In Exercises 59–94, solve each absolute value inequality.

3 ≤ |2...

Question

In Exercises 59–94, solve each absolute value inequality.

3 ≤ |2x - 1|

Accepted Answer

Hello. Everyone in this video we're going to be looking at this practice problem where we want to simplify the absolute value inequality, absolute value of X plus five greater than or equal to three. And we want to make sure to express the solution set in interval notation. So in this case we want to try to find the values of X that make the inequality true. And in this case you can see that X is inside an absolute value function. So the absolute value function makes whatever numbers inside positive. But in reality that number could be negative or positive. So in order to get a true range of X values that satisfy this inequality, I need to account for that. And I'm going to do that by solving the inequality twice. One for if the number is positive and another if the number is negative. So if my number was positive I would have positive X plus five greater than or equal to three. But if my number was negative I would have negative times X plus five greater than or equal to three. So I'm going to focus on solving for the positive number. And if I do that I want to isolate X. So I'm going to subtract five from both sides and I'm left with X greater than or equal to negative two. So you can see have isolated X. I'm going to do the same for if the number was negative and you can see I have this negative on the outside. So I'm gonna go ahead and divide by the negative and notice I'm dividing by a negative across an inequality. Still the inequality switches directions. So I'm left with X plus five less than or equal to negative three. And again I want to isolate X. I'm going to subtract by five and I'm left with X less than or equal to negative eight. So now that we've simplified both cases where we have X greater than or equal to negative two or X less than negative eight. Now we need to represent these inequalities in interval notation. So if we look at the first one, X greater than or equal to negative two, this is saying that we have to make X greater than negative two. So negative two is my lower limit and I have an equal to sign. So this is represented as a bracket in interval notation. If I had a parentheses, this would be simply the greater than or less than sign without the equal to. And the first inequality -2 is my lower bound and I have no upper limit. It just says X has to be greater than -2. So I can write my upper limit as positive infinity. My second inequality says X less than or equal to negative eight. So I want X to be less than negative eight, meaning that negative eight is my upper bound. It's the highest number that X can be and again I have a less than or equal to. So I can do a bracket to represent this. and I have no lower bound so I can write negative infinity. So these two intervals become the ranges that satisfy the inequality. So I can combine them using a union where I have parentheses negative infinity to negative eight bracket Union with Bracket -2 to positive infinity. And if we look at the answer choices, you can see that this aligns with answer choice D. And notice how all of the answer choices have the same values. And the only difference is the notation that they use. So you can see that D. Is the only option that has brackets On -8 and -2. All the other options have parentheses somewhere. So if you wanted to save time or just make sure that you selected the right answer, you can notice that we have an equal to at the start. So we really only care about answers that have brackets for both numbers. But solving it gets through the same answer anyways. But hopefully this video helped and see you next time.

In Exercises 59–94, solve each absolute value inequality. 3 ≤ |2x - 1|

Key Concepts

Absolute Value

Inequalities

Solving Absolute Value Inequalities

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