Solving absolute value inequalities 1

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About this video

Sal solves an inequality with |h| by first adding 19 1/2 to both sides to get |h| < 7 1/2. Thinking of absolute value as distance from 0, he finds that h must be between -7 1/2 and 7 1/2, and graphs it with open circles at both ends.

You'll be able to

  • isolate |h| in an inequality.
  • interpret |h| < 7 1/2 as h being less than 7 1/2 away from 0.
  • graph the solution to an absolute value inequality, using open circles for strict inequalities.

Key words

inequality
A statement that compares two amounts with <, >, <= or >= instead of =.
absolute value
The distance of a number from 0.
open circle
A circle on a number line showing an endpoint that is not included.
Transcript

We're told to graph all possible values for h on the number line. And this is a especially interesting inequality because we also have an absolute value here. So the way we're going to do it, we're going to solve this inequality in terms of the absolute value of h, and from there we can solve it for h.

So let's just get the absolute value of h on one side of the equation. So the easiest way to do this is to add 19 and 1/2 to both sides of this equation. I often like putting that as an improper fraction, but 1/2 is pretty easy to deal with, so let's add 19 and 1/2 to both sides of this inequality.

Did I just say equation? It's an inequality, not an equation, it's an inequality sign, not an equal sign. So plus 19 and 1/2. On the left-hand side, these guys obviously cancel out, that was the whole point, and we are left with the absolute value of h on the left-hand side is less than.

And then if we have 19 and 1/2, essentially minus 12, 19 minus 12 is 7, so it's going to be 7 and 1/2. So now we have that the absolute value of h is less than 7 and 1/2. So what does this tell us? This means that the distance, another way to interpret this-- remember, absolute value is the same thing as distance from 0-- so another way to interpret this statement is that the distance from h to 0 has to be less than 7 and 1/2.

So what values of h are going to have less than a distance from 7 and 1/2? Well, it could be less than 7 and 1/2 and greater than 0, or equal to 0. So let me put it this way. So h could be less than 7 and 1/2. But if it gets too far negative, if it goes to negative 3, we're cool, negative 4, negative 5, negative 6, negative 7, we're still cool, but then at negative 8, all of a sudden the absolute value isn't going to be less than this.

So it also has to be greater than negative 7 and 1/2. If you give me any number in this interval, its absolute value is going to be less than 7 and 1/2 because all of these numbers are less than 7 and 1/2 away from 0. Let me draw it on the number line, which they want us to do anyway.

So if this is the number line right there, that is 0, and we draw some points, let's say that this is 7, that is 8, that is negative 7, that is negative 8. What numbers are less than 7 and 1/2 away from 0? Well, you have everything all the way up to-- 7 and 1/2 is exactly 7 and 1/2 away, so you can't count that, so 7 and 1/2, you'll put a circle around it.

Same thing true for negative 7 and 1/2, the absolute value, it's exactly 7 and 1/2 away. We have to be less than 7 and 1/2 away, so neither of those points are going to be included, positive 7 and 1/2 or negative 7 and 1/2. Now, everything in between is less than 7 and 1/2 away from 0, so everything else counts.

Everything outside of it is clearly more than 7 and 1/2 away from 0. And we're done.

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