document.write( "Question 249008: I need help solving please:\r
\n" ); document.write( "\n" ); document.write( "When the length of a square is increased by 6 and the width is decreased by 4, the area remains unchanged. Find the dimensions of the square.\r
\n" ); document.write( "\n" ); document.write( "Thank you!!
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Algebra.Com's Answer #181404 by Earlsdon(6294)\"\" \"About 
You can put this solution on YOUR website!
Let the length of the side of the original square be x.
\n" ); document.write( "The area of the original square is:
\n" ); document.write( "\"A+=+x%5E2\"
\n" ); document.write( "The area of the new rectangle is:
\n" ); document.write( "\"A+=+%28x%2B6%29%2A%28x-4%29\"
\n" ); document.write( "\"A+=+x%5E2%2B2x-24\" But this is equal to the area of the original square, so...
\n" ); document.write( "\"x%5E2+=+x%5E2%2B2x-24\" Subtract \"x%5E2\" from both sides.
\n" ); document.write( "\"0+=+2x-24\" Add 24 to both sides.
\n" ); document.write( "\"2x+=+24\" Divide both sides by 2.
\n" ); document.write( "\"x+=+12\"
\n" ); document.write( "The dimensions of the square are: 12 X 12
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