CHAPTER 1:
Graphs, Functions,
and Models
1.1
1.2
1.3
1.4
1.5
Introduction to Graphing
Functions and Graphs
Linear Functions, Slope, and Applications
Equations of Lines and Modeling
Linear Equations, Functions, Zeros and
Applications
1.6 Solving Linear Inequalities
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1.5
Linear Equations, Functions, Zeros,
and Applications




Solve linear equations.
Solve applied problems using linear models.
Find zeros of linear functions.
Solve a formula for a given variable.
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Equations and Solutions
An equation is a statement that two expressions are
equal.
To solve an equation in one variable is to find all the
values of the variable that make the equation true.
Each of these numbers is a solution of the equation.
The set of all solutions of an equation is its solution
set.
Some examples of equations in one variable are
2x  3  5,
and
3x  1  4x  5,
x 2  3x  2  0.
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x3
 1,
x4
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Linear Equations
A linear equation in one variable is an equation
that can be expressed in the form mx + b = 0, where
m and b are real numbers and m ≠ 0.
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Equivalent Equations
Equations that have the same solution set are
equivalent equations.
For example, 2x + 3 = 5 and x = 1 are equivalent
equations because 1 is the solution of each equation.
On the other hand, x2 – 3x + 2 = 0 and x = 1 are not
equivalent equations because 1 and 2 are both
solutions of x2 – 3x + 2 = 0 but 2 is not a solution of
x = 1.
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Equation-Solving Principles
For any real numbers a, b, and c:
The Addition Principle:
If a = b is true, then a + c = b + c is true.
The Multiplication Principle:
If a = b is true, then ac = bc is true.
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Example
3
7
x 1 
Solve:
4
5
Solution: Start by multiplying both sides of the
equation by the LCD to clear the equation of fractions.
7
3

20  x  1  20 
4

5
3
20  x  20 1  28
4
15x  20  28
15x  20  20  28  20
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15x  48
15x 48

15 15
48
x
15
16
x
5
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Example (continued)
Check:
3
7
x 1 
4
5
3 16
 1 ?
4 5
12 5

