document.write( "Question 174583: Jim is 6 years older than Pam. Two years ago Jim was twice as old as Pam. How old is each now?\r
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document.write( "Milk that has 5% butterfat is mixed with milk that has 2% butterfat. How much of each is needed to obtain 60 gallons of milk that has 3% butterfar? \n" );
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Algebra.Com's Answer #129592 by Mathtut(3670)![]() ![]() ![]() You can put this solution on YOUR website! 1. lets call jim and pam's ages, j and p, respectively \n" ); document.write( ": \n" ); document.write( "j=p+6........eq 1 \n" ); document.write( "j-2=2(p-2)...eq 2 \n" ); document.write( ": \n" ); document.write( "take j's value of p+6 from eq 1 and plug it into eq 2 \n" ); document.write( ": \n" ); document.write( "p+6-2=2p-4 \n" ); document.write( ": \n" ); document.write( "p+4=2p-4 \n" ); document.write( ": \n" ); document.write( "pams age is \n" ); document.write( ": \n" ); document.write( "jim's age is \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "2)lets call x and y the number of gallons of 5% and 2% milk needed \n" ); document.write( ": \n" ); document.write( ".05x+.02y=.03(60) \n" ); document.write( ": \n" ); document.write( "x+y=60---->y=60-x......now plug that value in for y \n" ); document.write( ": \n" ); document.write( ".05x+.02(60-x)=1.8 \n" ); document.write( ": \n" ); document.write( ".05x+1.2-.02x=1.8 \n" ); document.write( ": \n" ); document.write( ".03x=.6 \n" ); document.write( ": \n" ); document.write( " \n" ); document.write( ": \n" ); document.write( " |