Transcript Section 1.5
1.5 Linear Equations, Functions, Zeros, and Applications
Solve linear equations.
Solve applied problems using linear models.
Find zeros of linear functions.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley
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 2
x
3 5, 3
x
1 and
x
2 3
x
2 0.
4
x
5,
x x
3 4 1,
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Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley
Linear Equations
A
linear equation in one variable
is an equation that can be expressed in the form
mx m
and
b
are real numbers and
m
≠ 0.
+
b
= 0, where Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley
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Equivalent Equations
Equations that have the same solution set are
equivalent equations
.
For example, 2
x
+ 3 = 5 and
x
= 1 are equivalent equations because 1 is the solution of each equation.
On the other hand,
x
2 – 3
x
+ 2 = 0 and
x
= 1 are not equivalent equations because 1 and 2 are both solutions of
x
2 – 3
x
+ 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
Solve: 3
x
1 4 7 5
Solution:
Start by multiplying both sides of the equation by the LCD to clear the equation of fractions.
20 3 4
x
1 20 7 5 20 3 4
x
20 1 28 15
x
20 28 15
x
20 20 28 20 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley 15
x
48 15 15
x
48
x
15 48 15
x
16 5
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Example
(continued)
Check
: 3
x
1 4 7 5 3 4 16 5 1 ?
12 5 5 5 7 5 7 5 7 5 TRUE The solution is 16 .
5 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley
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Example - Special Case
Some equations have
no
solution.
Solve: 24
x
7 17 24
x
Solution:
24
x
7 17 24 24
x
24
x
7 24
x
17 24
x
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.
Solve: 3 1 3
x
Solution:
1 3
x
3 1 3
x
1 3
x
3 1 3
x
1 3
x
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 Assign variables Find further information Organize into a chart or table 2.
Write a list Guess or estimate the answer
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 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 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
737-200 Dash 8-200 Distance
d d
Rate 500 302 Time
t t
+ 1 2.
Translate.
Using the formula
d
expressions for
d
: =
rt
, we get two
d
= 500
t
and
d
= 302(
t
+ 1).
Since the distance are the same, the equation is: 500
t
= 302(
t
+ 1)
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Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley
Example
(continued) 3.
Carry out.
We solve the equation.
500
t
= 302(
t
+ 1) 500
t
= 302
t
+ 302 198
t t
= 302 = 302/198 ≈ 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) Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley
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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.
5% loan Amount Borrowed Interest Rate Time
x
0.05
1 Amount of Interest 0.05
x
8% loan 12,000 –
x
0.08
Total 12,000 1 0.08(12,000 –
x
) 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.
0.05
x
+ 0.08(12,000
x
) = 750 3
. Carry out
. We solve the equation.
0.05
x
+ 0.08(12,000
x
) = 750 0.05
x
+ 960 0.03
x
0.08
+ 960 = 750 0.03
x x
= 750 = 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
) = 5
x
9.
Algebraic Solution
: 5
x
9 = 0 5
x
= 9
x
= 1.8
Visualizing the Solution
: The intercept of the graph is (9/5, 0) or (1.8, 0).
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