5 5
7
5
7
5
7
5
TRUE
16
The solution is
.
5
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Example - Special Case
Some equations have no solution.
Solve: 24x  7  17  24x
Solution:
24x  7  17  24
24x  24x  7  24x 17  24x
7  17
No matter what number we substitute for x, we get a
false sentence.
Thus, the equation has no solution.
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Example - Special Case
There are some equations for which any real number
is a solution.
1
1
Solve: 3  x   x  3
3
3
Solution:
1
1
1
1
x  3 x  x  x  3
3
3
3
3
3 3
Replacing x with any real number gives a true sentence.
Thus any real number is a solution. The equation has
infinitely many solutions. The solution set is the set of
real numbers, {x | x is a real number}, or (–∞, ∞).
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Applications Using Linear Models
Mathematical techniques are used to answer questions
arising from real-world situations.
Linear equations and functions model many of these
situations.
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Five Steps for Problem Solving
1. Familiarize yourself with the problem situation.
Make a drawing
Find further information
Assign variables
Organize into a chart or table
Write a list
Guess or estimate the answer
2. Translate to mathematical language or symbolism.
3. Carry out some type of mathematical manipulation.
4. Check to see whether your possible solution actually
fits the problem situation.
5. State the answer clearly using a complete sentence.
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The Motion Formula
The distance d traveled by an object moving at
rate r in time t is given by
d = rt.
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Example
America West Airlines’ fleet includes Boeing
737-200’s,each with a cruising speed of 500 mph,
and Bombardier deHavilland Dash 8-200’s, each
with a cruising speed of 302 mph (Source:
America West Airlines). Suppose that a Dash 8200 takes off and travels at its cruising speed. One
hour later, a 737-200 takes off and follows the
same route, traveling at its cruising speed. How
long will it take the 737-200 to overtake the Dash
8-200?
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Example (continued)
1. Familiarize. Make a drawing showing both the known and
unknown information. Let t = the time, in hours, that the
737-200 travels before it overtakes the Dash 8-200.
Therefore, the Dash 8-200 will travel t + 1 hours before
being overtaken. The planes will travel the same distance,
d.
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Example (continued)
We can organize the information in a table as follows.
d
=
r
•
t
Distance
Rate
Time
737-200
d
500
t
Dash 8-200
d
302
t+1
2. Translate. Using the formula d = rt , we get two
expressions for d:
d = 500t and d = 302(t + 1).
Since the distance are the same, the equation is:
500t = 302(t + 1)
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Example (continued)
3. Carry out. We solve the equation.
500t = 302(t + 1)
500t = 302t + 302
198t = 302
t ≈ 1.53
4. Check. If the 737-200 travels about 1.53 hours, it
travels about 500(1.53) ≈ 765 mi; and the Dash 8-200
travels about 1.53 + 1, or 2.53 hours and travels about
302(2.53) ≈ 764.06 mi, the answer checks.
(Remember that we rounded the value of t).
5. State. About 1.53 hours after the 737-200 has taken
off, it will overtake the Dash 8-200.
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Simple-Interest Formula
I = Prt
I = the simple interest ($)
P = the principal ($)
r = the interest rate (%)
t = time (years)
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Example
Jared’s two student loans total $12,000. One loan is
at 5% simple interest and the other is at 8%. After 1
year, Jared owes $750 in interest. What is the
amount of each loan?
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Example (continued)
Solution:
1. Familiarize. We let x = the amount borrowed at 5%
interest. Then the remainder is $12,000 – x,
borrowed at 8% interest.
Amount Interest
Borrowed
Rate
5% loan
x
8% loan 12,000 – x
Total
Time
Amount of Interest
0.05
1
0.05x
0.08
1
0.08(12,000 – x)
12,000
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750
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Example (continued)
2. Translate. The total amount of interest on the two
loans is $750. Thus we write the following equation.
0.05x + 0.08(12,000  x) = 750
3. Carry out. We solve the equation.
0.05x + 0.08(12,000  x) = 750
0.05x + 960  0.08x = 750
 0.03x + 960 = 750
0.03x = 210
x = 7000
If x = 7000, then 12,000  7000 = 5000.
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Example (continued)
4. Check. The interest on $7000 at 5% for 1 yr is
$7000(0.05)(1), or $350. The interest on $5000 at
8% for 1 yr is $5000(0.08)(1) or $400. Since $350 +
$400 = $750, the answer checks.
5. State. Jared borrowed $7000 at 5% interest and
$5000 at 8% interest.
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Zeros of Linear Functions
An input c of a function f is called a zero of the
function, if the output for the function is 0 when the
input is c. That is, c is a zero of f if f (c) = 0.
A linear function f (x) = mx + b, with m  0, has
exactly one zero.
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Example
Find the zero of f (x) = 5x  9.
Algebraic Solution:
5x  9 = 0
5x = 9
x = 1.8
Using a table in ASK mode
we can check the solution.
Enter y = 5x – 9 into the
equation editor, then enter
x = 1.8 into the table and it
yields y = 0. That means
the zero is 1.8.
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Example (continued)
Find the zero of f (x) = 5x  9.
Graphical Solution
or Zero Method:
Graph y = 5x – 9.
Use the ZERO feature
from the CALC
menu.
The x-intercept is 1.8,
so the zero of
f (x) = 5x – 9 is 1.8.
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Formulas
A formula is an equation that can be used to model a
situation.
We have used the motion formula, d = r • t.
We have used the simple-interest formula, I = Prt.
We can use the equation-solving principles to solve a
formula for a given variable.
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Example
Solve P = 2l + 2w for l.
Solution:
P  2l  2w
P  2w  2l
P  2w
l
2
P  2w
The formula l 
can be used to determine a
2
rectangle’s length if we know its perimeter and width.
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