document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=35169>Question 35169</A>: <I><SPAN STYLE=\"word-break: break-all;\"> .O6X+160An investor decides at the beginning of the year to invest some of his cash in an account paying 8% annual interest, and to put the rest in a stock fund that ends up earning 15%over the course of the year. He puts $2000 more in the first account than in the stock fund, and at the end of the year he finds he has earned $1,310 in interest. How much money was invested at each of the two rates?<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #21353 by </B><A HREF=/tutors/aboutme.mpl?userid=checkley71><B>checkley71(8403)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=checkley71><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=21353\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> (X+2000).08+.15X=1310 OR .O8X+160+.15X=1310 OR .23X=1310-160 OR .23X=1150 OR X=1150/23 OR X=5000 INVESTED @ 15% THE THE 8% INVESTMENT IS 5000+2000=7000\r<BR>\n" );
document.write( "\n" );
document.write( "PROOF .08*7000+.15*5000=1310 OR 560+750=1310 OR 1310=1310   </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );