document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=198175>Question 198175</A>: <I><SPAN STYLE=\"word-break: break-all;\"> a chemist has 2 alloys, one of which is 10%gold and 15% lead and the other is 30%gold and 40% lead.  how many grams of the 2 alloys should be used to make an alloy that contains 60g of gold and 88 g of lead.\r<BR>\n" );
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document.write( "This is from a pactive worksheet not a text book<BR>\n" );
document.write( "I tried 25X +70%y=148 as one euqation but not sure of the other equation to substitute x for\r<BR>\n" );
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document.write( "thanks </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #148684 by </B><A HREF=/tutors/aboutme.mpl?userid=scott8148><B>scott8148(6628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=scott8148><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=scott8148><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=148684\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> let x=\"grams of 10% Au alloy\", and y=\"grams of 30% Au alloy\"\r<BR>\n" );
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document.write( "(10%)x + (30%)y = 60 ___ multiplying by 3 ___ (30%)x + (90%)y = 180\r<BR>\n" );
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document.write( "(15%)x + (40%)y = 88 ___ multiplying by 2 ___ (30%)x + (80%)y = 176\r<BR>\n" );
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document.write( "subtracting ___ (10%)y = 4 ___ dividing by 10% ___ y = 40\r<BR>\n" );
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document.write( "substituting ___ (10%)x + (30%)(40) = 60 ___ (10%)x + 12 = 60 ___ (10%)x = 48\r<BR>\n" );
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document.write( "dividing by 10% ___ x = 480 </SPAN>\n" );
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