document.write( "Question 597587: Anne and Nancy use a metal alloy that is 13.2% copper to make jewelry. How many ounces of a 12% alloy must be mixed with a 15% alloy to form 55 ounces of the desired alloy? \n" ); document.write( "
Algebra.Com's Answer #378394 by ankor@dixie-net.com(22740)\"\" \"About 
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Anne and Nancy use a metal alloy that is 13.2% copper to make jewelry.
\n" ); document.write( " How many ounces of a 12% alloy must be mixed with a 15% alloy to form 55 ounces of the desired alloy?
\n" ); document.write( ":
\n" ); document.write( "Let x = amt of 15% alloy required
\n" ); document.write( "Resulting amt is to be 55 oz, therefore
\n" ); document.write( "(55-x) = amt of 12% alloy required
\n" ); document.write( ":
\n" ); document.write( "A typical mixture equation
\n" ); document.write( ".12(55-x) + .15x = .132(55)
\n" ); document.write( "6.6 - .12x + 15x = 7.26
\n" ); document.write( "-.12x + .15x = 7.26 - 6.6
\n" ); document.write( ",03x = .66
\n" ); document.write( "x = .66/.03
\n" ); document.write( "x = 22 oz of 15% alloy
\n" ); document.write( "then
\n" ); document.write( "55 - 22 = 33 oz 12% alloy
\n" ); document.write( ":
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\n" ); document.write( "Check solutions
\n" ); document.write( ".12(33) + .15(22) = .132(55)
\n" ); document.write( "3.96 + 3.3 = 7.26 \n" ); document.write( "
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