document.write( "Question 529692: A foundry needs to produce 75 tons of an alloy that is 34% copper. It has supplies of 9% copper and 84% copper alloy. How many tons of each must be mixed obtain he desired results \n" ); document.write( "
Algebra.Com's Answer #349756 by ankor@dixie-net.com(22740)\"\" \"About 
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A foundry needs to produce 75 tons of an alloy that is 34% copper. It has supplies of 9% copper and 84% copper alloy.
\n" ); document.write( " How many tons of each must be mixed obtain he desired results
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\n" ); document.write( "Let x = amt of 84% copper required
\n" ); document.write( "then the total is required to be 75 tons, therefore
\n" ); document.write( "(75-x) = amt of 9% copper required
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\n" ); document.write( "A typical mixture equation,
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\n" ); document.write( ".84x + .09(75-x) = .34(75)
\n" ); document.write( ".84x + 6.75 - .09x = 25.5
\n" ); document.write( ".84x - .09x = 25.5 - 6.75
\n" ); document.write( ".75x = 18.75
\n" ); document.write( "x = \"18.75%2F.75\"
\n" ); document.write( "x = 25 tons of 84% copper
\n" ); document.write( "then
\n" ); document.write( "75 - 25 = 50 tons of 9% copper
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\n" ); document.write( "Check this is
\n" ); document.write( ".84(25) + .09(50) = .34(75)
\n" ); document.write( "21 + 4.5 = 25.5 \n" ); document.write( "
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