document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=701319>Question 701319</A>: <I><SPAN STYLE=\"word-break: break-all;\"> British sterling silver is a copper-silver alloy that is 7.5% copper by weight. How many grams of pure copper and how many grams of British sterling silver should be used to prepare 200 grams of a copper-silver alloy that is 10% copper by weight? \r<BR>\n" );
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document.write( "Please start me off by helping with setting up the equation and defining the variables; I can take it from there.<BR>\n" );
document.write( "Thank you very much </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #432375 by </B><A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><B>ankor@dixie-net.com(22740)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><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=ankor@dixie-net.com><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=432375\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\">  British sterling silver is a copper-silver alloy that is 7.5% copper by weight.<BR>\n" );
document.write( " How many grams of pure copper and how many grams of British sterling silver should be used to prepare 200 grams of a copper-silver alloy that is 10% copper by weight?<BR>\n" );
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document.write( "let c = amt of pure copper required<BR>\n" );
document.write( "let s = amt of 7.5% copper which Br Sterling<BR>\n" );
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document.write( "c + s = 200<BR>\n" );
document.write( "s = (200-c)\r<BR>\n" );
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document.write( "A mixture equation based on amt of copper<BR>\n" );
document.write( "1c + .075s = .10(200)<BR>\n" );
document.write( "Replace s with (200-c)<BR>\n" );
document.write( "c + .075(200-c) = .10(200)<BR>\n" );
document.write( "c + 15 - .075c = 20<BR>\n" );
document.write( "c - .075c = 20 - 15<BR>\n" );
document.write( ".925c = 5<BR>\n" );
document.write( "c = 5/.925<BR>\n" );
document.write( "c = 5.4 grams of pure copper<BR>\n" );
document.write( "then find s<BR>\n" );
document.write( "200-5.4 = 194.6 grams of Br Sterling<BR>\n" );
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document.write( "I just noticed you only wanted the equations. Sorry!<BR>\n" );
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