document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=376445>Question 376445</A>: <I><SPAN STYLE=\"word-break: break-all;\"> HOw many oz. of a metal containing 66% silver must be combined with 3 oz. of a metal containing 90% sivler to form an alloy containing 85% silver? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #267825 by </B><A HREF=/tutors/aboutme.mpl?userid=Earlsdon><B>Earlsdon(6294)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Earlsdon><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=267825\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> How much silver is there in the alloys?<BR>\n" );
document.write( "Let x = the number of oz of 66% silver alloy.<BR>\n" );
document.write( "This contains (0.66)x oz of silver.<BR>\n" );
document.write( "The 3 oz of 90% silver alloy contain (0.9)(3) oz of silver.<BR>\n" );
document.write( "The sum of these is to equal (3+x)(0.85) oz of silver.<BR>\n" );
document.write( "We can write the equation to solve for x.<BR>\n" );
document.write( "0.66x+0.9(3) = 0.85(3+x) Simplify and solve for x.<BR>\n" );
document.write( "0.66x+2.7 = 2.55+0.85x  Subtract 0.66x from both sides.<BR>\n" );
document.write( "2.7 = 2.55+0.19x Subtract 2.55 from both sides.<BR>\n" );
document.write( "0.15 = 0.19x Finally, divide both sides by 0.19<BR>\n" );
document.write( "x = 0.789 oz of the 66% silver alloy will be required. </SPAN>\n" );
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