Course Title: Mathematics for Business &Economics
Ahram Canadian University
School of Business Administration
Dr. Wael Abdel Haleem (Groups B,D&E) & Dr. Nada Ashraf (Groups A&C)
Fall 2020
Tutorial # 1
• 100 points / 3 Credit Hours
30 Quizzes & Assignments
30 Midterm exam
40 Final exam
Evaluation
Grade
Process procedure
Quizzes
15
Three Quizzes
Assignments
15
Five Assignments
Mathematics for Business & Economics
Course objectives & syllabus
The course is designed to help students the use of mathematical tools
in solving business & economic problems. Linear equation and its
applications in economics such as breakeven, market price and
prediction. The main parts of this course are linear programming (the
simplex method), probability, derivatives, and integral calculus and its
applications in business and economics problems constitute.
Mathematics for Business &Economics
course contains:
• Mathematics Foundations
• Mathematical Functions
• Linear Equations
• Properties of Exponents
• Properties of Polynomials
• System of Linear Equations (formulation and solution)
• Properties of Functions
• Solving and Graphing Inequalities
• Graphing and Factoring Quadratic Equations
• Linear Programming Problems (formulation and solution)
Solving Linear Equations in One Variable
Form of linear equation
ax + b = c
a,b,c Constant values
X & y two variables
(X-1)⁴
One variables
2x +3y =11
4 can be 2, 3 or any other nu.. X⁴+1
NOT linear
A solution of a linear equation in one variable is a real number which, when substituted for the variable in the
equation, makes the equation true.
Ex 1 : 3 OR 4 is a solution of 2x + 4 = 12
Substitute 3 for x
2(3) + 4 = 12
6 + 4 not = 12
False equation
Substitute 4 for x
2(4) + 4 = 12
8 + 4 = 12
True equation
Properties of equations
• Addition Property of Equations (Addition & subtraction )
If a = b, then a + c = b + c and a  c = b  c
• Ex 2: Solve x  4 = 12
X-4 +4 = 12 +4
X=16
Check the answer: 16-4=12
true
• Multiplication Property of Equations (Multiplication and division)
If a = b and c  0, then ac = bc and
Ex 3: Solve 2x + 5 = 19
2x + 5  5 = 19  5
2x = 14
1
1
(2 x) 
(14)
2
2
X=7
Check the answer:
2(7)+5= 14+5=19
true
To solve a linear equation in one variable:
1. Simplify both sides of the equation.
2. Use the addition and subtraction properties to get all variable terms
on the left-hand side and all constant terms on the right-hand side.
3. Simplify both sides of the equation.
4. Divide both sides of the equation by the coefficient of the variable.
Ex 4: Solve x + 1 = 3(x  5).
x + 1 = 3(x  5)
Original equation
x + 1 = 3x  15
Simplify right-hand side.
x = 3x  16
Subtract 1 from both sides.
Subtract 3x from both sides.
 2x =  16
Divide both sides by 2.
x=8
The solution is 8.
Check the solution: (8) + 1 = 3((8)  5)  9 = 3(3) True
7
Ex5: Alice has a coin purse containing $5.40 in dimes and
quarters. There are 24 coins all together. How many dimes are
in the coin purse?
Let the number of dimes in the coin purse = d.
Then the number of quarters = 24  d.
10d + 25(24  d) = 540
Linear equation
10d + 600  25d = 540
Simplify left-hand side.
10d  25d =  60
15d =  60
d=4
Subtract 600.
Simplify right-hand side.
Divide by 15.
There are 4 dimes in Alice’s coin purse.
8
Ex6: The sum of three consecutive integers is 54. What
are the three integers?
Three consecutive integers can be represented as
n, n + 1, n + 2.
n + (n + 1) + (n + 2) = 54
3n + 3 = 54
3n = 51
n = 17
Linear equation
Simplify left-hand side.
Subtract 3.
Divide by 3.
The three consecutive integers are 17, 18, and 19.
17 + 18 + 19 = 54.
9
Check.
• EX 7:
A movie theater sells tickets for $8 each, with seniors receiving a discount of
$2. One evening the theater took in $3580 in revenue. If there were 525
tickets sold that evening. It is required to determine how many of each type
of ticket were sold that evening?
