document.write( "Question 45168: The lenght of a rectangle is 2 in. more than twice its width.if the perimeter of the rectangle is 34 in. find the dimensions of the rectangle. \n" ); document.write( "
Algebra.Com's Answer #29969 by aaaaaaaa(138)\"\" \"About 
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Some variables I'll use:\r
\n" ); document.write( "\n" ); document.write( "l = length
\n" ); document.write( "w = width
\n" ); document.write( "p = perimeter\r
\n" ); document.write( "\n" ); document.write( "First, we have that
\n" ); document.write( "\"l+=+2w+%2B+2\"\r
\n" ); document.write( "\n" ); document.write( "And the formula for a perimeter calculation, in general, is l+l+w+w, or:
\n" ); document.write( "\"p+=+2l+%2B+2w\"\r
\n" ); document.write( "\n" ); document.write( "Substituting 34 for p:
\n" ); document.write( "\"34+=+2l+%2B+2w\"\r
\n" ); document.write( "\n" ); document.write( "Substituting 2w + 2 for l:
\n" ); document.write( "\"34+=+2%282w+%2B+2%29+%2B+2w\"\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "And now we solve for the width:
\n" ); document.write( "\"34+=+4w+%2B+4+%2B+2w\"
\n" ); document.write( "\"34+=+6w+%2B+4\"
\n" ); document.write( "\"30+=+6w\"
\n" ); document.write( "\"5+=+w\"\r
\n" ); document.write( "\n" ); document.write( "If l is 2w+2, substituting w for 5 gives us:\r
\n" ); document.write( "\n" ); document.write( "\"l+=+2w+%2B+2+=+2%2A5+%2B+2+=+10+%2B+2+=+12\"\r
\n" ); document.write( "\n" ); document.write( "Therefore, the length is 12 inches and the width, 5 inches.
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