document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1047481>Question 1047481</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Larry Mitchell invested part of his $30000 advance at 3% annual interest and the rest at 6%anual simple interest .if his total yearly interest from both account is $930 find the amount invested in each rate  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #662989 by </B><A HREF=/tutors/aboutme.mpl?userid=addingup><B>addingup(3677)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=addingup><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=addingup><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=662989\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let the part of his 30000 invested at 3% be x. Then, the amount invested at 6% will be 30000-x:<BR>\n" );
document.write( "0.03x+0.06(30000-x) = 930<BR>\n" );
document.write( "0.03x+1800-0.06x = 930 <BR>\n" );
document.write( "0.03x-0.06x = 930-1800<BR>\n" );
document.write( "-0.03x = -870 Divide both sides by 0.03, and remember -/- = +<BR>\n" );
document.write( "x = 29000 are invested at 3% and 30000-29000 = 1000 at 6%<BR>\n" );
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document.write( "Check:<BR>\n" );
document.write( "29000*0.03 = 870<BR>\n" );
document.write( "1000*0.06 = = 60<BR>\n" );
document.write( "Total. . . . 930 Correct<BR>\n" );
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