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Section 1.5
Linear Equations,
Functions, Zeros, and
Applications
Copyright ©2013, 2009, 2006, 2001 Pearson Education, Inc.
Objectives



Solve linear equations.
Solve applied problems using linear models.
Find zeros of linear functions.
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.
x3
 1,
x4
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.
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.
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.
Example
3
7
x 1 
Solve:
4
5
Start by multiplying both sides of the equation by the
LCD to clear the equation of fractions.
15x  48
7
3

20  x  1  20 
15x 48
4

5

15
15
3
20  x  20 1  28
48
4
x
15
15x  20  28
16
x
15x  20  20  28  20
5
Example (continued)
Check:
3
7
x 1 
4
5
3 16
7

1 ?
4 5
5
12 5

5 5
7 7
TRUE
5 5
16
The solution is
.
5
Example - Special Case
Solve: 24x  7  17  24x
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.
Example - Special Case
Solve:
1
1
3 x   x  3
3
3
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
(–∞, ∞).
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.
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.
The Motion Formula
The distance d traveled by an object moving at rate r in
time t is given by
d = rt.
Example
An airline 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
Suppose that a Dash 8-200 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 8200?
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.
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)
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).
Example (continued)
5. State. About 1.53 hours after the 737-200 has taken
off, it will overtake the Dash 8-200.
Simple-Interest Formula
I = Prt
I = the simple interest ($)
P = the principal ($)
r = the interest rate (%)
t = time (years)
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?
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.
5% loan
Amount
Borrowed
Interest
Rate
Time
Amount of Interest
x
0.05
1
0.05x
0.08
1
0.08(12,000 – x)
8% loan 12,000 – x
Total
12,000
750
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.
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.
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.
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.
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.