document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=633045>Question 633045</A>: <I><SPAN STYLE=\"word-break: break-all;\"> On two investments totaling 10,500, Peter lost 4% on one and earned 6% on the other. If his net annual receipts were $185, how much was each investment? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #398627 by </B><A HREF=/tutors/aboutme.mpl?userid=graphmatics><B>graphmatics(170)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=graphmatics><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=graphmatics><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=398627\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let x be the investment amount of the 6% earned interest, then 10500-x is the investment amount of the 4% loss interest. The earned interest, .06x, minus the loss interest, .04(10500-x), is 185.    .06x- .04(10500-x)=185.   .1x-.04*10500=185.  .1x-420=185.    .1x=605.     x=605/.1.    x=6050.   So the 6% investment is $6050 and the 4% investment is $4450 </SPAN>\n" );
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