document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=190406>Question 190406</A>: <I><SPAN STYLE=\"word-break: break-all;\"> This problem is from a computer-based textbook. An alloy of tin and copper is 16 percent tin.  How much tin( to the nearest tenth) must be added to make an alloy that is 18 percent tin? \r<BR>\n" );
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document.write( "I'm really having trouble doing problems with 2 unknowns.  I think I need to know how much copper there is, but I'm not sure how to do that.  Please help.  Thank you.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #143040 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=143040\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <BR>\n" );
document.write( "Let A=amount of alloy that we have to start with (I think you left this out)<BR>\n" );
document.write( "and Let x=amount of tin that needs to be added\r<BR>\n" );
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document.write( "Now we know that the amount of copper before we add tin(0.84A) has to equal the amount of copper after we add tin ((0.82(A+x)).  So, our equation to solve is:\r<BR>\n" );
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document.write( "0.84A=0.82(A+x)  get rid of parens<BR>\n" );
document.write( "0.84A=0.82A+0.82x  subtract 0.82A from each side<BR>\n" );
document.write( "0.84A-0.82A=0.82x<BR>\n" );
document.write( "0.82x=0.02A  divide each side by 0.82<BR>\n" );
document.write( "x=0.024A\r<BR>\n" );
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document.write( "As you can see here, the amount of tin that needs to be added will depend upon the original amount of the alloy and you did not include that number.\r<BR>\n" );
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document.write( "Hope this helps---ptaylor </SPAN>\n" );
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