document.write( "Question 753059: I've been stuck on this problem since Friday. Help please. \r
\n" ); document.write( "\n" ); document.write( "The length of a rectangle is 4 inches shorter than twice the width. If the perimeter of the rectangle is 34 inches, find the length and the width of the triangle.\r
\n" ); document.write( "\n" ); document.write( "The answer on the paper is L=10 W=7
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Algebra.Com's Answer #458223 by MathTherapy(10810)\"\" \"About 
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\n" ); document.write( "I've been stuck on this problem since Friday. Help please. \r
\n" ); document.write( "\n" ); document.write( "The length of a rectangle is 4 inches shorter than twice the width. If the perimeter of the rectangle is 34 inches, find the length and the width of the triangle.\r
\n" ); document.write( "\n" ); document.write( "The answer on the paper is L=10 W=7\r
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\n" ); document.write( "\n" ); document.write( "Let width be W
\n" ); document.write( "Then length = 2W - 4\r
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\n" ); document.write( "\n" ); document.write( "Since perimeter = 34, then 2L + 2W = 34 ----- 2(L + W) = 2(17) ----- L + W = 17\r
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\n" ); document.write( "\n" ); document.write( "Since L = 2W - 4, then L + W = 17 becomes: 2W - 4 + W = 17\r
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\n" ); document.write( "\n" ); document.write( "3W = 17 + 4\r
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\n" ); document.write( "\n" ); document.write( "3W = 21\r
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\n" ); document.write( "\n" ); document.write( "W, or width = \"21%2F3\", or \"highlight_green%287%29\" inches\r
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\n" ); document.write( "\n" ); document.write( "Substituting 7 for W, L or length = 2(7) - 4, or \"highlight_green%2810%29\" inches \r
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\n" ); document.write( "\n" ); document.write( "You can do the check!!\r
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\n" ); document.write( "\n" ); document.write( "Send comments and “thank-yous” to “D” at MathMadEzy@aol.com
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