.Chapter7&8.

INTRODUCTION
In chapter 2 you will learn how to solve equations and inequalities involving a single variable. In this
chapter we are concerned with equation or inequalities form a system. In section 6.1 and 6.2 we solve
system of linear equations, whereas in section 6.3 we solve system involving quadratic equations.
Systems of linear inequalities are created in section 6.4, and in the same section there is an introduction to
linear programming.
6.1 SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES
Many applications of mathematics lead to more than one equation in several variables. The resulting
equations are called a system of equations. The solution set of a system of equations consists of all
solutions that are common to the equation in the system.
ax + by = c
We proved that the graph of an equation of the form is a line and that all ordered pairs (x , y) satisfying
the equation are coordinates of points in the line. A system of two linear equations in two variables x and
y can be written as.
{a1x + b1y = c1}
{a2x + b2y = c2}
Where a1,b1, c1, a2, b2,and c2 are real numbers. The left brace is used to indicate that the two equations
form a system. If an ordered pair (x , y) is to satisfy the system of two linear equations, the corresponding
points (x , y) must lie on the two lines that are the graph of the equations.
ILLUSTRATION 1
A particular system of two linear equations is.
{
2푥 + 푦 = 3
5푥 + 3푦 = 10
The solution set of each of the equations in the system is an infinite set of ordered pairs of real numbers,
and the graphs of these sets are lines. Recall that to draw a sketch of a line we need to find two points on
the line; usually we plot the points where the line intersects the coordinate axis. On the line 2x + y = 3 we
have the points (3/2 , 0) and (0,3) while on the line 5x + 3y = 10 we have the points (2,0) and(0, 10/3).
Shows on the same coordinate system sketches of two lines. It is apparent that two lines intersect at
exactly one point. This point, (-1 , 5) can be verified by substituting into the equations as follows:
2(-1) + 5 = 3
5(-1) + 3(5) = 10
The only ordered pair that is common to the solution sets of the two equations is (-1,5). Hence the
solution set of the system is {(-1,5)}.
ILLUSTRATION 2

Consider the system
{
6푥 − 3푦 = 5
2푥 − 푦 = 4
The lines having these equations appear to be parallel. It can easily to be proved that the lines are indeed
parallel by writing each of the equation, we have
6x - 3y = 5 2x – y = 4
-3y = -6x + 5 -y = -2x + 4
y = 2x – 5/3 y = 2x – 4
Example 1
Use the substitution method to find the solution set of the system.
2푥 + 푦 = 3
5푥 + 3푦 = 10
Illustration 1: {
Solution:
We solve the first equation for y and get the equivalent system
{
푦 = 3 − 2푥
5푥 + 3푦 = 10
We replace y in the second equation by its equal, 3 – 2 x, from the first equation. We then have the
equivalent system
{
푦 = 3 − 2푥
5푥 + 3(3 − 2푥 ) = 10
Simplifying the second equation, we have
푦 = 3 − 2푥
−푥 + 9 = 10
{
Solving the second equation for x, we get
푦 = 3 − 2푥
푥 = −1
{
Finally, we substitute the value of x from the second equation into the first equation and we have
{
푦 = 5
푥 = −1
This system is equivalent to the given one. Hence the solution set is ( -1 , 5)
Example 2

Use the elimination method to find the solution set of the system of equations in Example 1.
{
2푥 + 푦 = 3
5푥 + 3푦 = 10
Remember that our goal is to eliminate one of the variables. Observe that the coefficient
of y is 1 in the first equation and 3 in the second equation. To obtain an equation not involving y, we
therefore replace the second equation by the sum of the second equation and -3 times the first. We begin
by multiplying the first equation by -3 and writing the equivalent system.
{
−6 − 3푦 = −9
5푥 + 3푦 = 10
Adding the equations given the following computations:
−6−3푦 = −9
5푥 +3푦 =10
−푥 =1
With this equation and the first equations in the given system, we can write the following equivalent
system
{
2푥 + 푦 = 3
−푥 = 1
If we now multiply both sides of the second equation by -1, we have the equivalent system
{
2푥 + 푦 = 3
푥 = −1
We next substitute -1 for x in the first equation to obtain
{
2(−1) + 푦 = 3
푥 = −1
푦 = 5
푥 = −1
{
Therefore the solution set (-1,5), which agrees with the result of example 1.
Exercises 6.1 Show the solutions
푥 − 푦 = 8
2푥 + 푦 = 1 2.) {
1.) {
푦 = 8 + 2푥
6푥 + 3푦 = 0
2푥 + 푦 = 6
8푥 = 6푦 + 9 4.) {
3.) {
9푥 − 3푦 = 7
푦 = 3푥 − 5/2
5.) {
푦 = 2푥 − 4
6푥 − 3푦 − 12 = 0 6.) {
2푥 − 3푦 = −1
5푥 − 4푦 = 8
4푥 − 2푦 − 7 = 0
푥 =
7.) {
1
2
푦 + 5
8.) {
3푥 − 푦 = 1
6푥 + 5푦 = 2

2푥 + 6푦 = −11
4푥 − 3푦 = −2 10.) {
9.) {
푦 = 3푥 − 5
6푥 − 2푦 = 10
6.2 SYSTEMS OF LINEAR EQUATIONS IN THREE VARIABLES
So far the linear (first degree) equations we have discussed have contained at most two variables. In this
section we introduce systems of linear equations in three variables.
Consider the equation
2푥 − 푦 + 4푧 = 10
For which the replacement set of each of the three variables x, y and z is the set R of real numbers. This
equation is linear in the three variables. A solution of a linear equation in the three variables x, y and z is
the ordered triple of real numbers (r,s,z) such that if x is replaced by r, y by s, and z by t, the resulting
statement is true. The set of all solutions is the solution set of the equation
Illustration 1
For the equation
2푥 − 푦 + 4푧 = 10
The ordered triple pair (3,4,2) is a solution because
2(3) – 4 + 4 (2) = 10
Some other ordered triples that satisfy this equation are (-1, 8, 5) , (2, -6, 0), (1,0,2), (5,0,0) , (0,-6,1) , (8,
2 1) ,and (7,2 - ½). It appears that the solution set is infinite.
The graph of an equation in three variables is a set of points represented by ordered triples of real
numbers. Such points appear in a three dimensional coordinate system, which we do not discuss. You
should, however, be aware that the graph of a linear equation in a three variables is a plane.
Suppose that we have the following system of linear equations in the variables x, y and z.
{
푎1푥 + 푏1푦 + 푐1푧 = 푑1
푎2푥 + 푏2푦 + 푐2푧 = 푑2
푎3푥 + 푏3푦 + 푐3푧 = 푑3
The solution set of this system is the intersection of the solution sets of the three equations. Because the
graph of each equation is a plane, the solution set can be interpreted geometrically as the intersection of
three planes. When this intersection consist and independent.
Algebraic methods for finding the solution set of a system of three linear equations in three variables are
analogous to those used to solve linear systems in two variables. The following examples shows the
substitution method.
Example Find the solution set of the system

{
푥 − 푦 − 4푧 = 3
2푥 − 3푦 + 2푧 = 0
2푥 − 푦 + 2푧 = 2
Solution, we solve the first set of the system
{
푥 = 푦 + 4푧 + 3
2푥 − 3푦 + 2푧 = 0
2푥 − 푦 + 2푧 = 2
We now substitute the value of x from the first equation into the other two equations , and we obtain the
equivalent system
{
푥 = 푦 + 4푧 + 3
2(푦 + 4푧 + 3) − 3푦 + 2푧 = 0
2(푦 + 4푧 + 3) − 푦 + 2푧 = 2
{
푥 = 푦 + 4푧 + 3
−푦 + 10푧 = −6
푦 + 10푧 = −4
We next solve the second equation for y and get
푥 = 푦 + 4푧 + 3
푦 = 10푧 + 6
푦 + 10푧 = −4
Substituting the value of y from the second equation into the third gives the equivalent system.
{
푥 = 푦 + 4푧 + 3
푦 = 10푧 + 6
(10푧 + 6 +) + 10 = −4
푥 = 푦 + 4푧 + 3
푦 = 10푧 + 6
20푧 = −10
{
{
푥 = 푦 + 4푧 + 3
푦 = 10푧 + 6
푧 = −1/2
Substituting the value of z from the third equation into the second equation, we obtain
푥 = 푦 + 4푧 + 3
{
푦 = 1
푧 = −1/2
Substituting the values of y and z from the second and third equations into the first equation, we get

{
푥 = 2
푦 = 1
푧 = 1/2
The latter system is equivalent to the given system. Hence the solution set of the given system is (2,1
,1/2).The solution can be checked by substituting into each of the given equations. Doing this we have
{
2 − 1 + 2 = 3
4 − 3 − 1 = 0
4 − 1 − 1 = 2
The equations of the given system are consistent and independent.
Exercise
4푥 + 3푦 + 푧 = 15
푥 − 푦 − 2푧 = 2
2푥 − 2푦 + 푧 = 4
1.) {
2푥 + 3푦 + 푧 = 8
5푥 + 2푦 + 3푧 = −13
푥 − 2푦 + 5푧 = 15
2.) {
푥 − 푦 + 3푧 = 2
2푥 + 2푦 − 푧 = 5
5푥 + 2푧 = 7
3.) {
4.) {
3푥 + 2푦 − 푧 = 4
3푥 + 푦 + 3푧 = −2
6푥 − 3푦 − 2푧 = −6
2푥 − 3푦 − 5푧 = 4
푥 + 7푦 + 6푧 = −7
7푥 + 2푦 − 9푧 = 6
5.) {
6.) {
3푥 − 2푦 + 4푧 = 4
7푥 − 5푦 − 푧 = 9
푥 + 9푦 − 9푧 = 1
3푥 − 5푦 + 2푧 = −2
7.) {
2푥 + 3푧 = −3
4푦 − 3푧 = 8
8.) {
푥 − 푦 = 2
3푦 + 푧 = 1
푥 − 2푧 = 7
3푥 − 2푦 = 1
푧 − 푦 = 5
푧 − 2푥 = 5
9.) {
10.) {
푥 − 푦 + 5푧 = 2
4푥 − 3푦 + 5푧 = 3
3푥 − 2푦 + 4푧 = 1
6.3 SYSTEMS INVOLVING QUADRATIC EQUATIONS
In sections 6.1 and 6.2 our discussions of systems of equations was confined to linear systems. However,
a number of applications lead to nonlinear systems as illustrated in exercises 25 through 36. The word
problems in these exercises use concepts presented previously, but the resulting systems involve at least
one quadratic equation. In this section we discussed methods of solving such systems of two equations in
two variables.
We consider first a system that contains a linear equation and quadratic equation. In this case the system
can be solved for one variable in terms of the other, and the resulting expression can be substituted into
the quadratic equation, as shown in the following example.

Example 1
Find the solution set of the system.
푦2 = 4푥
푥 + 푦 = 3
{
Solution
We solve the second equation for x and obtain the equivalent system.
푦2 = 4푥
푥 = 3 − 푦
{
Replacing x in the first equation by its equal from the second, we have the equivalent system
{
푦2 = 4(3 − 푦)
푥 = 3 − 푦
푦2 + 4푦 − 12 = 0
{
푥 = 3 − 푦
We now solve the first equation.
(푦 − 2)(푦 + 6) = 0
푦 − 2 = 0 푦 + 6 = 0
y= 2 푦 = −6
Because the first equation of system (II) is equivalent to the equations 푦 = 2 and 푦 = −6, system (II) is
equivalent to the systems
{
푦 = 2
푥 = 3 − 푦 and {
푦 = −6
푥 = 3 − 푦
In each of the latter two systems we have substitute into the second equation the value of y from the first ,
and we have
푦 = 2
푥 = 1
{
and {
푦 = −6
푥 = 9
These two systems are equivalent to system (I). Thus, the solution set of (I) is (1,2) , (9,-6).
Exercises
Find the solution set of the system.
푥2 + 푦2 = 25
푥 − 푦 + 1 = 0
1.) {
2.) {
푥2 + 푦2 = 25
푥 − 2푦 = −2
푥2 − 푦 = 1
푥2 − 2푦 = −1
3.) {
4.) {
푥2 − 푦2 = 9
2푥 + 푦 = 6

푥2 − 푦2 = 9
푥 + 푦 = 5
5.) {
4푥2 + 푦2 = 25
2푥 + 푦 + 1 = 0
6.) {
푥2 − 푦 − 4 = 0
푥 − 푦 − 3 = 0
7.) {
8.) {
4푥2 + 푦 − 3 = 0
8푥 + 푦 − 7 = 0
푥2 − 2푦2 = 2
푥 + 2푦 = 2
9.) {
10.) {
4푥2 + 푦2 = 17
푥2 + 푦 = 5
6.4 SYSTEMS OF LINEAR INEQUALITIES AND INTRODUCTION TO LINEAR
PROGRAMMING
Systems of linear inequalities are important in economics, business, statistics, science, engineering, and
other fields. With electronic computers performing most of the computation, large numbers of inequalities
with many unknowns are usually involved. In this section we briefly discuss how to solve system of
linear inequalities. We then give an introduction to linear programming, a related approach to decision
making problems.
Statement of the form
퐴푥 + 퐵푦 + 퐶 > 0 퐴푥 + 퐵푦 + 퐶 < 0
퐴푥 + 퐵푦 + 퐶 ≥ 0 퐴푥 + 퐵푦 + 퐶 ≤ 0
Where A,B and C are constants, A and B are not both zero, are inequalities of first degree in two
variables. By the graph of such an inequality, we mean the (x, y) in the rectangular Cartesian coordinate
system for which (x, y) is an ordered pair satisfying the inequality.
Every line in a plane divides the plane into two regions, one on each side of the line. Each of these
regions is called a half plane. The graphs of inequalities of the forms.
퐴푥 + 퐵푦 + 퐶 > 0 and 퐴푥 + 퐵푦 + 퐶 < 0
Are half planes. We shall show this for the particular inequalities
2푥 − 푦 − 4 > 0 푎푛푑 퐴푥 + 퐵푦 + 퐶 < 0
Let L be the line having the equation 2푥 − 푦 − 4 = 0. If we solve this equation for y, we obtain 푦 =
2푥 − 4. If (x, y) is any point in the plane, exactly one of the following statements holds:
푦 = 2푥 − 4 푦 > 2푥 − 4 푦 < 2푥 − 4
Now, 푦 > 2푥 − 4 if and only if the point (x, y) is any point (x, 2x - 4) on line L; Furthermore, 푦 < 2푥 −
4 if and only if the point (x, y) is below the point (x, 2x - 4) on L; therefore the line L divides the plane
into two regions. One region is the half plane above L, which is the graph of inequality 푦 > 2푥 − 4, and
the other region is the half plane above L, which is the graph of the inequalities 푦 > 2푥 − 4, and the
region is the half plane below L, which is the graph of inequality 푦 < 2푥 − 4. A similar discussion holds
for any line L having an equation of the form 퐴푥 + 퐵푦 + 퐶 = 0 푤ℎ푒푟푒 퐵 ≠ 0.

