Just as we saw happen with linear equations, sometimes linear inequalities will also have fractions, variables on both sides of the symbol, or maybe even both at the same time, but we're actually going to address them the same way we did with linear equations by getting rid of those fractions using the least common denominator and then getting all of our variables to one side. So let's go ahead and take a look at an example. So here I have 14 × x + 2 ≥ 112 - 13 x. Now the very first thing we want to do is get rid of those fractions, which we can do by multiplying the entire inequality by the least common denominator. Now I have 3 denominators here. I have 4, 12, and 3. So my least common denominator is going to be 12. Let's multiply our entire inequality by 12. Remember that whenever we're multiplying by our least common denominator, we need to make sure to carry that into every single term in our inequality. Let's expand this out. This becomes 12 ÷ 4 × x + 2 ≥ - 12 ÷ 12 + - 12 ÷ 3 × x. Let's simplify. 12 ÷ 4 = 3, so this becomes 3 × (x + 2) ≥ - 1 - 4 × x. Now let's distribute the 3. This is going to become 3x + 6 ≥ - 1 - 4x. Since we have variables on both sides, we need to move all of our variables to one side, all of our constants to the other, the same way we did with linear equations. So I want to move this 4x to my left side and move my - 6 to my right side. And I can do this by adding 4x to both sides and subtracting 6 from both sides. 3x + 4x = 7x and - 1 - 6 = - 7. Isolate x by dividing both sides by 7. Now I am simply left with x ≥ - 1, and this is my solution. But remember, we want to express our solution set in interval notation and graph it. So let's graph it first so that we can better visualize this solution. My solution is x ≥ - 1. I know that since it's greater than or equal to, I need to use a solid circle, a closed circle, for my endpoint of negative one. And since it's anything that's greater than or equal to negative 1, I'm just going to use an arrow to indicate that it goes all the way to infinity. Now that I can visualize it better, what is this in interval notation? Well, interval notation for this, since I know that the negative one endpoint is included, I'm going to use a square bracket to enclose that endpoint of negative one, and then that goes all the way to anything greater than negative one which is just infinity, which I always enclose with a parenthesis. And this is my final answer and my final graph. Let me know if you have any questions.
Table of contents
- 0. Review of Algebra4h 16m
- 1. Equations & Inequalities3h 18m
- 2. Graphs of Equations43m
- 3. Functions2h 17m
- 4. Polynomial Functions1h 44m
- 5. Rational Functions1h 23m
- 6. Exponential & Logarithmic Functions2h 28m
- 7. Systems of Equations & Matrices4h 6m
- 8. Conic Sections2h 23m
- 9. Sequences, Series, & Induction1h 19m
- 10. Combinatorics & Probability1h 45m
1. Equations & Inequalities
Linear Inequalities
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Textbook Question
In Exercises 59–94, solve each absolute value inequality.
5|2x + 1| - 3 ≥ 9
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Textbook Question
In Exercises 59–94, solve each absolute value inequality.
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Linear Inequalities practice set
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