document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1033185>Question 1033185</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Sandy needs 14 lb of metal with 50 % silver. If Sandy combines one metal with 16 % silver, and another with 78 % silver, how much of each metal does Sandy need? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #647770 by </B><A HREF=/tutors/aboutme.mpl?userid=fractalier><B>fractalier(6550)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=fractalier><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=fractalier><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=647770\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Call the number of pounds of 16% used, x.<BR>\n" );
document.write( "Then the number of pounds of 78% used must be 14-x.<BR>\n" );
document.write( "The setup looks like this:<BR>\n" );
document.write( ".16x + .78(14-x) = .50(14)<BR>\n" );
document.write( "Now solve for x...<BR>\n" );
document.write( ".16x + 10.92 - .78x = 7<BR>\n" );
document.write( "-.62x = -3.92<BR>\n" );
document.write( "x = 6.32 lb of 16%<BR>\n" );
document.write( "which means Sandy needs<BR>\n" );
document.write( "14-x = 7.68 lb of 78% silver </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );