document.write( "Question 392187: Originally a rectangle was twice as long as its wide. When 5 m was subtracted from its length and 3 m was subtracted from its width, the new rectangle had an area of 55m. Find the dimensions of the NEW rectangle. ( not the old dimensions )\r
\n" ); document.write( "\n" ); document.write( "Please show work clearly for me to understand.
\n" ); document.write( "Thanks :D\r
\n" ); document.write( "\n" ); document.write( "This is what I tried:
\n" ); document.write( "Let length = 2x-5
\n" ); document.write( "Let width = x-3
\n" ); document.write( "( 2x-5) ( x- 3 ) = 55
\n" ); document.write( "2x^-6x-5x+15=55
\n" ); document.write( "2x^-11x+15=55
\n" ); document.write( "2x^-11x-40=0
\n" ); document.write( "Now im stuck on this part. Please help. \n" ); document.write( "

Algebra.Com's Answer #278353 by ewatrrr(24785) $\"\"$ $\"About$

You can put this solution on YOUR website!


\n" );
document.write( "Hi
\n" );
document.write( "Let length = 2x-5
\n" );
document.write( "Let width = x-3
\n" );
document.write( "( 2x-5) ( x- 3 ) = 55m^2
\n" );
document.write( "2x^-6x-5x+15=55m^2
\n" );
document.write( "2x^-11x+15=55
\n" );
document.write( "2x^-11x-40 = 0   |good work thus far! Note m^2 on Area
\n" );
document.write( "Our best friend in this case: the Quadratic formula
\n" );
document.write( "
\n" );
document.write( "
\n" );
document.write( "
\n" );
document.write( " x = 8 or x = -2.5   |Tossing out negative solution
\n" );
document.write( " x = 8m, the original width, Original length 16m
\n" );
document.write( "NEW Dimensions: width = 5m, length = 11m\r
\n" );
document.write( "\n" );
document.write( "CHECKING our Answer
\n" );
document.write( " 5m*11m = 55m^2 \n" );
document.write( "

\n" );