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Chapter 2
Section 1
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
2.1
1
2
3
The Addition Property of Equality
Identify linear equations.
Use the addition property of equality.
Simplify and then use the addition property of
equality.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Objective 1
Identify linear equations.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2.1 - 3
Identify linear equations.
A linear equation in one variable can be written in the form
Ax  B  C,
for real numbers A, B, and C, with A ≠ 0.
4 x  9  0,
2x  3  5 , and
x7
Linear Equations
are linear equations in one variable (x). The final two can be
written in the specified form with the use of properties developed
in this chapter.
x  2 x  5,
2
2 x  6  0, and
1
6
x
Nonlinear Equations
are not linear equations.
Although x and y are typically used, other letters can be used
for variables in equations.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2.1 - 4
Objective 2
Use the addition property of
equality.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2.1 - 5
Use the addition property of equality.
To solve an equation, add the same number to each side. The
addition property of equality justifies this step.
If A, B, and C are real numbers, then the equations
A  B and
AC  B C
are equivalent equations.
That is, we can add the same number to each side of an
equation without changing the solution.
Equations can be thought of in terms of a
balance. Thus, adding the same quantity to
each side does not affect the balance.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2.1 - 6
EXAMPLE 1
Using the Addition Property of
Equality
Solve x 12  3.
Solution:
Check:
x 12 12  3 12
x 9
x 12  3
9 12  3
3  3
The solution set is 9 .
Do NOT write the solution set as {x = 9}. This is incorrect notation.
Simply write {9}.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2.1 - 7
EXAMPLE 2
Using the Addition Property of
Equality
Solve m  4.1  6.3.
Solution:
m  4.1  4.1  6.3  4.1
m  2.2
Check:
m  4.1  6.3
2.2  4.1  6.3
6.3  6.3
The solution set is {2.2}.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2.1 - 8
Use the addition property of equality. (cont’d)
The addition property of equality says that the same number
may be added to each side of an equation.
In Section 1.5, subtraction was defined as addition of the
opposite. Thus, we can also use the following rule when solving
an equation.
The same number may be subtracted from each side of an
equation without changing the solution.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2.1 - 9
EXAMPLE 3
Using the Addition Property of
Equality
Solve 22  x  16.
Solution:
22 16  x 16 16
38  x
Check:
22  x  16
22  38 16
22  22
The solution set is {38}.
The final line of the check does not give the solution to the
problem, only a confirmation that the solution found is correct.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2.1 - 10
EXAMPLE 4
Subtracting a Variable
Expression
7
9
Solve m  1  m.
2
2
Solution:
7
7
9
7
m 1 m  m  m
2
2
2
2
1 m
Check:
7
9
m 1  m
2
2
7
2 9
1   1
2
2 2
9 9

2 2
The solution set is {1} .
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2.1 - 11
Objective 3
Simplify and then use the addition
property of equality.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2.1 - 12
EXAMPLE 5
Simplifying an Equation before
Solving
Solve 9r  4r  6  2  9r  4  3r.
Solution:
13r  4  12r  4
13r  4 12r  4  12r  4 12r  4
r 0
Check: 9r  4r  6  2  9r  4  3r
9(0)  4(0)  4  9(0)  4  3(0)
44
The solution set is {0}.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2.1 - 13
EXAMPLE 6
Using the Distributive Property
to Simplify an Equation
Solve 4  x  1   3x  5  1 .
4x  4  3x  5  1
Solution:
x 1  1  1  1
x2
Check:
4  x  1   3x  5  1
4(2  1)  (3  2  5)  1
4  3   6  5  1
12  11  1
11
The solution set is {2}.
Be careful to apply the distributive property correctly, or a sign
error may result.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2.1 - 14