1. The document discusses equations, inequalities, and systems of equations for modeling real-world situations. It covers topics such as linear equations, quadratic equations, simultaneous equations, and inequalities. 2. Examples are provided to demonstrate modeling problems using different equation and inequality types. Variables are identified and equations or inequalities are set up to represent the relationships in the word problems. 3. The solutions show solving the equations or systems of equations to find the values of the variables that satisfy the constraints, thus addressing the questions asked.

1. CHAPTER 2
Equation and Inequalities

2. Equation
 Equation are the basic mathematical
tool for solving real-world problem
 To solve the problem, we must know
how to construct equation that model
real-life situations.

3. Equation
 Linear Equation
 Quadratic Equation
 Polinomial equation
0=+ bax
02
=++ cbxax
01
2
2
1
1 ... axaxaxaxa n
n
n
n +++++ −
−

4. Solving the Linear Equation
Solve the equation below
1.
Solution
8347 +=− xx 2.
Solution
x
x
4
3
3
2
6
=+
3
124
1237
=
=
+=
x
x
xx
7
8
4842
544812
4
3
18
123
=
=
=+
=
+
x
x
xx
x
x

5. Modeling with Equation
Guideline for modeling with equation
 Identify the variable
 Express all unknown quantities in
term of the variable
 Set up the model
 Solve the equation and check your
answer

6. Example 1
A square garden has a walkway 3m wide around its outer
edge. If the area of the entire garden, including the
walkway, is 18,000m2
, what are the
dimensions of planted area?
Solution:
We are asked to find the length and width of the
planted area. So we let
x = the length of the planted area

7. Next, translate the information into the language of algebra
We now set up the model.
area of entire garden = 18, 000m2
In word In Algebra
Length of planted area x
Length of entire garden x + 6
Area of entire garden (x + 6)2
( )
128
6000,18
000,186
000,186
2
≈
−=
=+
=+
x
x
x
x

8. Example 2
A manufacturer of soft drinks advertise their orange soda as
‘natural flavored’although it contains only 5% orange juice.
A new federal regulation stipulated that to be called ‘natural’
a drink must contain at least 10% fruit juice. How much
pure orange juice must this manufacturer add to 900 gal of
orange soda to conform to the new regulation?
Solution
The problem asks for the amount of pure orange juice to be
added. So let
x = the amount (in gallons) of pure orange juice to be
added

9. Next, translate the information into the language of algebra
To set up the model, we use the fact that the total amount of
orange juice in the mixture is equal to the orange juice in the
first two vats.
Amount of amount of amount of
Orange juice + orange juice = orange juice
In first vat in second vat in mixture
The manufacturer should add 50 gal of pure orange juice to the soda
In word In Algebra
Amount of orange juice to be added x
Amount of the mixture 900 + x
Amount of orange juice in the first vat (0.05)(900) = 45
Amount of orange juice in the second vat (1)(x) = x
Amount of orange juice in the mixture 0.10(900 + x)
50
9.0
45
459.0
1.09045
)900(1.045
==
=
+=+
+=+
x
x
xx
xx

10. Solving Quadratic Equations
 By factoring
1. Solve the equation
Solution:
 By completing the square
2. Solve the equation
Solution:
2452
=+ xx
( )( )
8@3
083
02452
−==
=+−
=−+
xx
xx
xx
06123 2
=+− xx
( )
( ) )43(6443
643
6123
2
2
2
⋅+−=+−
−=−
−=−
xx
xx
xx ( )
( )
22
22
22
623
2
2
+±=
±=−
=−
=−
x
x
x
x

11. Modeling with Quadratic Equations
A farmer has rectangular garden plot surrounded by 200m
of fence. Find the length and width of the garden if its
area is 2400m2
.
Solution:
We are asked to find the length and width of the garden. So
let w = width of the garden
Now set up the model.
(width of garden).(length of garden) = area of garden
The dimension of the garden is 60m by 40m
In word In Algebra
Width of garden w
Length of garden (200–2w)/2 = 100-w
( )
( )( )
40@60
04060
02400100
2400100
2
==
=−−
=+−
=−
ww
ww
ww
ww

12. Polynomial Equations Polynomial equation can be solve by change it into
quadratic equation.
 Solve by substituting
If then
When Then so or
When Then so
0122 234
=++−+ xxxx
x
xu
1
+=
.
x
xu
1
+= 2
22 1
2
x
xu +=−
0122 234
=++−+ xxxx
( ):2
x÷ 0
12
12 2
2
=++−+
xx
xx
( )( )
1@3
013
032
0122
01
1
2
1
2
2
2
2
=−=
=−+
=−+
=−+−
=−





++





+
uu
uu
uu
uu
x
x
x
x
3−=u 3
1
−=+
x
x
2
53 ±−
=x
1=u 1
1
=+
x
x
2
31 i
x
±
=

15. Application
 Energy Expended in Bird Flight
Ornithologist have determined that some species of birds
tend to avoid flights over large bodies of water during
daylight hours, because air generally rises over land and
falls over water in the daytime, so flying over water
requires more energy. A bird is released from point A on
an island, 5 mi from B, the nearest point on a straight
shoreline. The bird flies to a point C on the shoreline and
then flies along the shoreline to its nesting area D.
Suppose the bird has 170 kcal of energy reserves. It
uses 10 kcal/mi flying over land and 14 kcal/mi flying
over water .
(a) Where should the point C be located so that the bird
use exactly 170kcal of energy during its flight?
(b) Does the bird have enough energy reserves to fly
directly from A to D?

