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Video: Solving Equations with the Absolute Value of an Unknown

Tim Burnham

This video explains how to solve equations that involve finding the absolute value of a variable or the simple linear expression involving that variable but do not also include that variable on the other side of the equation (e.g., |𝑥 + 2| = 7).

10:03

Video Transcript

In this video, we’ll be looking at some absolute value linear equations and solving them. Taking the absolute value means just taking the positive version of the number. So if your expression gave you a value of negative seven for example, then you’d ignore the negative and just use the seven. In fact, you’d take the negative of the negative seven to make it a positive seven.

Now we’re gonna limit ourselves to examples that take the absolute value of an unknown variable, but don’t also include that variable elsewhere in the equation. For example, the absolute value of 𝑥 is equal to five or the absolute value of 𝑥 plus two is equal to seven. But not equations like the absolute value of 𝑥 plus three is equal to 𝑥 plus five, which have the unknown 𝑥 both inside and outside the absolute value.

Right then. Let’s get on and look at some examples.

Number one: Solve the absolute value of 𝑥 is equal to five.

Well we’ve got two possible scenarios. Either 𝑥 is greater than or equal to zero, in which case the value of 𝑥 is already positive so the absolute value of 𝑥 is just gonna be 𝑥. So we could just write the left-hand side in this case as 𝑥. And so, the equation will be 𝑥 is equal to five.

Now the other possibility is that the value of 𝑥 is less than zero. Now to turn that into a positive number we have to take the negative of that negative number. And so we’d say the negative of 𝑥 is equal to five. Now because it’s such a simple expression, 𝑥, actually in this case we don’t need the parenthesis. And the other thing is, if 𝑥 was actually equal to zero, the negative of zero is also zero. So this would work for zero as well.

So what we’re saying there is, if the value inside this absolute sign here is equal to zero, then it doesn’t actually matter which branch we go down. We’re gonna get the same answer. So in this simple case, we’ve already got our answer on the left-hand side: 𝑥 is equal to five. That’s the solution. And on the other side, we’ve got the negative of 𝑥 is equal to five. And so if I multiply both sides of that equation by negative one, on the left I’m gonna just get 𝑥 and on the right I’m gonna get negative five. So 𝑥 is equal to negative five is another solution.

And just from a common sense point of view, this does seem to make sense as well. The question said: The absolute value of 𝑥 is equal to five. So we’re saying what values of 𝑥, when I take the absolute value of them, give me an answer of five. Well if 𝑥 was five, the absolute value of five is five. And if 𝑥 was negative five, the absolute value of negative five is also five. So we’re pretty sure we’ve got the right answer.

So number two: Solve the absolute value of 𝑥 plus two is equal to seven.

Now if the value of 𝑥 plus two was already zero or above, when we take the absolute value of it, we’re not actually having to do anything; we’ve already got an answer which is zero or above. So in that case, the equation we’re trying to solve is 𝑥 plus two equals seven. So if I subtract two from each side, I get a solution of 𝑥 equals five. But we just need to check something. Is that actually valid for that side of our calculation? Well up here we said that 𝑥 plus two had to be greater than or equal to zero. So if I take away two from both sides of that inequality, I’ve got 𝑥 is greater than or equal to negative two. So we can only follow this region when the value of 𝑥 is greater than or equal to negative two. And the solution we’ve got is 𝑥 equals five; and that is greater than or equal to two- to negative two. So yep, it looks like that’s a valid solution.

Now when would we follow the right-hand side? Well that would be when 𝑥 plus two is less than or equal to zero. And in those cases, to turn that value 𝑥 plus two into a positive answer, we’d have to take the negative of that negative answer. So the equation we’re going to solve is the negative of 𝑥 plus two is equal to seven. So let’s just multiply out the parentheses. That means negative one times 𝑥 plus two so negative 𝑥 take away two. And now I can solve that. So adding 𝑥 to each side and then subtracting seven, gives us the solution 𝑥 is equal to negative nine. So we just need to check that that’s a valid answer given our original criteria, which was that 𝑥 plus two has to be less than or equal to zero. So if I take away two from both sides of that inequality, I’d need 𝑥 to be less than or equal to negative two. And 𝑥 is equal to negative nine is less than negative two. So again that’s a valid answer.

