This document provides an overview and outline of topics covered in a textbook on introductory mathematical analysis. The first chapter focuses on applications of algebra, including using equations and inequalities to model real-world scenarios, solving linear and quadratic equations, using summation notation to write sums, and working with absolute values. Sample problems are provided with step-by-step solutions to illustrate key concepts like setting up equations to determine optimal pricing or amounts of materials needed.

1. INTRODUCTORY MATHEMATICALINTRODUCTORY MATHEMATICAL
ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences
©2007 Pearson Education Asia
Chapter 1Chapter 1
Applications and More AlgebraApplications and More Algebra

2. ©2007 Pearson Education Asia
INTRODUCTORY MATHEMATICAL
ANALYSIS
0. Review of Algebra
1. Applications and More Algebra
2. Functions and Graphs
3. Lines, Parabolas, and Systems
4. Exponential and Logarithmic Functions
5. Mathematics of Finance
6. Matrix Algebra
7. Linear Programming
8. Introduction to Probability and Statistics

3. ©2007 Pearson Education Asia
9. Additional Topics in Probability
10. Limits and Continuity
11. Differentiation
12. Additional Differentiation Topics
13. Curve Sketching
14. Integration
15. Methods and Applications of Integration
16. Continuous Random Variables
17. Multivariable Calculus
INTRODUCTORY MATHEMATICAL
ANALYSIS

4. ©2007 Pearson Education Asia
• To model situations described by linear or
quadratic equations.
• To solve linear inequalities in one variable and
to introduce interval notation.
• To model real-life situations in terms of
inequalities.
• To solve equations and inequalities involving
absolute values.
• To write sums in summation notation and
evaluate such sums.
Chapter 1: Applications and More Algebra
Chapter ObjectivesChapter Objectives

5. ©2007 Pearson Education Asia
Chapter 1: Applications and More Algebra
Chapter OutlineChapter Outline
Applications of Equations
Linear Inequalities
Applications of Inequalities
Absolute Value
Summation Notation
1.1)
1.2)
1.3)
1.4)
1.5)

6. ©2007 Pearson Education Asia
• Modeling: Translating relationships in the
problems to mathematical symbols.
Chapter 1: Applications and More Algebra
1.1 Applications of Equations1.1 Applications of Equations
A chemist must prepare 350 ml of a chemical
solution made up of two parts alcohol and three
parts acid. How much of each should be used?
Example 1 - Mixture

7. ©2007 Pearson Education Asia
Solution:
Let n = number of milliliters in each part.
Each part has 70 ml.
Amount of alcohol = 2n = 2(70) = 140 ml
Amount of acid = 3n = 3(70) = 210 ml
70
5
350
3505
35032
==
=
=+
n
n
nn
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 1 - Mixture

8. ©2007 Pearson Education Asia
• Fixed cost is the sum of all costs that are
independent of the level of production.
• Variable cost is the sum of all costs that are
dependent on the level of output.
• Total cost = variable cost + fixed cost
• Total revenue = (price per unit) x
(number of units sold)
• Profit = total revenue − total cost
Chapter 1: Applications and More Algebra
1.1 Applications of Equations

9. ©2007 Pearson Education Asia
The Anderson Company produces a product for
which the variable cost per unit is $6 and the fixed
cost is $80,000. Each unit has a selling price of
$10. Determine the number of units that must be
sold for the company to earn a profit of $60,000.
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 3 – Profit

10. ©2007 Pearson Education Asia
Solution:
Let q = number of sold units.
variable cost = 6q
total cost = 6q + 80,000
total revenue = 10q
Since profit = total revenue − total cost
35,000 units must be sold to earn a profit of $60,000.
( )
q
q
qq
=
=
+−=
000,35
4000,140
000,80610000,60
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 3 – Profit

11. ©2007 Pearson Education Asia
A total of $10,000 was invested in two business
ventures, A and B. At the end of the first year, A and
B yielded returns of 6%and 5.75 %, respectively, on
the original investments. How was the original
amount allocated if the total amount earned was
$588.75?
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 5 – Investment

12. ©2007 Pearson Education Asia
Solution:
Let x = amount ($) invested at 6%.
$5500 was invested at 6%
$10,000−$5500 = $4500 was invested at 5.75%.
( ) ( )( )
5500
75.130025.0
75.5880575.057506.0
75.588000,100575.006.0
=
=
=−+
=−+
x
x
xx
xx
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 5 – Investment

13. ©2007 Pearson Education Asia
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 7 – Apartment Rent
A real-estate firm owns the Parklane Garden
Apartments, which consist of 96 apartments. At
$550 per month, every apartment can be rented.
However, for each $25 per month increase, there
will be three vacancies with no possibility of filling
them. The firm wants to receive $54,600 per month
from rent. What rent should be charged for each
apartment?

