In Exercises 59–94, solve each absolute value inequality.5|2x + 1...

Question

In Exercises 59–94, solve each absolute value inequality.5|2x + 1| - 3 ≥ 9

Accepted Answer

Hello everyone in this video we're going to be looking at this practice problem where we want to simplify the inequality five times the absolute value of X plus one minus two greater than or equal to three. And when we find the solution set for this inequality, we want to make sure to express that in interval notation. So already, just by looking at the answer choices, you can see that there's answer choices with parentheses or brackets and that's really the only thing that changes in between the answer choices. So I'm going to keep that in mind as I solve this problem. So already starting off I can see that I have an equal to sign. So I know that that indicates brackets. So I could start to eliminate answers that have parentheses. For example we have answer choice A as a parentheses there, answer choice B has a parentheses next to the two and then answer choice D. Has double parentheses. So C really is the only one that makes sense but let's go ahead and solve it to make sure. So in this case I'm dealing with an absolute value inequality and that means that the number inside could be positive or negative. So in order to account for that I'm going to solve this inequality twice one if the number was positive And the 2nd 1, if the number was negative. So if I write the inequality as if the number was positive I would have five times X plus one minus two greater than or equal to three. And if I wrote it as if the number was negative, I would have five negative X plus one -2 greater than or equal to three. So let's go ahead and just look at if the number was positive first. So in this case I want to isolate X. I'm going to add to and I'll do the multiplication. So I have five times X. Is five X. Five times positive, one is positive five and I added two. So these cancel out and my right hand side becomes five. So I'm left with five X plus five greater than or equal to +55. Subtract by five. I'm left with five X. Greater than or equal to zero. I divide by five I have X greater than or equal to zero. So you can see I've simplified if the number was positive as much as I can. So now I'm going to go ahead and simplify if the number was negative and see what inequality I get. So I'm gonna do the same add to and we'll keep this five on the outside but I have this negative now. So if I distribute it I have negative X and negative one. I added two. So this cancels out and I'm left with greater than or equal to five And I'm gonna complete this multiplication. So I have negative five x and five times negative one is negative five greater than or equal to five. And if I continue simplifying I have negative five x greater than or equal to 10. And if I divide this negative five across inequality recall that I'm dividing by negative across the inequality. So my inequality switches direction. So instead of greater than I'm left with less than and my inequality becomes X Less than or equal to -2. So you can see that I have these two simplified inequalities and these represent the ranges that allow for the original inequality five absolute value of X plus one minus two, greater than equal to three to be true. So these inequalities represent my solution set. So now let's write them in interval notation. So my first range is saying that X has to be greater than zero and it says greater than or equal to. So I know I have a bracket and zero is my lower bound in this case because I want X to be greater than zero. And since we're given or not given any upper bound, I'm just going to say positive infinity is my upper bound. And now we can look at the second inequality is saying X less than or equal to negative two. So in this case I have the upper limit of negative two. I noticed that I'm using a bracket because it's still an equal to and in this case I want X to be lower and there's no lower bound. So I'm going to write negative infinity. So these become the two ranges that represent the solution set for the inequality. And if I combine them, I have from negative infinity to negative two, union with 0 to positive infinity. And notice how, as we discussed earlier, these remain brackets and remember that goes down to the original inequality where we have an equal to here. So our answer is, in fact, see hopefully this video helped and see you next time.

Struggling with College Algebra?

In Exercises 59–94, solve each absolute value inequality.5|2x + 1| - 3 ≥ 9

Watch next

Linear Inequalities practice set