document.write( "Question 403285: The perimeter of a rectangle is 76m. If the width were doubled and the length were increased by 5 m, the perimeter would be 120m. What are the length and width of the original rectangle?
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Algebra.Com's Answer #285226 by Tatiana_Stebko(1539) $\"\"$ $\"About$

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If L is the length, W is the width
\n" ); document.write( "Perimeter of the rectangle is $\"P=2%28L%2BW%29\"$ , so $\"2%28L%2BW%29=76\"$ .
\n" ); document.write( "If the width be double (2W) and the length ne increase by 5 m (L+5)? the perimeter will be $\"2%282W%2B%28L%2B5%29%29=120\"$
\n" ); document.write( "Equation system
\n" ); document.write( " $\"2%28L%2BW%29=76\"$
\n" ); document.write( " $\"2%282W%2B%28L%2B5%29%29=120\"$
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\n" ); document.write( " $\"L%2BW=38\"$
\n" ); document.write( " $\"L%2B2W%2B5=60\"$
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\n" ); document.write( " $\"L%2BW=38\"$
\n" ); document.write( " $\"L%2B2W=55\"$ From the second equation we subtract the first equation
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\n" ); document.write( " $\"L%2BW=38\"$
\n" ); document.write( " $\"W=17\"$
\n" ); document.write( "If $\"W=17\"$ from the first equation will have $\"L=38-W=38-17=21\"$
\n" ); document.write( "So the length is 21m and the width is 17m \n" ); document.write( "

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