If B= 0, an equation of line L is 퐴푥 + 퐶 = 0, and L is a vertical line whose equation 푥 = 4. Then if
(x, y) is any point in the plane, exactly one of the following statements is true:
The point (x, y) is to the right of the point (4, y) if and only if x >4. Showing the graph of inequality x >4
as the half plane lying to the right of the line x = 4. Similarly, the graph of x < 4 if, and only if the point
(x, y) is to the left of the point (4, y). The discussion can be extended to any line having an equation of
the form Ax +퐶 = 0.
By generalizing the above arguments to any line, we can prove this theorem.
THEOREM
(I) the graph of y> 푚푥 + 푏 is the half plane lying above the line y= 푚푥 + 푏.
(II) the graph of y< 푚푥 + 푏 is the half plane lying below the line y= 푚푥 + 푏.
(III) the graph of (y< 푚푥 + 푏) x> 푎 is the half plane lying to the right of line x= 푎.
(IV) The graph of x< 푎 is the half plane lying to the left of the line x= 푎.
Example 1
Draw a sketch of the graph of the inequality
2푥 − 4푦 + 5 > 0
Solution
The given inequality is equivalent to
−4푦 > −2푥 − 5
푦 >
1
2
+ 5/4
The graph of inequality is the half plane below the line having the equation 푦 = 1/2푥 + 5/4. A sketch of
this graph is the shaded half plane.
A closed half plain is a half plane together with the line bounding it is the graph of an inequality of the
form.
퐴푥 + 퐵푦 + 퐶 = 0 표푟 퐴푥 + 퐵푦 + 퐶 ≤ 0
Illustration
The inequality
4푥 + 5푦 − 20 ≥ 0
Is equivalent to
5푦 ≥ −4푥 + 20

푦 ≥ −4/5푥 + 4
Therefore the graph of this inequality is the closed half plane consisting of the line 푦 = −3/5푥 + 4 and the
half plane above it. A sketch of the graph.
Two intersecting lines divide the points of the plane into four regions. Each of these regions is the
intersection of two half planes and is defined by a system of two linear inequalities.
Exercise
Draw the sketch of the graph of inequality
1. (a) x> 2; (푏)푥 ≤ −3 2. (푎)푦 ≥ −5; (푏)푦 < 6
3. (푎)푥 ≥ 4; (푏)푥 ≤ 7 4. (푎)푦 > −1; (푏)푦 ≤ 8
5. 푦 < 4푥 − 2 6. 푦 ≥ 2푥 − 3
7. 2푥 − 7 ≥ 0 8. 푦 + 8 < 0
9. 5푥 + 6 > 2푦 10. 2푦 − 8푥 + 5 > 0
CHAPTER 8
STATISTICS AND PROBABILITY
APPLICATION IN PROBABILITY
Probability is a way to measure the chances that something will occur in relation to the possible
alternatives. For example, the probability is not a guarantee. A couple might have six children and all are
boys, or they might have six children and all are girls.
Now you might think that a couple with six girls would not expect to have another girl if they decided to
have a seventh child. In fact, the probability that the seventh child is a girl is still ½ , since the gender of this
child is not affected by the gender of the previous six children.
What if you know that a family has seven children and six of them are girls? Is the probability that the
seventh child is a girl still1/2? There are eight possibilities: all seven children are girls, or either the first,
second, third, fourth, fifth, sixth, or seventh child is a boy. Since in only one of these cases the seventh child
is a girl, the probability is 1/8.
14-1 STATISTICS AND LINE PLOTS
Objectives After studying this lesson, you should be able to:
 Interpret numerical data from a table, and
 Display and interpret statistical data on a line plot.
Each day when you read news papers or magazines, watch television, or listen to the radio, you are
bombarded with numerical information about the national economy, sports, politics, and so on.

Interpreting this numerical information, or data, is important to your understanding of the world around
you. A branch of mathematics called statistics helps provide you with methods for collecting, organizing
and interpreting data.
Statistical data can be organized and
presented in numerous ways. One of the most
common ways is to use a table or chart. The
chart at the right shows the hourly wages
earned by the principal wage earner in ten
families.
Using tables or charts like the one at the right should enable you to more easily analyze the given data.
Example 1
Use the information in the chart above to answer each question.
a. What are the maximum and minimum hourly wages of the principal wage earner for the ten families.
The families.
b. What percent of the families have a principal wage earner that makes less than $ 10.00 per hour.
In some instances, statistical data can be presented on a number line. Numerical information displayed on a
number line is called a line plot. For example, the data in the table above can presented in a line plot.
For example 1, you know that the data in the chart range from $ 8.00 per hour to $ 20.25 per hour. In order
to represent each hourly wage on a number line, the scale used must includes these values. A “w” is use to
represent each hourly wage. When more than one “w” s has the same location on the number line, additional
“w” are placed one above the other. A line plot for the hourly wages is shown below.
Exercises
Practice
1. Use the line plot at the right to answer each question.
a. What was the highest score on the test?
b. What was the lowest score on the test?
c. How many students took the test?
d. How many students scored in the 40’s?
e. What score was received by the most students?
Family Hourly wage
A $ 8.00
B $ 10.50
C $ 20.25
D $ 9.40
E $11.00
F $ 13.75
G $ 8.50
H $ 10.50
I $ 9.00
J $ 11.00
20-25 3
25-30 10
30-35 8
35-40 2
40-45 7
45-50 5
50 1

14.2 STEM AND LEAF PLOTS
Objective : After studying this lesson, you should be able to:
 Display and interpret data on a stem-leaf-plot.
Application: Mr.Juaez wants to study the distribution of the scores for a 100-point unit exam given in his first-period
Biology class. The scores of the 35 students in the class are listed below.
82 77 49 84 44 98 93 71 76 65 89 95 78 69 89 64
88 54 96 87 92 80 44 85 93 89 55 62 79 90 86 75
74 99 62
He can organize and display the scores in a compact way using a stem-leaf-plot.
In a stem-leaf-plot, the greatest common place value of the data is used to form the stems. The numbers in the
next common place-value position are then used to form the leaves. In the list above, the greatest place value is
tens. Thus, the number 82 would have stem 8 and leaf 2.
To make the stem-and-leaf-plot, first make
a vertical list of the stems. Since the test
scores range from 44-99, the stems range
from 4-9. Then, plot each number by
placing leaf 2 to the right of the stem 8. The
right.
A second stem-and-leaf-plot can be made
To arrange the leaves in numerical order from
Least to greatest as shown at the right. This
will make it easier for Mr. Juarez to analyze
the data.
Example 1
STEM LEAF
4 9 4 4
5 4 5
6 5 9 4 2 2
7 7 1 6 8 9 5 4
8 2 4 9 9 8 7 0 5 9 6
9 8 3 5 6 2 3 0 9
8/2 Represents a score of 82
STEM LEAF
4 4 4 9
5 4 5
6 2 2 4 5 9
7 1 4 5 6 7 8 9
8 0 2 4 5 6 7 8 9 9 9
9 0 2 3 3 5 6 8 9
Use the information in the stem-and-leaf-plots above to answer each question.
a. What were the highest and lowest scores on the test? 99 and 44
b. Which test scores occurred most frequently? 89 (3times)
c. In which 10-point interval did the most students score?
80-89 (10 students)
d. How many students received a score of 70 or better? 25 students

Sometimes the data for a stem-and-leaf-plot are numbers that have more than two digits. Before plotting these
numbers, they may need to be rounded or truncated to determine each stem and leaf. Suppose you wanted to
plot 356 using the hundreds digit for the stem.
ROUNDED TRUNCATED
Rounded 356 to 360. Thus, you To truncate means to cut off, so
Would plot 356 using stem 3 truncate 356 as 350. Thus, you
And leaf 6. What would be the would plot 356 using stem 3 and leaf
Stem and leaf of 499? 5 and 0 5. What would be the stem and leaf of 499? 4 and 9
A back-to-back stem-and-leaf plot is sometimes used to compare two sets of data or rounded and truncated
values of the same set of data. In a back-to-back plot, the same stem is used for the leaves of both plot.
Exercises
Applications
1. Football The stem-and-leaf plot below gives the number of catches of the NFL’s leading pass receiver for
each season through 1990.
a. What was the greatest number of
catches during the season?
b. What was the least number of
catches during the season?
c. How many seasons are listed?
d. What number of catches occurred
most frequently?
e. How many times did the leading
pass receiver have at least 90
catches?
STEM LEAF
6 0 1 2 6 7
7 1 1 1 2 2 3 3 3 4 5 8
8 0 2 6 8 8 9
9 0 1 2 2 5
10 0 0 1 6
9/2 Represents 92 catches

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