16. Solution:
(a) We are asked to find the location of C. So let
x = distance from B to C
From the fact that
energy used = energy per mile X miles flown
We determine the following:
Now we set up the model.
total energy used = energy used over water + energy used over land
To solve this equation, we eliminate the square root by first bringing
all other terms to the left of the equal sign and then squaring each
side
In word In Algebra
Distance from B to C x
Distance flown over water (from A to C)
Distance flown over land (from C to D) 12 – x
Energy used over water 14
Energy used over land 10(12-x)
( )xx −++= 12102514170 2
252
+x
252
+x

17. Point C should be either 6(2/3)mi or 3(3/4)mi from B so
that the bird uses exactly 170 kcal of energy during its
flight.
(b) By the Pythagorean Theorem, the length of the route
directly from A to D is √52
+122
= 13mi, so the energy the
bird requires for that route is 14 x 13 = 182 kcal. This is
more energy than the bird has available, so it can’t use
this route.
( )
( ) ( )
4
3
3@
3
2
6
02400100096
490019610010002500
25141050
25141050
25141210170
2
22
222
2
2
==
=+−
+=++
+=+
+=+
+=−−
xx
xx
xxx
xx
xx
xx

18. Simultaneous Equation
 Simultaneous equation has two or more equation that has
similar set of solution. Linear equation can be solved by
using substitution and elimination method.

23. Modeling with Linear Systems
 Guideline for modeling with systems
of equations
1. Identify the variable.
2. Express all unknown quantities in
terms of variables.
3. Set up a system of equations.
4. solve the system and interpret the
results.

24. Example
A researcher performs an experiment to test the hypothesis that
involves the nutrients niacin and retinol. She feeds one group of
laboratory rats a daily diet of precisely 32 units of niacin and
22000 units of retinol. She uses two types of commercial pallet
foods. Food A contains 0.12 unit of niacin and 100 units of retinol
per gram. Food B contains 0.20 unit of niacin and 50 units of
retinol per gram. How many grams of each food does she feed this
group of rats each day?
Solution:
For food A 200 grams and for food B 40 grams
200
40
56014
)4(2640612
)3(32002012
)2(000,2250100
)1(3220.012.0
=
=
=
−−−−−−−−−−−=+
−−−−−−−−−−=+
−−−−−−−−−=+
−−−−−−−−−=+
x
y
y
yx
yx
yx
yx

25. Example
A farmer has 1200 acres of land on which he grows corn,
banana and watermelon. It costs RM45 per acre to grow
watermelon, RM60 for corn and RM50 for banana. With
RM63 750 how many acres of each crop can be planted if
the acreage of corn planting twice as watermelon?
Solution:
)1(1200 −−−−−−−−−−=++ zyx
)2(63750506045 −−−−−=++ zyx
)3(2 −−−−−−−−= yx
)4(12003 −−−−−−−−=+ zx
)5(6375050165 −−−−−=+ zx
)6(6000050150 −−−−−=+ zx
375015 =x
250=x
500)250(2 ==y
450)500250(1200 =+−=z

26. Partial Fraction
 The denominators of the algebraic fractions encountered
will be of three basic types:

34. Inequalities
 Linear inequality with one variable

35.  Two linear inequality with one variable

36.  Rational inequality

39.  Quadratic inequality

41. Modeling with Inequalities
Example
Students in Animal Husbandry Science from Faculty of Agro
Industry and Natural Resources, Universiti Malaysia
Kelantan planning to held an Animal Carnival. One of the
activities in the Carnival is riding the horse. The
students who handle this activity has two plans for tickets
Plan A: RM5 entrance fee and 25sen each ride
Plan B: RM2 entrance fee and 50sen each ride
How many rides would you have to take for plan A to be
less expensive than plan B?
Solution: cost plan A < cost plan B
x = number of rides
Cost with plan A = 5 + 0.25x
Cost with plan B = 2 + 0.50x
So if you plan to take more than 12 rides, plan A is less
expensive
12
325.0
5.0225.05
>
>
+<+
x
x
xx

42. Example
A ticket for a Biotechnology exhibition is RM50 per person. A
reduction of 10 cent per ticket will be given to students who
come by group. A group of UMK students decides to attend
the exhibition. The cost of chartering the bus is RM450,
which is to be shared equally among the students. How
many students must be in the group for the total cost (bus
fare and exhibition ticket) per student to be less than RM54?
Solution:
We were asked for the number of students in that group. So
let x = number of students in the group. The information in
the problem maybe organized as follows:
In words In algebra
Numbers of students in a group x
Bus cost per student 450/x
Ticket cost per student 50 – 0.1x

43. Now we set up the model;
Bus cost per student + Ticket cost per student < 54
-90 0 50
Set of solution:
There is no negative number of student so that the group
must have more than 50 students so that the total
( )( ) 0
5090
0
404500
54)10.050(
450
2
<
−+
<
−−
<−+
x
xx
x
xx
x
x
( )
x
xx )50(90 −+
(90 + x) - + + +
(50 - x) + + + -
x - - + +
+ - + -
( ) ),50(0,90 ∞∪−