14. ©2007 Pearson Education Asia
Solution 1:
Let r = rent ($) to be charged per apartment.
Total rent = (rent per apartment) x
(number of apartments rented)
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 7 – Apartment Rent

15. ©2007 Pearson Education Asia
Solution 1 (Con’t):
Rent should be $650 or $700.
( )
( )
( ) ( )( )
( )
25675
6
500,224050
32
000,365,13440504050
0000,365,140503
34050000,365,1
25
34050
600,54
25
165032400
600,54
25
5503
96600,54
2
2
±=
±
=
−−±
=
=+−
−=





 −
=





 +−
=





 −
−=
r
rr
rr
r
r
r
r
r
r
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 7 – Apartment Rent

16. ©2007 Pearson Education Asia
Solution 2:
Let n = number of $25 increases.
Total rent = (rent per apartment) x
(number of apartments rented)
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 7 – Apartment Rent

17. ©2007 Pearson Education Asia
Solution 2 (Con’t):
The rent charged should be either
550 + 25(6) = $700 or
550 + 25(4) = $650.
( )( )
( )( )
4or6
046
02410
0180075075
75750800,52600,54
39625550600,54
2
2
2
=
=−−
=+−
=+−
−+=
−+=
n
nn
nn
nn
nn
nn
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 7 – Apartment Rent

18. ©2007 Pearson Education Asia
Chapter 1: Applications and More Algebra
1.2 Linear Inequalities1.2 Linear Inequalities
• Supposing a and b are two points on the real-
number line, the relative positions of two points
are as follows:

19. ©2007 Pearson Education Asia
• We use dots to indicate points on a number line.
• Suppose that a < b and x is between a and b.
• Inequality is a statement that one number is less
than another number.
Chapter 1: Applications and More Algebra
1.2 Linear Inequalities

20. ©2007 Pearson Education Asia
• Rules for Inequalities:
1. If a < b, then a + c < b + c and a − c < b − c.
2. If a < b and c > 0, then ac < bc
and a/c < b/c.
3. If a < b and c < 0, then a(c) > b(c) and a/c > b/c.
4. If a < b and a = c, then c < b.
5. If 0 < a < b or a < b < 0, then 1/a > 1/b .
6. If 0 < a < b and n > 0, then an
< bn
.
If 0 < a < b, then .
Chapter 1: Applications and More Algebra
1.2 Linear Inequalities
nn
ba <

21. ©2007 Pearson Education Asia
Chapter 1: Applications and More Algebra
1.2 Linear Inequalities
• Linear inequality can be written in the form
ax + b < 0
where a and b are constants and a ≠ 0
• To solve an inequality involving a variable is to
find all values of the variable for which the
inequality is true.

22. ©2007 Pearson Education Asia
Chapter 1: Applications and More Algebra
1.2 Linear Inequalities
Example 1 – Solving a Linear Inequality
Solve 2(x − 3) < 4.
Solution:
Replace inequality by equivalent inequalities.
( )
5
2
10
2
2
102
64662
462
432
<
<
<
+<+−
<−
<−
x
x
x
x
x
x

23. ©2007 Pearson Education Asia
Chapter 1: Applications and More Algebra
1.2 Linear Inequalities
Example 3 – Solving a Linear Inequality
Solve (3/2)(s − 2) + 1 > −2(s − 4).
( ) ( )
( ) ( )[ ]
( )[ ] ( )
7
20
207
16443
442232
42212
2
3
2
4212
2
3
>
>
+−>−
−−>+−
−>





+−
−−>+−
s
s
ss
ss
s-s
ss
The solution is ( 20/7 ,∞).
Solution:

24. ©2007 Pearson Education Asia
Chapter 1: Applications and More Algebra
1.3 Applications of Inequalities1.3 Applications of Inequalities
Example 1 - Profit
• Solving word problems may involve inequalities.
For a company that manufactures aquarium
heaters, the combined cost for labor and material is
$21 per heater. Fixed costs (costs incurred in a
given period, regardless of output) are $70,000. If
the selling price of a heater is $35, how many must
be sold for the company to earn a profit?

25. ©2007 Pearson Education Asia
Solution:
profit = total revenue − total cost
( )
5000
000,7014
0000,702135
0costtotalrevenuetotal
>
>
>+−
>−
q
q
qq
Let q = number of heaters sold.
Chapter 1: Applications and More Algebra
1.3 Applications of Inequalities
Example 1 - Profit

26. ©2007 Pearson Education Asia
After consulting with the comptroller, the president
of the Ace Sports Equipment Company decides to
take out a short-term loan to build up inventory.
The company has current assets of $350,000 and
current liabilities of $80,000. How much can the
company borrow if the current ratio is to be no less
than 2.5? (Note: The funds received are
considered as current assets and the loan as a
current liability.)
Chapter 1: Applications and More Algebra
1.3 Applications of Inequalities
Example 3 – Current Ratio

