document.write( "Question 54306: Here is another question I need assistance with. You have a rectangle that is 2 cm more than twice its width and the perimeter is 52 cm, what are the dimensions of this rectangle? \n" ); document.write( "
Algebra.Com's Answer #36560 by fanks(6)\"\" \"About 
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The rectangle has such edges: a, b, a, b where \"b+=+2a+%2B+2\"
\n" ); document.write( "So the edges go like this: a, 2a + 2, a, 2a + 2\r
\n" ); document.write( "\n" ); document.write( "Their sum is \"a+%2B+2a+%2B+2+%2B+a+%2B+2a+%2B+2+=+6a+%2B+4\" which is the same as the perimeter: \"6a+%2B+4+=+52\"\r
\n" ); document.write( "\n" ); document.write( "Thus \"6a+=+48\" and \"a+=+8\", and \"b+=+2a+%2B+2+=+18\"\r
\n" ); document.write( "\n" ); document.write( "One edge is 8cm, the other one is 18cm.\r
\n" ); document.write( "\n" ); document.write( "(Sorry, I misread the problem. Do I understand it correctly now - the rectangle's height is 2cm more than twice its width; that is rectangle's height is two times its width plus 2cm?)\r
\n" ); document.write( "\n" ); document.write( "The rectangle has such edges: a, b, a, b where \"b+=+a+%2B+2\"
\n" ); document.write( "So the edges go like this: a, a + 2, a, a + 2\r
\n" ); document.write( "\n" ); document.write( "Their sum is \"a+%2B+a+%2B+2+%2B+a+%2B+a+%2B+2+=+4a+%2B+4\" which is the same as the perimeter: \"4a+%2B+4+=+52\"\r
\n" ); document.write( "\n" ); document.write( "Thus \"4a+=+48\" and \"a+=+12\", and \"b+=+a+%2B+2+=+14\"\r
\n" ); document.write( "\n" ); document.write( "One edge is 12cm, the other one is 14cm.
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