document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1072191>Question 1072191</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A metal alloy is 35% copper. Another metal alloy is 50% copper. How much of each alloy should be used to make 600 grams of a metal alloy that is 45% copper? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #687086 by </B><A HREF=/tutors/aboutme.mpl?userid=jorel1380><B>jorel1380(3719)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jorel1380><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=jorel1380><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=687086\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let m be your first alloy, and n be your second. Then:<BR>\n" );
document.write( "m+n=600<BR>\n" );
document.write( "n=600-m<BR>\n" );
document.write( "and<BR>\n" );
document.write( ".35m+.5(600-m)=.45(600)<BR>\n" );
document.write( ".35m+300-.5m=270<BR>\n" );
document.write( ".15m=30<BR>\n" );
document.write( "m=200 grams of the first alloy<BR>\n" );
document.write( "600-m=400 grams of the second alloy. &#9786;&#9786;&#9786;&#9786;<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );