document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=108532>Question 108532</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I've figured this out (trial and error) but need help with setting up the formula.\r<BR>\n" );
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document.write( "Mr. Jones has $7000 to invest.  He invests part at 8% and the rest at 8.5%.  The amount at 8.5% makes $65 less at the end of the year than the amount invested at 8%.  What amount has he invested in each account?<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #79193 by </B><A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><B>ankor@dixie-net.com(22740)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><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=ankor@dixie-net.com><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=79193\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Mr. Jones has $7000 to invest. He invests part at 8% and the rest at 8.5%. The amount at 8.5% makes $65 less at the end of the year than the amount invested at 8%. What amount has he invested in each account?<BR>\n" );
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document.write( "Let x = amt invested at 8%<BR>\n" );
document.write( "Then<BR>\n" );
document.write( "(7000-x) = amt invested at 8.5%<BR>\n" );
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document.write( "Interest @ 8% = Interest @ 8.5% + $65<BR>\n" );
document.write( ".08x = .085(7000-x) + 65<BR>\n" );
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document.write( "Find x </SPAN>\n" );
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