27. ©2007 Pearson Education Asia
Solution:
Let x = amount the company can borrow.
Current ratio = Current assets / Current liabilities
We want,
The company may borrow up to $100,000.
( )
x
x
xx
x
x
≥
≥
+≥+
≥
+
+
000,100
5.1000,150
000,805.2000,350
5.2
000,80
000,350
Chapter 1: Applications and More Algebra
1.3 Applications of Inequalities
Example 3 – Current Ratio

28. ©2007 Pearson Education Asia
• On real-number line, the distance of x from 0 is
called the absolute value of x, denoted as |x|.
DEFINITION
The absolute value of a real number x, written |x|,
is defined by



<−
≥
=
0if,
0if,
xx
xx
x
Chapter 1: Applications and More Algebra
1.4 Absolute Value1.4 Absolute Value

29. ©2007 Pearson Education Asia
Chapter 1: Applications and More Algebra
1.4 Absolute Value
Example 1 – Solving Absolute-Value Equations
a. Solve |x − 3| = 2
b. Solve |7 − 3x| = 5
c. Solve |x − 4| = −3

30. ©2007 Pearson Education Asia
Solution:
a. x − 3 = 2 or x − 3 = −2
x = 5 x = 1
b. 7 − 3x = 5 or 7 − 3x = −5
x = 2/3 x = 4
c. The absolute value of a number is never
negative. The solution set is ∅.
Chapter 1: Applications and More Algebra
1.4 Absolute Value
Example 1 – Solving Absolute-Value Equations

31. ©2007 Pearson Education Asia
Absolute-Value Inequalities
• Summary of the solutions to absolute-value
inequalities is given.
Chapter 1: Applications and More Algebra
1.4 Absolute Value

32. ©2007 Pearson Education Asia
Chapter 1: Applications and More Algebra
1.4 Absolute Value
Example 3 – Solving Absolute-Value Equations
a. Solve |x + 5| ≥ 7
b. Solve |3x − 4| > 1
Solution:
a.
We write it as , where ∪ is the union
symbol.
b.
We can write it as .
212
75or75
≥−≤
−≥+−≤+
xx
xx
] [( )∞−∞− ,212, 
3
5
1
143or143
><
>−−<−
xx
xx
( ) 





∞∪∞− ,
3
5
1,

33. ©2007 Pearson Education Asia
Properties of the Absolute Value
• 5 basic properties of the absolute value:
• Property 5 is known as the triangle inequality.
baba
aaa
abba
b
a
b
a
baab
+≤+
≤≤−
−=−
=
⋅=
.5
.4
.3
.2
.1
Chapter 1: Applications and More Algebra
1.4 Absolute Value

34. ©2007 Pearson Education Asia
Chapter 1: Applications and More Algebra
1.4 Absolute Value
Example 5 – Properties of Absolute Value
( )
( ) 323251132g.
222f.
5
3
5
3
5
3
e.
3
7
3
7
3
7
;
3
7
3
7
3
7
d.
77c.
24224b.
213737-a.
+−=+=≤==+−
≤≤
−
=
−
−
=
−
−
=
−
−
=
−
−
=
−
=
−
−=−
=−=−
=⋅−=⋅
-
xxx
xx
Solution:

35. ©2007 Pearson Education Asia
Chapter 1: Applications and More Algebra
1.5 Summation Notation1.5 Summation Notation
DEFINITION
The sum of the numbers ai, with i successively
taking on the values m through n is denoted as
nmmm
n
mi
i aaaaa ++++= ++
=
∑ ...21

36. ©2007 Pearson Education Asia
Evaluate the given sums.
a. b.
Solution:
a.
b.
Chapter 1: Applications and More Algebra
1.5 Summation Notation
Example 1 – Evaluating Sums
( )∑=
−
7
3
25
n
n ( )∑=
+
6
1
2
1
j
j
( ) ( )[ ] ( )[ ] ( )[ ] ( )[ ] ( )[ ]
115
3328231813
27526525524523525
7
3
=
++++=
−+−+−+−+−=−∑=n
n
( ) ( ) ( ) ( ) ( ) ( ) ( )
97
3726171052
1615141312111 222222
6
1
2
=
+++++=
+++++++++++=+∑=j
j

37. ©2007 Pearson Education Asia
• To sum up consecutive numbers, we have
where n = the last number.
( )
2
1
1
+
=∑=
nn
i
n
i
Chapter 1: Applications and More Algebra
1.5 Summation Notation

38. ©2007 Pearson Education Asia
Evaluate the given sums.
a. b. c.
Solution:
a.
b.
c.
( ) ( ) 550,251003
2
101100
53535
100
1
100
1
100
1
=+




 ⋅
=+=+ ∑∑∑ === kkk
kk
300,180,24
6
401201200
999
200
1
2
200
1
2
=




 ⋅⋅
== ∑∑ == kk
kk
Chapter 1: Applications and More Algebra
1.5 Summation Notation
Example 3 – Applying the Properties of Summation Notation
28471444
71
1
100
30
=⋅== ∑∑ == ij
( )∑=
+
100
1
35
k
k ∑=
200
1
2
9
k
k∑=
100
30
4
j