document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1111255>Question 1111255</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Urban Community College is planning to offer courses in Finite Math, Applied Calculus, and Computer Methods. Each section of Finite Math has 40 students and earns the college $40,000 in revenue. Each section of Applied Calculus has 40 students and earns the college $60,000, while each section of Computer Methods has 10 students and earns the college $27,000. Assuming the college wishes to offer a total of seven sections, accommodate 220 students, and bring in $294,000 in revenues, how many sections of each course should it offer? <BR>\n" );
document.write( "Finite Math	 <BR>\n" );
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document.write( "Applied Calculus	 <BR>\n" );
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document.write( "Computer Methods	 <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 #726313 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=726313\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> let F = number of finite match sections.<BR>\n" );
document.write( "let A = number of applied calculus sections.<BR>\n" );
document.write( "let C = number of computer method sections.\r<BR>\n" );
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document.write( "there will be a total of 7 sections, therefore:\r<BR>\n" );
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document.write( "F + A + C = 7\r<BR>\n" );
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document.write( "each section of F will have 40 students, each section of A will have 40 students, each section of C will have 10 students, and the total number of students will be 220, therefore:\r<BR>\n" );
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document.write( "40F + 40A + 10C = 220\r<BR>\n" );
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document.write( "each section of F will generate 40,000 in revenue, each section of A will generate 60,000 in revenue, and each section of C will generate 27,000 in revenue for a total of 294,000 in revenue, therefore:\r<BR>\n" );
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document.write( "40,000 * F + 60,000 * A + 27,000 * C = 294,000\r<BR>\n" );
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document.write( "you have 3 equations that need to be solved simultaneously.\r<BR>\n" );
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document.write( "they are:\r<BR>\n" );
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document.write( "F + A + C = 7 (first original equation)<BR>\n" );
document.write( "40F + 40A + 10C = 220 (second original equation)<BR>\n" );
document.write( "40000F + 60000A + 27000C = 294000 (third original equation)\r<BR>\n" );
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document.write( "in the first equation, solve for F to get:\r<BR>\n" );
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document.write( "F = 7 - A - C\r<BR>\n" );
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document.write( "in the second equation, replace F with (7 - A - C) to get:\r<BR>\n" );
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document.write( "40F + 40A + 10C = 220 becomes 40 * (7 - A - C) + 40A + 10C = 220\r<BR>\n" );
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document.write( "simplify to get:\r<BR>\n" );
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document.write( "40 * 7 - 40A - 40C + 40A + 10C = 220\r<BR>\n" );
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document.write( "combine like terms and simplify further to get:\r<BR>\n" );
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document.write( "280 - 30C = 220\r<BR>\n" );
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document.write( "in the third equation, replace F with (7 - A - C) to get:\r<BR>\n" );
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document.write( "40000F + 60000A + 27000C = 294000 becomes 40000 * (7 - A - C) + 60000A + 27000C = 294000\r<BR>\n" );
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document.write( "simplify to get:\r<BR>\n" );
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document.write( "40000 * 7 -40000A - 40000C + 60000A + 27000C = 294000\r<BR>\n" );
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document.write( "combine like terms and simplify further to get:\r<BR>\n" );
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document.write( "280000 + 20000A - 13000C = 294000\r<BR>\n" );
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document.write( "you now have 2 equations in 2 variables that need to be solved simultaneously.\r<BR>\n" );
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document.write( "they are:\r<BR>\n" );
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document.write( "280 - 30C = 220 (first reduced equation)<BR>\n" );
document.write( "280000 + 20000A - 13000C = 294000 (second reduced equation)\r<BR>\n" );
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document.write( "in the first reduced equation, solve for C as follows:\r<BR>\n" );
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document.write( "start with 280 - 30C = 220<BR>\n" );
document.write( "add 30C to both sides of this equation and subtract 220 from both sides of this equation to get:<BR>\n" );
document.write( "60 = 30C<BR>\n" );
document.write( "solve for C to get:<BR>\n" );
document.write( "C = 2\r<BR>\n" );
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document.write( "now that you know that C = 2, replace C in the second reduced equation and solve for A as follows:\r<BR>\n" );
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document.write( "start with 280000 + 20000A - 13000C = 294000<BR>\n" );
document.write( "replace C with 2 to get:<BR>\n" );
document.write( "280000 + 20000A - 13000 * 2 = 294000<BR>\n" );
document.write( "simplify to get:<BR>\n" );
document.write( "280000 + 20000A - 26000 = 294000<BR>\n" );
document.write( "combine like terms to get:<BR>\n" );
document.write( "254000 + 20000A = 294000<BR>\n" );
document.write( "subtract 254000 from both sides of this equation to get:<BR>\n" );
document.write( "20000A = 40000<BR>\n" );
document.write( "solve for A to get:<BR>\n" );
document.write( "A = 2\r<BR>\n" );
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document.write( "you now have:\r<BR>\n" );
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document.write( "A = 2<BR>\n" );
document.write( "C = 2\r<BR>\n" );
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document.write( "go back to your 3 original equations.\r<BR>\n" );
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document.write( "they are:\r<BR>\n" );
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document.write( "F + A + C = 7 (first original equation)<BR>\n" );
document.write( "40F + 40A + 10C = 220 (second original equation)<BR>\n" );
document.write( "40000F + 60000A + 27000C = 294000 (third original equation)\r<BR>\n" );
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document.write( "in the first original equation, replace A with 2 and C with 2 to get:\r<BR>\n" );
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document.write( "F + 2 + 2 = 7<BR>\n" );
document.write( "solve for F to getL<BR>\n" );
document.write( "F = 3\r<BR>\n" );
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document.write( "you now have:\r<BR>\n" );
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document.write( "F = 3<BR>\n" );
document.write( "A = 2<BR>\n" );
document.write( "C = 2\r<BR>\n" );
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document.write( "evaluate all 3 original equations with these values of F and A and C to determine whether all 3 original equations are true.\r<BR>\n" );
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document.write( "the 3 original equations are, once again:\r<BR>\n" );
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document.write( "F + A + C = 7 (first original equation)<BR>\n" );
document.write( "40F + 40A + 10C = 220 (second original equation)<BR>\n" );
document.write( "40000F + 60000A + 27000C = 294000 (third original equation)\r<BR>\n" );
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document.write( "first original equation becomes:\r<BR>\n" );
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document.write( "3 + 2 + 2 = 7 which becomes 7 = 7, which is true.\r<BR>\n" );
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document.write( "second original equation becomes:\r<BR>\n" );
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document.write( "40 * 3 + 40 * 2 + 10 * 2 = 220 which becomes 120 + 80  20 = 220 which  becomes 220 = 220, which is true.\r<BR>\n" );
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document.write( "third original equation becomes:\r<BR>\n" );
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document.write( "40000 * 3 + 60000 * 2 + 27000 * 2 = 294000 becomes 120000 + 120000 + 54000 = 294000 which becomes 294000 = 294000, which is true.\r<BR>\n" );
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document.write( "all 3 original equations are true when F = 3 and A = 2 and C = 2, therefore the solution can be assumed to be good.\r<BR>\n" );
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document.write( "your solution is:\r<BR>\n" );
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document.write( "there are 3 sections of finite math and 2 sections of applied calculus and 2 sections of computer methods that should be offered.<BR>\n" );
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