document.write( "Question 632119: How much of an alloy this is 20% should be mixed with 500 ounces of alloy that is 80% copper in order to get and alloy that is 30% copper? \n" ); document.write( "

Algebra.Com's Answer #398048 by josmiceli(19441) $\"\"$ $\"About$

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Let $\"+x+\"$ = ounces of 20% alloy needed
\n" ); document.write( " $\"+.2x+\"$ = ounces of copper in $\"+x+\"$ ounces
\n" ); document.write( " $\"+.8%2A500+=+400+\"$ ounces of copper in 80% alloy
\n" ); document.write( "---------------
\n" ); document.write( " $\"+%28+.2x+%2B+400+%29+%2F+%28+x+%2B+500+%29+=+.3+\"$
\n" ); document.write( " $\"+.2x+%2B+400+=+.3%2A%28+x+%2B+500+%29+\"$
\n" ); document.write( " $\"+.2x+%2B+400+=+.3x+%2B+150+\"$
\n" ); document.write( " $\"+.1x+=+400+-+150+\"$
\n" ); document.write( " $\"+.1x+=+250+\"$
\n" ); document.write( " $\"+x+=+2500+\"$
\n" ); document.write( "2500 ounces of 20% alloy are needed
\n" ); document.write( "check:
\n" ); document.write( " $\"+%28+.2%2A2500+%2B+400+%29+%2F+%28+2500+%2B+500+%29+=+.3+\"$
\n" ); document.write( " $\"+%28+500+%2B+400+%29+%2F+3000+=+.3+\"$
\n" ); document.write( " $\"+900+=+.3%2A3000+\"$
\n" ); document.write( " $\"+900+=+900+\"$
\n" ); document.write( "OK \n" ); document.write( "

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