<PRE><B>Question 1111255</B>
let F = number of finite match sections.
let A = number of applied calculus sections.
let C = number of computer method sections.


there will be a total of 7 sections, therefore:


F + A + C = 7


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:


40F + 40A + 10C = 220


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:


40,000 * F + 60,000 * A + 27,000 * C = 294,000


you have 3 equations that need to be solved simultaneously.


they are:


F + A + C = 7 (first original equation)
40F + 40A + 10C = 220 (second original equation)
40000F + 60000A + 27000C = 294000 (third original equation)


in the first equation, solve for F to get:


F = 7 - A - C


in the second equation, replace F with (7 - A - C) to get:


40F + 40A + 10C = 220 becomes 40 * (7 - A - C) + 40A + 10C = 220


simplify to get:


40 * 7 - 40A - 40C + 40A + 10C = 220


combine like terms and simplify further to get:


280 - 30C = 220


in the third equation, replace F with (7 - A - C) to get:


40000F + 60000A + 27000C = 294000 becomes 40000 * (7 - A - C) + 60000A + 27000C = 294000


simplify to get:


40000 * 7 -40000A - 40000C + 60000A + 27000C = 294000


combine like terms and simplify further to get:


280000 + 20000A - 13000C = 294000


you now have 2 equations in 2 variables that need to be solved simultaneously.


they are:


280 - 30C = 220 (first reduced equation)
280000 + 20000A - 13000C = 294000 (second reduced equation)


in the first reduced equation, solve for C as follows:


start with 280 - 30C = 220
add 30C to both sides of this equation and subtract 220 from both sides of this equation to get:
60 = 30C
solve for C to get:
C = 2


now that you know that C = 2, replace C in the second reduced equation and solve for A as follows:


start with 280000 + 20000A - 13000C = 294000
replace C with 2 to get:
280000 + 20000A - 13000 * 2 = 294000
simplify to get:
280000 + 20000A - 26000 = 294000
combine like terms to get:
254000 + 20000A = 294000
subtract 254000 from both sides of this equation to get:
20000A = 40000
solve for A to get:
A = 2


you now have:


A = 2
C = 2


go back to your 3 original equations.


they are:


F + A + C = 7 (first original equation)
40F + 40A + 10C = 220 (second original equation)
40000F + 60000A + 27000C = 294000 (third original equation)


in the first original equation, replace A with 2 and C with 2 to get:


F + 2 + 2 = 7
solve for F to getL
F = 3


you now have:


F = 3
A = 2
C = 2


evaluate all 3 original equations with these values of F and A and C to determine whether all 3 original equations are true.


the 3 original equations are, once again:


F + A + C = 7 (first original equation)
40F + 40A + 10C = 220 (second original equation)
40000F + 60000A + 27000C = 294000 (third original equation)


first original equation becomes:


3 + 2 + 2 = 7 which becomes 7 = 7, which is true.


second original equation becomes:


40 * 3 + 40 * 2 + 10 * 2 = 220 which becomes 120 + 80  20 = 220 which  becomes 220 = 220, which is true.


third original equation becomes:


40000 * 3 + 60000 * 2 + 27000 * 2 = 294000 becomes 120000 + 120000 + 54000 = 294000 which becomes 294000 = 294000, which is true.


all 3 original equations are true when F = 3 and A = 2 and C = 2, therefore the solution can be assumed to be good.


your solution is:


there are 3 sections of finite math and 2 sections of applied calculus and 2 sections of computer methods that should be offered.
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