document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1073596>Question 1073596</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A chemist has two alloys, one of which is 15% gold and 10% lead and the other which is 20% gold and 30% lead. How many grams of each of the two alloys should be used to make an alloy that contains 82.5 g of gold and 65 g of lead?<BR>\n" );
document.write( "g (first alloy)<BR>\n" );
document.write( "g (second alloy) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #688422 by </B><A HREF=/tutors/aboutme.mpl?userid=ptaylor><B>ptaylor(2198)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ptaylor><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=ptaylor><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=688422\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let x=grams of alloy1 (15% gold, 10%lead)\r<BR>\n" );
document.write( "\n" );
document.write( "And let y=grams of alloy2(20% gold, 30% lead)\r<BR>\n" );
document.write( "\n" );
document.write( "Total of amount of gold:<BR>\n" );
document.write( "0.15x+0.20y=82.5  Eq1\r<BR>\n" );
document.write( "\n" );
document.write( "Total amount of lead:<BR>\n" );
document.write( "0.10x+0.30y=65  Eq2\r<BR>\n" );
document.write( "\n" );
document.write( "Sooooo<BR>\n" );
document.write( "0.15x+0.20y=82.5<BR>\n" );
document.write( "0.10x+0.30y=65\r<BR>\n" );
document.write( "\n" );
document.write( "Multiply eq1 by 2 and eq2 by3 and we get:<BR>\n" );
document.write( "0.30x+0.40y=165 eq1a<BR>\n" );
document.write( "0.30x+0.90y=195 eq2a\r<BR>\n" );
document.write( "\n" );
document.write( "subtract eq1a from eq2a and we get<BR>\n" );
document.write( "0.50y=30<BR>\n" );
document.write( "y=60 grams amount of alloy 1<BR>\n" );
document.write( "substitute 60 gms for y in eq1a<BR>\n" );
document.write( "0.30x+24=165<BR>\n" );
document.write( "0.30x=141<BR>\n" );
document.write( "x=470 grams amount of alloy 2<BR>\n" );
document.write( "CK<BR>\n" );
document.write( "total amount of gold:<BR>\n" );
document.write( "60*0.20+470*0.15=<BR>\n" );
document.write( "12+70.5=82.5gms---check<BR>\n" );
document.write( "total amount of lead:<BR>\n" );
document.write( "60*0.30+470*0.10=<BR>\n" );
document.write( "18+47=65gms---check<BR>\n" );
document.write( "Hope this helps---ptaylor\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );