document.write( "Question 417356: I'm totally lost. My problem: A rectangle is 4 times as long as it is wide. A second rectangle is 5cm longer and 2cm wider than the first. The area of the second rectangle is 530 square centimeters greater than the first. What are the dimensions of the original rectangle? P.S. I really stink at these kind of problems.Any help would be really appreciated. \n" ); document.write( "
Algebra.Com's Answer #292180 by mananth(16946)\"\" \"About 
You can put this solution on YOUR website!
I rectangle\r
\n" ); document.write( "\n" ); document.write( "width = x
\n" ); document.write( "length = 4x\r
\n" ); document.write( "\n" ); document.write( "..
\n" ); document.write( "II rectangle
\n" ); document.write( "width = x+2
\n" ); document.write( "length = 4x+5
\n" ); document.write( "...
\n" ); document.write( "Area II = Area I +530\r
\n" ); document.write( "\n" ); document.write( "(x+2)(4x+5)= 4x^2+530
\n" ); document.write( "4x^2+5x+8x+10=4x^2+530
\n" ); document.write( "13x+10=530
\n" ); document.write( "13x=520
\n" ); document.write( "/13
\n" ); document.write( "x=40 cm width of I
\n" ); document.write( "length of I = 160 cm\r
\n" ); document.write( "\n" ); document.write( "
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