Now there’s a slight problem with the way we’ve been doing it so far cause we’ve ended up with the region 𝑥 is less than or equal to negative two and 𝑥 is greater than or equal to negative two. So if it’s equal to negative two, which branch do we go around?

Well if we wanna be really strict about it, the absolute thing, taking the negative of the negative value to turn into positive, only happens when 𝑥 plus two is genuinely less than zero. So although it’s a coincidence that when-when that was equal to zero we’d get the same answer, if we wanna be really strict about it we would use less than, strictly less than, signs on this side. And, like I was saying, practice. It doesn’t make much difference which way you use it. But I think from a clarity point of view and knowing which region you’re in, you probably should use just strictly less than on the right-hand side there.

So number three then. Five times the absolute value of 𝑥 minus thirty-five is equal to zero. Solve for 𝑥.

But with this one, we can do a little bit of manipulation before we have to split off into two separate parts. So I could quite happily, for example, add thirty-five to both sides of that equation, and then divide both sides by five. And obviously thirty-five divided by five is seven.

So now I can make some decisions about how we solve this. So this is much like the first question we looked at. Either 𝑥 is greater than or equal to zero, in which case we don’t need to do anything about taking negatives of the function to-to give our absolute values. So 𝑥 is just equal to seven, which straightaway out pops our answer. And that is in the region. It is greater than or equal to zero. So that is our first answer.

Or if 𝑥 is less than zero, then we’re going to have to take the negative of that negative number to turn into a positive number. So we’ll have the negative of 𝑥 is equal to seven. Well that’s such a simple function; we don’t need the parentheses. And we can multiply both sides by negative one to get 𝑥 is equal to negative seven. And that’s our other answer, because 𝑥 is negative seven is in the region 𝑥 is less than zero. So we’ve got our two solutions.

So onto our last example. Solve three times the absolute value of 𝑥 minus two minus nine is equal to zero. Again I can do a bit of manipulation before I need to think about the absolute value bit of the equation.

So adding nine to either side gives me three times the absolute value of 𝑥 minus two is equal to nine. And then dividing both sides by three gives me the absolute value of 𝑥 minus two is equal to three. So now I need to think about the two branches. Either 𝑥 minus two is greater than or equal to zero, which amounts to 𝑥 being greater than or equal to two. Or the 𝑥 minus two bit was less than zero, and we’ve got to take the negative of the value we get, in order to turn it into a positive number. And that amounts to the situations where 𝑥 itself was less than two.

So as we said, down the left-hand branch the 𝑥 minus two value is already greater than or equal to zero, so we don’t need to do anything to it. So we can just say that 𝑥 minus two is equal to three. And when we add two to each side of that equation, we get 𝑥 is equal to five. And if 𝑥 is equal to five, that is greater than or equal to two; so that is a valid solution.

However on the right-hand side we were saying that 𝑥 minus two gave us a negative answer, so we’ve got to take the negative of that negative answer to turn it into a positive answer. So we’re saying the negative of 𝑥 minus two would be equal to three in that case. And just using the distributive property to put the minus one times 𝑥 and minus one times minus two, we’ve got negative 𝑥 plus two is equal to three. So adding 𝑥 to both sides gives us two is equal to three plus 𝑥. And then subtracting three from both sides gives us negative one is equal to 𝑥. And again when 𝑥 is equal to negative one, that is less than two; so it is in the region. So it’s a valid solution.

So to summarise what we’ve been up to then, sometimes you can do a bit of rearranging of your equation at the beginning to get your absolute value part of the equation on its own. And sometimes that was already the case anyway.

Secondly, we had to think about the situations where the absolute value function kicks in. When do we have to change the number to a positive number; when do we not have to change the number to a positive number. So it’s a question of working out those two regions.

And thirdly, we can just work out, if the absolute value function doesn’t need to do anything cause the result was already positive, we can work out a solution and just check that it’s in the correct region.

And then finally, we look at this case where we had a negative value and we had to change it into a positive value. But remember, in both cases we had to check that the solution we got in each of those regions matched up with the criterion that we set at the beginning for each of those regions.