectangle is 2in more than twice its width. If the perimeter of the rectangle is 34in, find the dimensions of the rectangle. \n" ); document.write( "
Algebra.Com's Answer #88511 by rapaljer(4671)\"\" \"About 
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Let x = width
\n" ); document.write( "2x+2= length\r
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\n" ); document.write( "\n" ); document.write( "2W + 2L = P
\n" ); document.write( "2(x)+2(2x+2)=34
\n" ); document.write( "2x+4x+4=34
\n" ); document.write( "6x+4=34
\n" ); document.write( "6x=30
\n" ); document.write( "x=5 in Width
\n" ); document.write( "2x+2=2(5)+2=12 in Length\r
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\n" ); document.write( "\n" ); document.write( "Check:
\n" ); document.write( "2*5+2*12
\n" ); document.write( "=10+24
\n" ); document.write( "=34 in\r
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\n" ); document.write( "\n" ); document.write( "For additional explanations on Word Problems, especially coin problems, please see my website by clicking on my tutor name \"rapaljer\" anywhere in algebra.com. Click on the first link, which is \"Basic and Intermediate Algebra\". Look for Basic Algebra, Chapter 1, Section 1.09 and 1.10. For solutions in \"LIVING COLOR\", look for my \"MATH IN LIVING COLOR\" page, and click on Basic Algebra, Chapter 1, Sections 1.10. \r
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\n" ); document.write( "\n" ); document.write( "R^2\r
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