document.write( "Question 903644: The dimensions of a rectangle are such that its length is
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document.write( "9
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document.write( "in. more than its width. If the length were doubled and if the width were decreased by
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document.write( "4
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document.write( "in., the area would be increased by
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document.write( "168 \r
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document.write( "in.2.
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document.write( "What are the length and width of the rectangle? \n" );
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Algebra.Com's Answer #548193 by lwsshak3(11628)![]() ![]() ![]() You can put this solution on YOUR website! The dimensions of a rectangle are such that its length is \n" ); document.write( "9in. more than its width. If the length were doubled and if the width were decreased by 4in., the area would be increased by 168 in^2.What are the length and width of the rectangle? \n" ); document.write( "*** \n" ); document.write( "let x=original width of rectangle \n" ); document.write( "x+9=original length of rectangle \n" ); document.write( "2(x+9)=new length of rectangle \n" ); document.write( "x-4=new width of rectangle \n" ); document.write( "area=length*width \n" ); document.write( ".. \n" ); document.write( "2(x+9)(x-4)-x(x+9)=168 \n" ); document.write( "2(x^2+5x-36)-x^2-9x=168 \n" ); document.write( "2x^2+10x-72-x^2-9x=168 \n" ); document.write( "x^2+x-240=0 \n" ); document.write( "(x+16)(x-15)=0 \n" ); document.write( "x=-16 (reject) \n" ); document.write( "or \n" ); document.write( "x=15 \n" ); document.write( "x+9=24 \n" ); document.write( "width of rectangle=15 in \n" ); document.write( "length of rectangle=24 in \n" ); document.write( " |