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 linear inequalities.
- Solve compound inequalities.
- Solve an applied problem using an inequality.


An inequality is a sentence with `<`, `>`, `<=`, or `>=`.

Multiplication principle for inequalities. When both sides of an inequality are multiplied by a negative number, the inequality sign must be reversed.



Solve the linear inequality, and graph the solution set.

`3x-2 > x+6`

So let's start by adding 2 to both sides of our inequality. We'll get `3x > x+8`. Now subtract both sides by x, so we can have all of our x values on one side of our inequality. This will give us `2x > 8`. Now divide both sides by 2 and we get `x > 4`. The solution in set notation is `{x| x > 4}`. Then in interval notation, we have `(4, oo)`, since parentheses indicate that we don't have an equal sign in our inequality, or that we're not including the value 4.

We can graph the solution set by drawing a number line, labeling 4, then drawing our open parenthesis to the right, and then shading to the right.



Solve the linear inequality and graph the solution set.

`x+5 < 4x-5`

So let's subtract both sides by 5, and we get `x < 4x-10`. Then subtract 4x on both sides, and we get `-3x < -10`. Now we divide both sides by -3. So we get `x > 10/3`. Notice that we flipped the sign here, because we divided by a negative number.

So the solution to our linear inequality in set notation is `{x| x > 10/3}`. And then in interval notation, we have `(10/3, oo)`, since our inequality does not include 10/3.

Now the graph to our solution set is a number line with 10/3 in open parenthesis to the right, shaded to the right.



Find the domain of the function which has a square root.

`h(x) = sqrt(x-8)`

So the domain will include all values of x for which `x-8 >= 0`. Since you can only take the square root of a positive number or zero. So `x-8 >= 0`. If you add 8 to both sides, this gives you `x >= 8`. So the domain in set notation is `{x | x >= 8}`, and then in interval notation, we have `[8, oo)`. Notice the bracket, because our inequality includes the equal signs, or includes the value 8.



Find the domain of the function which has a square root.

`g(x) = x/sqrt(4+x)`

So our x value in the numerator can be any real number, but in the denominator, it's restricted. `4+x >= 0`, since you can only take the square root of a positive number. And then since the radical is in the denominator, the `sqrt(4+x) != 0`, since zero in the denominator would give us an undefined number.

So the domain must follow both of these requirements. So we have to solve the inequality `4+x > 0`. First, we'll subtract 4 from both sides, and this will give us `x > -4`.

In set notation, our domain is `{x| x > -4}`. And then in interval notation, our domain is `(-4, oo)`, with parentheses, because our domain does not include -4.



Solve the compound inequality involving a conjunction and write in interval notation. Then graph the solution set.

`-1 <= x+2 < 5`

So let's subtract 2 on all three sides. We'll get `-3 <= x < 3`. In interval notation, this is written `[-3, 3)`. The bracket indicates that we include -3, and the parenthesis indicates that we don't include 3.

Now let's graph the solution set by drawing a number line, labeling -3 and 3, putting a bracket over -3, and a parenthesis over 3, and shading in between.



Solve the compound inequality involving a disjunction and write in interval notation. Then graph the solution set.

`2x <= -6` or `x-1 > 1`

So for the first portion of our compound inequality, let's divide both sides by 2. So we get `x <= -3`. Then for the second part, add 1 to both sides. So we get `x > 2`.

This can be written in interval notation as `(-oo, -3]` (bracket since we're including -3) union `(2, oo)`.

Now let's graph the solution set. Draw a number line, label -3 and 2, then draw an open parenthesis to the right at 2 and a bracket opening to the left at -3. Now shade to the right of 2 and shade to the left of -3.



Dave's Painters charges $300 plus $50 an hour to paint a home. Joe's Painters charges $80 an hour. For what lengths of time does it cost less to hire Joe's Painters?

So if we let `x = "the amount of time in hours"`, then the amount it costs to hire Dave's Painters is `300 + 50x`. And the amount it costs to hire Joe's Painters is 80x. Then since we want to know what length of time it costs less to hire Joe's Painters, set `80x < 300 + 50x`.

Now solve for x. So if we subtract 50x on both sides, we get `30x < 300`. And then divide both sides by 30, we can cancel the zero, and then 3 goes into 30, 10 times. So `x < 10`.

The cost is less to hire Joe's Painters for less than 10 hours.