document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=316179>Question 316179</A>: <I><SPAN STYLE=\"word-break: break-all;\"> two investments in companies total $2000. If one investment earns 3% annually and the other earns 2% annually. Find the amount invested in each, if the total interest earned in $54. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #226142 by </B><A HREF=/tutors/aboutme.mpl?userid=mananth><B>mananth(16946)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=mananth><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=mananth><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=226142\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> let $ x be invested at 3 %<BR>\n" );
document.write( "2000-x will be invested at 2%<BR>\n" );
document.write( "..<BR>\n" );
document.write( "interest equation<BR>\n" );
document.write( "0.03x+0.02(2000-x)= 54<BR>\n" );
document.write( "0.03x+40-0.02x=54<BR>\n" );
document.write( "0.01x= 14<BR>\n" );
document.write( "x=14/0.01<BR>\n" );
document.write( "x= 1400 invested at2%<BR>\n" );
document.write( "balance 600 at 2%<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );