# A State University’s Mathematics Department ollers three courses: Calculus Linea

A State University’s Mathematics Department ollers three courses: Calculus Linear Algebra, and Discrete Mathematics, and the chairperson is trying to decide how many sections of each to offer this semester. The department is allowed to offer 45 sections total, there are 5000 students who would like to take a course, and there are 60 teaching assistants to teach them. Sections of Calculus have 200 students each, sections of Discrete Mathematics have 100 students each, and sections of Linear Algebra have 50 students each. Calculus sections are taught by a team of 2 teaching assistants, while Discrete Mathematics and Linear Algebra need only 1 teaching assistant per section. How many sections of each course should the chair schedule in order to offer all the sections that they are allowed to, accommodate all of the students, and give one teaching assignment to each teaching assistant?

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Egreane61
Step 1
Let A be the number of sections in Calculus.
Let B be the number of sections in Discrete Mathematics.
Let C be the number of sections in Linear algebra.
It is given that a total of 45 sections are allowed. That is, $$\displaystyle{A}+{B}+{C}={45}$$.
The number of students in each section of Calculus = 200
The number of students in each section of Discrete Mathematics= 100
The number of students in each section of Linear algebra= 50
Since the total number of students is 5000, $$\displaystyle{200}{A}+{100}{B}+{50}{C}={5000}$$. That is, $$\displaystyle{4}{A}+{2}{B}+{C}={100}$$.
Step 2
The number of professors needed for each section of Calculus = 2
The number of professors needed for each section of Calculus = 1
The number of professors needed for each section of Calculus = 1
Total number of professors = 60
That is, $$\displaystyle{2}{A}+{B}+{C}={60}$$.
Thus, the three equations in three variable are obtained as $$\displaystyle{A}+{B}+{C}={45}$$, $$\displaystyle{4}{A}+{2}{B}+{C}={100}$$ and $$\displaystyle{2}{A}+{B}+{C}={60}$$.
Step 3
Use elimination method to get values of A, B and C as follows:
Subtract $$\displaystyle{A}+{B}+{C}={45}$$ from $$\displaystyle{2}{A}+{B}+{C}={60}$$ as follows:
$$\displaystyle{2}{A}+{B}+{C}−{\left({A}+{B}+{C}\right)}={60}−{45}$$ $$\displaystyle{A}={15}$$ Substitute $$\displaystyle{A}={15}$$ in A+B+C=45 and $$\displaystyle{4}{A}+{2}{B}+{C}={100}$$:
$$\displaystyle{15}+{B}+{C}={45}$$ $$\displaystyle{B}+{C}={30}$$ $$\displaystyle{4}{\left({15}\right)}+{2}{B}+{C}={100}$$ $$\displaystyle{2}{B}+{C}=={40}$$
Step 4
Subtract $$\displaystyle{B}+{C}={30}$$ from $$\displaystyle{2}{B}+{C}=={40}$$:
$$\displaystyle{2}{B}+{C}−{\left({B}+{C}\right)}=={40}−{30}{B}={10}$$ Substitute $$\displaystyle{A}={15}$$,$$\displaystyle{B}={10}$$ in $$\displaystyle{A}+{B}+{C}={45}$$: $$\displaystyle{15}+{10}+{C}={45}$$ $$\displaystyle{C}={45}−{25}$$ $$\displaystyle{C}={20}$$
Therefore, the number of sections in Calculus is 15, the number of sections in Discrete Mathematics is 10 and the number of sections in Linear algebra is 20.