• Let suppose that the number of senior's tickets = S
• Then the number of Normal tickets= N = (525-S)
• 6S + 8(525-S) = 3580
Convert data to a linear equation
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•
•
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6S + 4200- 8S= 3580
Simplify left-hand side
6S-8S=-4200+3580
-2S=-620
S=310
There are 310 senior's tickets AND 215 normal tickets.
• Ex 8: Four less than twice some number is 16.
Solution:
First, choose a variable for the unknown number.
Let x represents the unknown indicated by “some number.”
2𝑥 − 4 = 16
2𝑥 − 4 + 4 = 16 + 4
2𝑥 = 20
𝑥 = 10
• Ex 9: When 7 is subtracted from 3 times the sum of a number and
12, the result is 20.
• Solution:
• Let n represent the unknown number.
• 3 𝑛 + 12 − 7 = 20
• 3 𝑛 + 12 = 27
• 𝑛 + 12 = 9
• 𝑛 = −3
• Ex 10: The sum of two consecutive even integers is 46. Find the integers.
• Solution:
• Let x represents the first even integer and x+2 represents the next even
integer
• Add the even integers and set them equal to 46.
• 𝑥 + 𝑥 + 2 = 46
• 2𝑥 + 2 = 46
• 2𝑥 = 44
• 𝑥 = 22
• Use x + 2 to find the next even integer.
• 𝑥 + 2 = 22 + 2 = 24
• The consecutive even integers are 22 and 24.
• Ex 11: The sum of two consecutive odd integers is 36. Find the integers.
• Solution:
• Let x represents the first odd integer and x+2 represents the next odd
integer.
• Add the two odd integers and set the expression equal to 36.
• 𝑥 + 𝑥 + 2 = 36
• 2𝑥 + 2 = 36
• 2𝑥 = 34
• 𝑥 = 17
• Use x + 2 to find the next odd integer.
• 𝑥 + 2 = 17 + 2 = 19
• The consecutive odd integers are 17 and 19.
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Ex 12: The sum of three consecutive integers is 24. Find the integers.
Solution:
Let x represents the first integer
Let x+1 represents the second integer
Let x+2 represents the third integer
Add the 3 integers and set the sum equal to 24.
𝑥 + 𝑥 + 1 + 𝑥 + 2 = 24
3𝑥 + 3 = 24
3𝑥 = 21
𝑥=7
Substitute to find the other two integers.
𝑥+1=7+1=8
𝑥+2=7+2=9
The three consecutive integers are 7, 8 and 9.
• Ex 13: A larger integer is 2 less than 3 times a smaller integer. The sum of the two
integers is 18. Find the integers.
• Solution:
• Let x represents the smaller integer and 3x-2 represents the larger integer.
• Add the expressions that represent the two integers and set the resulting expression
equal to 18.
• 𝑥 + 3𝑥 − 2 = 18
• Solve the equation to obtain the smaller integer x.
• 4𝑥 − 2 = 18
• 4𝑥 = 20
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𝑥=5
Use the expression 3x−2 to find the larger integer.
3𝑥 − 2 = (3 ∗ 5) − 2 = 13
The two integers are 5 and 13.
• Ex 14: Solve for the variable
• 2(x + 5) - 7 = 3(x - 2)
• 2𝑥 + 10 − 7 = 3𝑥 − 6
• 2𝑥 + 3 = 3𝑥 − 6
• −𝑥 = −9
•𝑥=9
Ex 15: Example 8: Solve for the variable.
𝟓
𝟏
𝟏
𝒙 + = 𝟐𝒙 −
𝟒
𝟐
𝟐
To get rid of the fractions multiply both sides by 4
5
1
1
4 𝑥 + = 4 2𝑥 −
4
2
2
5𝑥 + 2 = 8𝑥 − 2
−3𝑥 = −4
4
𝑥=
3
Review tutorials on the same day.
Thursday 12 Nov 2020, Quiz #1 will be available on
Teams from 08:00 pm to 10:00 pm.
The duration of quiz#1 is 30 min.
Any question?