chapter 2 section 1 copyright © 2008 pearson education, inc. publishing as pearson addison-wesley
TRANSCRIPT
Chapter Chapter 22Section Section 11
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
The Addition Property of Equality
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2.12.12.12.1Identify 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 11
Slide 2.1 - 3
Identify linear equations.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Identify linear equations.
Slide 2.1 - 4
A linear equation in one variable can be written in the form
,
for real numbers A, B, and C, with A ≠ 0.
, , and 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.
Ax B C
4 9 0x 2 3 5x 7x
, , and
are not linear equations.
2 2 5x x 1
6x
2 6 0x
Linear Equations
Nonlinear 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
Objective 22
Slide 2.1 - 5
Use the addition property of equality.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
If A, B, and C are real numbers, then the equations
and are equivalent equations.
That is, we can add the same number to each side of an equation without changing the solution.
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.
Slide 2.1 - 6
A B A C B C
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
EXAMPLE 1
Slide 2.1 - 7
Solution:
Using the Addition Property of Equality
12 3x
122 3 121x 9x
9The solution set is .
Do NOT write the solution set as {x = 9}. This is incorrect notation. Simply write {9}.
Check:
12 3x 9 12 3
3 3
Solve .
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 2
Slide 2.1 - 8Slide 2.1 - 8
Solution:
Using the Addition Property of Equality
4.14 . 13 4.1 .6m 2.2m
The solution set is .{ 2.2}
4.1 6.3m
Check: 4.1 6.3m 4.1 6.2 3.2 6.3 6.3
Solve .
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
The addition property of equality says that the same number may be added to each side of an equation.
Slide 2.1 - 9
Use the addition property of equality. (cont’d)
The same number may be subtracted from each side of an equation without changing the solution.
In Section 1.5, subtraction was defined as addition of the opposite. Thus, we can also use the following rule when solving an equation.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 3
Slide 2.1 - 10
Solution:
Using the Addition Property of Equality
Solve .22 16x
22 116 166x
38 x
22 16x Check:
3822 16 22 22
The final line of the check does not give the solution to the problem, only a confirmation that the solution found is correct.
The solution set is .{ 38}
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 4
Slide 2.1 - 11
Solution:
Subtracting a Variable Expression
Solve .7 9
12 2
m m
1 m
7 91
2 2m m
Check:
7 2 9
2 2 21 1
9 9
2 2
The solution set is .{1}
7 91
7
22
7
22mm mm
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Objective 33
Slide 2.1 - 12
Simplify and then use the addition property of equality.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 5
Slide 2.1 - 13
Solution:
Simplifying an Equation before Solving
Solve .
113 4 12 12 424 4r rr r
9( ) 4( ) 4 9(0 0 0 0) 4 3( ) Check:
4 4
The solution set is .{0}
13 4 12 4r r
0r
9 4 6 2 9 4 3r r r r
9 4 6 2 9 4 3r r r r
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 6
Slide 2.1 - 14
Solution:
Using the Distributive Property to Simplify an Equation
Check:
The solution set is .{2}
4 4 3 5 1x x 1 11 1x
2x
4( 1) ( ) 12 3 2 5 4 3 6 5 1
12 11 1
Solve . 4 1 3 5 1x x
4 1 3 5 1x x
1 1
Be careful to apply the distributive property correctly, or a sign error may result.