document.write( "Question 107112This question is from textbook Beginning Algebra
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document.write( ": The Length of a rectangular label is three times its width. If the length is decreased by 1 while the width stayed the same, the area of the new label would be 44 square centimeters. Find the original length and the width of the label. \n" );
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Algebra.Com's Answer #78079 by ptaylor(2198)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "Let W=width of label \n" ); document.write( "Then Length(L)=3W \n" ); document.write( "Length decreased by DISABLED_event_one= 3W-1 \n" ); document.write( "Area(A)=L*W \n" ); document.write( "In this problem, W=W and L=(3W-1), So:\r \n" ); document.write( "\n" ); document.write( "A=W(3W-1)=44 get rid of parens (distributive law) \n" ); document.write( "3W^2-W=44 subtract 44 from both sides\r \n" ); document.write( "\n" ); document.write( "3W^2-W-44=44-44 collect like terms\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "(Note: How did I know that it could be factored??? First, I calculated \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "and:\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "CK \n" ); document.write( "4*(12-1)=44 \n" ); document.write( "4*11=44 \n" ); document.write( "44=44\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Hope this helps----ptaylor \n" ); document.write( " \n" ); document.write( " |