In advanced algebra, one of the concepts that you must learn is absolute value inequalities. Watch the video below to begin learning this important topic.
Absolute Value Inequalities
This is a relatively rare topic from the test. So it doesn’t often appear but may appear especially on one of the hard problems. If you haven’t seen the introductory material on absolute value in the lesson “Positive and Negative Numbers II” in the arithmetic and fraction module, I strongly suggest watching that lesson before you watch this one.
So from that lesson, recall the distance definition of absolute value. Absolute value is the distance of x from zero. And of course distance is always positive, which is why the absolute value is always positive. The absolute value of x- 5 is the distance of x from 5. Absolute value of x + 3 is the distance of x from -3.
This way of thinking about absolute value is enormously helpful in understanding inequalities that involve absolute values.
Absolute Value Inequalities: Practice Question One
Here’s a practice question, pause the video and then we’ll talk about this.
Okay, express the absolute value of x- 7 is less than or equal to -3 as an ordinary inequality, an equality not involving the absolute value. Well, we start off from 7 on the number line, and we can go a distance of 3 from that starting point.
We can go a distance of 3 in either direction. So when we go down 3 we get to 4, when we go up 3 we get to 10, those are the two endpoints. Because it’s less than or equal to, that means that the endpoints are included. So x can be 4, and x can be 10. Those are the endpoints, and so everything in between them is allowed.
And so this would just be 4 is less than or equal to x is less than or equal to 10. So that re-expresses that absolute value inequality as an ordinary inequality.
Practice Problem Two
Here is another practice question. Pause the video and then we’ll talk about this.
Express the region on the number line as an absolute value inequality. So as an ordinary inequality that would be easy. As an ordinary inequality this would just x is greater than 20 and less than 90. But that’s how we’re being asked. We’re not asked for an ordinary inequality when an absolute value inequality.
The first thing we need to do is find the center of that region. The middle of the region is the average of the endpoints, 20 and 90, that middle is 55.
So 55 is the center. Now the question is how far do we move away from that center? Well, 90 is 55 + 35, 20 is 55- 35. So we start at that center 55, and from 55 we can go a distance of 35 in both directions.
Notice also that the endpoints are not included, so we cannot go as far as 35. We have to go a distance that is < 35. So in other words, the distance from 55 is < 35. And that’s the absolute value in equality that exactly expresses that particular region.
Practice Problem Three
Here’s another practice question, pause the video and then we’ll talk about this.
Okay, once again the first thing we have to do is find the middle of that region, we find the middle of the region by averaging. So negative 3 plus positive 11 is positive 8 divided by 2 is 4. That’s the middle point. So if the midpoint is 4, how far do we go in either direction? Notice that x can go as far as 7 above 4, which is 11, or 7 below 4, which is -3.
Notice also that those endpoints are included, so the distance can be less than 7, the distance can also be equal to 7. It can go a distance of 7 away from 4, and that’s still included. So the distance from 4 is less than or equal to 7, and that’s the absolute inequality expression of that same region. We demonstrated how the distance understanding of the absolute value makes inequalities involving absolute values very easy to solve.