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 rational equations.
- Solve radical equations.


Solve rational equations by multiplying both sides of the equation by the LCD, or least common denominator. This will get rid of all of the fractions.



`1/(5x+15)-1/(x^2-9) = 5/(x-3)`. So if we start by factoring all of our denominators, we have `1/(5(x+3))-1/((x-3)(x+3)) = 5/(x-3)`. Now we need to multiply both sides by our LCD. `5(x+3)(x-3)`. So now we can start canceling. The `x+3`'s and the 5's cancel on the first fraction, and then the `(x-3)(x+3)`'s' cancel on our second fraction, and the `(x-3)`'s cancel in our third fraction.

So that we're left with `x-3-5 = 5*5(x+3)`, so that we have `25(x+3)`, which is `25x+75` on the right and `x-8` on the left. So if we subtract x from both sides, we get 24x on the right, and if we subtract 75 from both sides, this gives us -83.

This means that `x = -83/24` is a possible solution. You have to plug back into our x-values in our equation to see if it is a solution.

So, `1/(5(-83/24)+15)-1/((-83/24)^2-9) = 5(-83/24-3)`. This gives us `-24/31` on the left, and `-24/31` on the right. `-83/24` checks, so it is a solution.



The Principle of Powers:

If n is positive and `a = b`, then `a^n = b^n`.



Solve the following equation for x.

We're given `sqrt(x)+sqrt(4+x) = 3`. So using the Principle of Powers, we can go ahead and square both sides. This will give us `sqrt(x)^2+2(sqrt(x))(sqrt(4+x))+sqrt(4+x)^2 = 9`.

This is `x+2sqrt(x(4+x))+4+x = 9`. So `x+2sqrt(4x+x^2)+4+x = 9`. Now let's combine our x's and subtract 4 on both sides, so that we get `2x + 2sqrt(4x+x^2) = 5`. And then subtract 2x from both sides, and we get `2sqrt(4x+x^2) = 5-2x`. And then by the principle of powers, we can square both sides again and get `4(4x+x^2) = 25-2(5)(2x)+4x^2`. This gives us `16x+4x^2 = 25-20x+4x^2`. And now the `4x^2`'s cancel on both sides, and we can add 20x to both sides, so that we get `36x = 25`.

`x = 25/36`. So now we need to plug `25/36` back in, to make sure it satisfies our equation. So we have `sqrt(25/36)+sqrt(4+25/36) = 3`. So that gives us `5/6+sqrt(169/36) = 3`, or `5/6+13/6 = 3`. That's `18/6 = 3`, or `3 = 3`, which is a true statement.

So that means that `x = 25/36` is a solution.