document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=58158>Question 58158</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A metallurgist has one alloy containing 20% copper and another containing 70% copper. How many pounds of alloy must he use to make 50 pounds of the a third alloy containing 50% copper?\r<BR>\n" );
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document.write( "Thank you for taking the time to help! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #39943 by </B><A HREF=/tutors/aboutme.mpl?userid=Nate><B>Nate(3500)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Nate><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=39943\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> x = amount of 20% copper<BR>\n" );
document.write( "y = amount of 70% copper<BR>\n" );
document.write( "*Remember: x = 50 - y<BR>\n" );
document.write( "(0.2x + 0.7y)/50 = 0.5<BR>\n" );
document.write( "0.2x + 0.7y = 25<BR>\n" );
document.write( "0.2(50 - y) + 0.7y = 25<BR>\n" );
document.write( "10 - 0.2y + 0.7y = 25<BR>\n" );
document.write( "0.5y = 15<BR>\n" );
document.write( "y = 30<BR>\n" );
document.write( "x = 50 - y = 50 - 30 = 20<BR>\n" );
document.write( "30 pounds of 70% copper<BR>\n" );
document.write( "20 pounds of 20% copper </SPAN>\n" );
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