document.write( "Question 149381: The length of a rectange is 6in more than 3 times its width. If the width were increased by 4 and the length decreased by 10, the area of the new rectangle would equal the area of the orignal rectangle. Find the length and width of the original rectangle. \n" ); document.write( "

Algebra.Com's Answer #109591 by jojo14344(1513) $\"\"$ $\"About$

You can put this solution on YOUR website!
Let $\"+A%5Bold%5D+=+old+Area+\"$ , with following conditions:
\n" ); document.write( " $\"L%5Blength%5D+=+3W%2B6+\"$
\n" ); document.write( " $\"+W%5Bwidth%5D+=+W+\"$ ,
\n" ); document.write( "So, $\"+A%5Bold%5D+=+L%2AW+\"$ ----> $\"A%5Bold%5D+=+%283W%2B6%29%28W%29\"$
\n" ); document.write( " $\"+A%5Bold%5D=+3W%5E2+%2B+6W+\"$
\n" ); document.write( ".
\n" ); document.write( "Let $\"+A%5Bnew%5D+=+new+Area+\"$ with following conditions:
\n" ); document.write( " $\"+L%5Blength%5D+=+3W+%2B+6+-+10+=+3W+-4+\"$
\n" ); document.write( " $\"+W%5Bwidth%5D+=+W%2B4+\"$
\n" ); document.write( "So, $\"+A%5Bnew%5D+=+L%2AW+\"$ ----> $\"A%5Bnew%5D+=+%283W-4%29%28W%2B4%29\"$
\n" ); document.write( " $\"+A%5Bnew%5D+=+3W%5E2+%2B+8W+-16+\"$
\n" ); document.write( ".
\n" ); document.write( "Equating the 2 Areas because of the conditions, becoming $\"A%5Bold%5D+=+A%5Bnew%5D\"$
\n" ); document.write( " $\"+cross%283W%5E2%29+%2B+6W+=+cross%283W%5E2%29+%2B+8W+-16+\"$ , rearranging thereafter:
\n" ); document.write( " $\"+8W+-+6W+=16+\"$ ----> $\"2W+=+16\"$
\n" ); document.write( " $\"+W+=+8+\"$
\n" ); document.write( "Going back to the old condition for $\"+L=+3W+%2B+6\"$ ---> $\"L=+%283%2A8%29%2B6\"$
\n" ); document.write( " $\"+L=+30\"$
\n" ); document.write( "In doubt? Go back $\"A%5Bold%5D+=+%28%283%2A8%29%2B6%29%288%29=+240\"$
\n" ); document.write( "Also, $\"A%5Bnew%5D+=+%28%283%2A8%29-4%29%288%2B4%29=240\"$
\n" ); document.write( " $\"A%5Bold%5D+=+A%5Bnew%5D\"$
\n" ); document.write( "Thank you,
\n" ); document.write( "Jojo \n" ); document.write( "

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