document.write( "Question 1057349: In a rectangle, the length is 12cm longer than its width. If the perimeter of the rectangle 84cm, what are the dimensions?
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Algebra.Com's Answer #672404 by math_helper(2461)\"\" \"About 
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Let l = length
\n" ); document.write( "Let w = width\r
\n" ); document.write( "\n" ); document.write( "Need to find l and w. Two unknowns so look for two equations.\r
\n" ); document.write( "\n" ); document.write( "(1) l = w+12 (\"length is 12cm longer than its width\")
\n" ); document.write( "(2) 2l + 2w = 84 (\"perimeter is 84cm\")\r
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\n" ); document.write( "Eq (1) allows us to subs \"w+12\" for \"l\" in (2):\r
\n" ); document.write( "\n" ); document.write( " 2(w+12) + 2w = 84 (eq (2) w/substitution made)
\n" ); document.write( " 2w + 24 + 2w = 84 (multiplied 2 into parens)
\n" ); document.write( " 4w + 24 = 84 (combined like terms)
\n" ); document.write( " 4w = 60 (subtracted 24 from both sides)
\n" ); document.write( " w = 60/4 (divided both sides by 4)
\n" ); document.write( " w = 15cm\r
\n" ); document.write( "\n" ); document.write( "From (1), l = 15+12 = 27cm
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\n" ); document.write( "\n" ); document.write( "Ans: The rectangle has dimensions 27cm x 15cm
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\n" ); document.write( "\n" ); document.write( "Check:
\n" ); document.write( " We can plug into (2) to check:
\n" ); document.write( " 2(27) + 2(15) = 54 + 30 = 84 (ok)
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