document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1173226>Question 1173226</A>: <I><SPAN STYLE=\"word-break: break-all;\"> a metalworker has a metal alloy that is 20​% copper and another alloy that is 60% copper. How many kilograms of each alloy should the metalworker combine to create 100 kg of a 44 ​% copper​ alloy? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #798415 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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/cgi-bin/show-face.mpl?user=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=798415\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> x kg of 0.20 Cu<BR>\n" );
document.write( "100-x kg of 0.60 Cu<BR>\n" );
document.write( "that has to make 100 kg of 0.44 copper, which is 44 kg pure copper.\r<BR>\n" );
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document.write( "0.20x+0.60(100-x)=44<BR>\n" );
document.write( "0.20x+60-0.60x=44<BR>\n" );
document.write( "-0.40x=-16<BR>\n" );
document.write( "x=40 kg of 20% (8 gm of pure)<BR>\n" );
document.write( "100-x=60 kg of 60% (36 gm of pure)\r<BR>\n" );
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document.write( " </SPAN>\n" );
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