document.write( "Question 389447: a rectangle with a perimeter of 34cm has a length that is 3cm more than twice its width. What are the dimensions of the rectangle? Show a complete solution,including \"let\" and \"therefore\" statements. \n" ); document.write( "

Algebra.Com's Answer #275976 by gwendolyn(128) $\"\"$ $\"About$

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Let W be the width of the rectangle.
\n" ); document.write( "Let L be the length of the rectangle.\r
\n" ); document.write( "\n" ); document.write( "The length is 3 cm more than twice its width. Therefore:
\n" ); document.write( "L = 2W + 3\r
\n" ); document.write( "\n" ); document.write( "The perimeter of the rectangle is 34 cm. Therefore:
\n" ); document.write( "2W + 2L = 34\r
\n" ); document.write( "\n" ); document.write( "We can substitute the value of L from the first equation into the second:\r
\n" ); document.write( "\n" ); document.write( "2W + 2(2W + 3) = 34
\n" ); document.write( "Distributing the 2 on the right:
\n" ); document.write( "2W + 2*2W + 2*3 = 34
\n" ); document.write( "2W + 4W + 6 = 34
\n" ); document.write( "6W + 6 = 34
\n" ); document.write( "Subtracting 6 from both sides:
\n" ); document.write( "6W + 6 - 6 = 34 - 6
\n" ); document.write( "6W = 28
\n" ); document.write( "Dividing both sides by 6:
\n" ); document.write( " $\"6W%2F6+=+28%2F6\"$
\n" ); document.write( " $\"W+=+28%2F6\"$
\n" ); document.write( " $\"W+=+14%2F3\"$ or,
\n" ); document.write( "W = 4 and 2/3 \r
\n" ); document.write( "\n" ); document.write( "We can substitute this value of W back into the first equation
\n" ); document.write( "to find the value of L:\r
\n" ); document.write( "\n" ); document.write( "L = 2*W + 3
\n" ); document.write( " $\"L+=+2%2A%2814%2F3%29+%2B+3\"$
\n" ); document.write( " $\"L+=+28%2F3+%2B+3\"$ or,
\n" ); document.write( "L = 9 and 1/3 + 3 so,
\n" ); document.write( "L = 12 and 1/3\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

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