document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=366078>Question 366078</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Jack invests $1500 at a certain annual interest rate, and he invests another $3000 at an annual rate that is one-half percent higher. If he receives a total of $285 interest in 1 year, at what rate is the $1500 invested?<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 #260917 by </B><A HREF=/tutors/aboutme.mpl?userid=fractalier><B>fractalier(6550)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=fractalier><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=fractalier><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=260917\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let r be the rate at which the $1500 is invested. Thus r + .005 is the rate at which the $3000 is invested at (be careful on the decimal point)...thus we have<BR>\n" );
document.write( "1500r + 3000(r + .005) = 285<BR>\n" );
document.write( "Multiply through and get<BR>\n" );
document.write( "1500r + 3000r + 15 = 285 <BR>\n" );
document.write( "Collect and get<BR>\n" );
document.write( "4500r + 15 = 285<BR>\n" );
document.write( "Subtract 15 and get<BR>\n" );
document.write( "4500r = 270<BR>\n" );
document.write( "Now divide by 4500 and get<BR>\n" );
document.write( "r = .06 = 6% </SPAN>\n" );
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