document.write( "Question 159128: Ziggy's famous yogurt blends regular yogurt that is 3% fat with its no fat yogart to obtain 1%fat. How many pounds of no fat yogurt should be mixed to obtain 60 pounds of low fat yogurt?\r
\n" ); document.write( "\n" ); document.write( "0.03x + 0.0y= 60\r
\n" ); document.write( "\n" ); document.write( "0.03x=60\r
\n" ); document.write( "\n" ); document.write( "x=60/0.03\r
\n" ); document.write( "\n" ); document.write( "x= 20 pounds\r
\n" ); document.write( "\n" ); document.write( "I am not sure if this is the way to set up the problem, please help \n" ); document.write( "
Algebra.Com's Answer #117258 by nerdybill(7384)\"\" \"About 
You can put this solution on YOUR website!
Yes, your method is fine! BUT, on the right side of your equation, it's suppose to be the amount of fat from 60 pounds -- which is 1% of 60:
\n" ); document.write( ".
\n" ); document.write( "You created TWO variables:
\n" ); document.write( "Let x = amount of 3% yogurt
\n" ); document.write( "and y = amount of 0% yogurt
\n" ); document.write( ".
\n" ); document.write( "\"fat from 3% yogurt\" + \"fat from no-fat yogurt\" = \"fat from total mixture\"
\n" ); document.write( "0.03x + 0.0y= .01(60)
\n" ); document.write( "0.03x = 0.6
\n" ); document.write( "x=0.6/0.03
\n" ); document.write( "x= 20 pounds (amount of 3% yogurt)
\n" ); document.write( ".
\n" ); document.write( "Then, since we knew that the total mixture was 60 pounds:
\n" ); document.write( "x+y = 60
\n" ); document.write( "20+y= 60
\n" ); document.write( "y = 60-20
\n" ); document.write( "y = 40 pounds (amount of 0% yogurt)
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