document.write( "Question 1057753: how much of an alloy that 30% copper should be mixed with 300 ounces of an alloy that is 90% copper in order to get an alloy that is 40% copper ? \n" ); document.write( "

Algebra.Com's Answer #672757 by josmiceli(19441) $\"\"$ $\"About$

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Let $\"+x+\"$ = ounces of 30% copper alloy needed
\n" ); document.write( " $\"+%28+.3x+%2B+.9%2A%28+300+%29+%29+%2F+%28+x+%2B+300+%29++=+.4+\"$
\n" ); document.write( " $\"+%28+.3x+%2B+270+%29+%2F+%28+x+%2B+300+%29+=+.4+\"$
\n" ); document.write( " $\"+.3x+%2B+270+=+.4%2A%28+x+%2B+300+%29+\"$
\n" ); document.write( " $\"+.3x+%2B+270+=+.4x+%2B+120+\"$
\n" ); document.write( " $\"+.1x+=+150+\"$
\n" ); document.write( " $\"+x+=+1500+\"$
\n" ); document.write( "1,500 ounces of 30% copper alloy are needed
\n" ); document.write( "-------------------------
\n" ); document.write( "check:
\n" ); document.write( " $\"+%28+.3x+%2B+.9%2A%28+300+%29+%29+%2F+%28+x+%2B+300+%29++=+.4+\"$
\n" ); document.write( " $\"+%28+.3%2A1500+%2B+.9%2A%28+300+%29+%29+%2F+%28+1500+%2B+300+%29++=+.4+\"$
\n" ); document.write( " $\"+%28+450+%2B+270+%29+%2F+1800+%29+=+.4+\"$
\n" ); document.write( " $\"+720+%2F+1800+=+.4+\"$
\n" ); document.write( " $\"+720+=+.4%2A1800+\"$
\n" ); document.write( " $\"+720+=+720+\"$
\n" ); document.write( "OK \n" ); document.write( "

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