document.write( "Question 310839: Anne & Nancy use a metal alloy that is 23.371% copper to make jewelry. How many ounces of a 10% alloy must be mixed with a 28% alloy to form 105 ounces of the desired alloy? \n" ); document.write( "
Algebra.Com's Answer #222319 by ankor@dixie-net.com(22740)\"\" \"About 
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Anne & Nancy use a metal alloy that is 23.371% copper to make jewelry.
\n" ); document.write( "How many ounces of a 10% alloy must be mixed with a 28% alloy to form 105 ounces of the desired alloy?
\n" ); document.write( ":
\n" ); document.write( "Let x = amt of 28% alloy required
\n" ); document.write( "It give the resulting total to be 105 oz, therefore
\n" ); document.write( "(105-x) = amt of 10% alloy
\n" ); document.write( ":
\n" ); document.write( ".28x + .10(105-x) = .23371(105)
\n" ); document.write( ":
\n" ); document.write( ".28x + 10.5 - .10x = 23.48955
\n" ); document.write( ":
\n" ); document.write( ".28x - .10x = 23.48955 - 10.5
\n" ); document.write( ":
\n" ); document.write( ".18x = 12.98955
\n" ); document.write( "x = \"12.98955%2F.18\"
\n" ); document.write( "x = 72.164 oz of 28% alloy
\n" ); document.write( "then
\n" ); document.write( "105 - 72.164 = 32.836 oz of 10% alloy \n" ); document.write( "
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