College Algebra

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Previous Lessons
Open Chapter Ch. 1: Graphs, Functions, and Models
Lesson #1 Graphing
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Lesson #2 Graphing Functions
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Lesson #3 Slope of Linear Functions
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Lesson #4 Linear Equations and Modeling
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Lesson #5 Linear Equations, Functions, Zeros and Applications
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Lesson #6 Linear Inequalities
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Open Chapter Ch. 2: More on Functions
Lesson #7 Analyzing Functions
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Exam Exam 1
Lesson #8 Algebra for Functions
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Lesson #9 The Composition of Functions
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Lesson #10 Symmetry
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Lesson #11 Transformations
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Lesson #12 Equations of Variation
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Open Chapter Ch. 3: Quadratic Functions and Equations; Inequalities
Lesson #13 Complex Numbers
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Lesson #14 Quadratic Equations, Functions, Zeros and Models
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Lesson #15 Analyzing Graphs of Quadratic Functions
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Lesson #16 Rational and Radical Equations
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Exam Midterm Exam
Lesson #17 Absolute Value Equations and Inequalities
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Open Chapter Ch. 4: Polynomial Functions and Rational Functions
Lesson #18 Polynomial Functions and Models
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Lesson #19 Graphing Polynomial Functions
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Lesson #20 Polynomial Division
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Lesson #21 The Zeros of Polynomial Functions
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Lesson #22 Rational Functions
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Lesson #23 Polynomial and Rational Inequalities
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Open Chapter Ch. 5: Exponential Functions and Logarithmic Functions
Lesson #24 Inverse Functions
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Lesson #25 Exponential Functions and Graphs
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Lesson #26 Logarithmic Functions and Graphs
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Exam Exam 3
Lesson #27 Properties of Logarithmic Functions
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Lesson #28 Solving Exponential Equations and Logarithmic Equations
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Lesson #29 Applications and Models: Growth and Decay; Compound Interest
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Open Chapter Ch. 6: Systems of Equations
Lesson #30 Systems of Equations in Two Variables
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Lesson #31 Systems of Inequalities and Linear Programming
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Lesson #32 Nonlinear Systems of Equations and Inequalities
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Lesson #33 Sequences and Series
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Exam Final Exam

Assignments:

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Lesson Objectives:

- Solve absolute value equation.
- Solve absolute value inequality.


The absolute value of a number is its distance from 0 on a number line.

So for any positive value a, the `abs(x) = a` means that `x = a` or `x = -a`.



Solve `abs(x-1) + 4 = 10`.

So start by subtracting 4 from both sides. This gives us `abs(x-1) = 6`.

So `x-1 = 6` or `x-1 = -6`. So if we add 1 to both sides, we have `x = 7` or `x = -5`.

So let's check `x = 7`. `abs(7-1)+4 = 10`. That means `6+4 = 10` or `10 = 10`. So, this is a true statement.

Now let's check `x = -5`. We get `abs(-5-1)+4 = 10`, which is `6+4 = 10` or `10 = 10`. Another true statement.

So the solutions are 7 and -5.



Absolute Value Inequalities.

If `a > 0` and `abs(x) < a`, then `-a < x < a`.
If `a > 0` and `abs(x) > a`, then `x < -a` or `x > a`.
Similarly, if `a > 0` and `abs(x) <= a`, then `-a <= x <= a`.
And if `a > 0` and `abs(x) >= a`, then `x <= -a` or `x >= a`.



Solve and give your solution set in interval notation and then graph the solution set.

`abs(x+12) >= 15`. This means that `x+12 >= 15` or `x+12 <= -15`. So if we subtract 12 from both sides, we get `x >= 3` or if we subtract 12 over here, we get `x <= -27`.

We can graph the solution set by having a closed circle at -27 and a closed circle at 3. And `x <= -27`, so we shade to the left, and `>= +3`, so we shade to the right.

In interval notation, this is `(-oo, -27] cup [3, oo)`.