document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=833974>Question 833974</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A woman invests $8000<BR>\n" );
document.write( "-Part of it at 11% and the remainder at 12%<BR>\n" );
document.write( "-Her yearly total interest for the two investments is $930<BR>\n" );
document.write( "How much did she invest at 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 #502773 by </B><A HREF=/tutors/aboutme.mpl?userid=DrBeeee><B>DrBeeee(684)</B></A><img src=\"/images/stars/stars-11.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=DrBeeee><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=502773\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Use the formula for simple interest<BR>\n" );
document.write( "(1) I = prt<BR>\n" );
document.write( "where I is the interest (only) earned on an investment (or bank deposit) at the annual rate of r (decimal, not a %) for a period of t years.<BR>\n" );
document.write( "For this problem we have two investment that we will call x and y. The total interest earned on the total for one year is the sum of each or<BR>\n" );
document.write( "(2) x*0.11*1 + y*0.12*1 = 930<BR>\n" );
document.write( "We are also given that the total investment (x+y) is $8000 whiuch gives us a second equation<BR>\n" );
document.write( "(3) x + y = 8000<BR>\n" );
document.write( "We need to solve (2) and (3) as simultaneous equations. You can use substitution or elimination. Let's do both.<BR>\n" );
document.write( "To use substitution, solve (3) for x and put it into (2). This gives us<BR>\n" );
document.write( "(4) x = 8000 - y and (2) becomes<BR>\n" );
document.write( "(5) 0.11*(8000 - y) + 0.12*y = 930 or<BR>\n" );
document.write( "(6) 880 - 0.11*y + 0.12*y = 930 or<BR>\n" );
document.write( "(7) 0.01*y = 930 - 880 or<BR>\n" );
document.write( "(8) y = 100*50 or<BR>\n" );
document.write( "(9) y = 5000<BR>\n" );
document.write( "Then from (3) we get<BR>\n" );
document.write( "(10) x = 3000<BR>\n" );
document.write( "Using elimination (one technique used by computers), we \"fix\" the coefficient of one of the variables to be the same in both of the given equations. Let's work with x. My first step is to multiply (2) by 100 to get<BR>\n" );
document.write( "(11) 11*x + 12*y = 93000 (Look mom, no decimals)<BR>\n" );
document.write( "and multiply (3) by 11 to get<BR>\n" );
document.write( "(12) 11*x + 11*y = 88000<BR>\n" );
document.write( "Now subtract (12) from (11) to get<BR>\n" );
document.write( "(13) 0 + y = 5000 or<BR>\n" );
document.write( "(14) y = 5000 and from (3) we get<BR>\n" );
document.write( "(15) x = 3000<BR>\n" );
document.write( "Let's use (2) to check our answers.<BR>\n" );
document.write( "Is (3000*0.11*1 + 5000*0.12*1 = 930)?<BR>\n" );
document.write( "Is (330 + 600 = 930)?<BR>\n" );
document.write( "Is (930 = 930)? Yes<BR>\n" );
document.write( "Answer: $3000 was invested at 11% and $5000 was invested at 12%.\r<BR>\n" );
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