document.write( "Question 242342: a rectangle is 5 meters longer than its width. If thee length is shortened by 2 meters, and the width is increased by one meter, the area will remain the same. Find the length and the area. \n" ); document.write( "

Algebra.Com's Answer #177392 by ptaylor(2198) $\"\"$ $\"About$

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Let w=width of original rectangle
\n" ); document.write( "Then w+5=length of original rectangle
\n" ); document.write( "(w+5)-2=length shortened by 2 meters
\n" ); document.write( "And w+1=width increased by 1 meter\r
\n" ); document.write( "\n" ); document.write( "Now we know that the area of a rectangle=Length times Width or A=lw\r
\n" ); document.write( "\n" ); document.write( "Area of original rectangle=w*(w+5)
\n" ); document.write( "Area of modified rectangle=(w+1)*(w+5-2)=(w+1)*(w+3)
\n" ); document.write( "We are told that the above two areas are equal, so our equation to solve is:\r
\n" ); document.write( "\n" ); document.write( "w(w+5)=(w+1)(w+3) get rid of parens by expanding each side
\n" ); document.write( "w^2+5w=w^2+4w+3 subtract w^2 and also 5w from each side and we get:
\n" ); document.write( "w^2-w^2+5w-5w=w^2-w^2+4w-5w+3 collect like terms
\n" ); document.write( "-w+3=0 subtract 3 from each side
\n" ); document.write( "-w=-3 or
\n" ); document.write( "w=3 meters width of original rectangle
\n" ); document.write( "w+5=3+5=8 length of original rectangle\r
\n" ); document.write( "\n" ); document.write( "Area of original rectangle=8*3=24 sq meters\r
\n" ); document.write( "\n" ); document.write( "CK
\n" ); document.write( "Area of modified rectangle=(3+1)(3+3)=4*6=24 sq meters\r
\n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor \n" ); document.write( "

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