Lesson Video: Absolute Value Equations Mathematics

In this video, we will learn how to solve equations involving the absolute value.

19:14

Video Transcript

In this video, we will learn how to solve equations involving the absolute value. We’ll recap how to solve absolute value expressions, for example, the absolute value of negative four. We’ll then look at how we solve absolute value equations. For example, we’ll work through the problem where we have to solve the absolute value of π‘₯ plus three is equal to the absolute value of two π‘₯ minus six.

But first, let’s recap absolute value. The absolute value is the distance of a number from zero on a number line. So let’s say, for example, we’re considering the value four. The value four would be at a distance of four units from zero. Next, let’s consider negative four. When we are considering distance, distance will be a nonnegative value. And we know that negative four will also be at a distance of four units from zero.

We write the absolute value of an expression like this. It’s got the expression within two vertical lines. So for the absolute value of four, well, that’s four units from zero. The absolute value of negative four will be four units. Both of these values would be the same distance of four from zero on the number line. We can also write the absolute value a little more formally. We can say that the absolute value of a value 𝑝 equals 𝑝 if 𝑝 is greater than or equal to zero and the absolute value of 𝑝 equals negative 𝑝 if 𝑝 is less than zero.

For example, we can say that the absolute value of five equals five because five is greater than or equal to zero. And we could say that the absolute value of negative five equals negative negative five because negative five is less than zero. And of course, we know that negative negative five is equivalent to five.

We can now use this information about absolute values to help us solve equations. Let’s take a look at our first question.

What is the solution set of the equation the absolute value of π‘₯ equals 94?

We should begin by noting that these two vertical lines on either side of the π‘₯ indicate the absolute value. We recall that the absolute value of π‘₯ is equal to π‘₯ if π‘₯ is greater than or equal to zero or it equals negative π‘₯ if π‘₯ is less than zero. So when we have the absolute value of π‘₯ is equal to 94, we really have two different options for π‘₯. When π‘₯ is greater than or equal to zero, we’ll have π‘₯. So the equation would be π‘₯ is equal to 94. When π‘₯ is less than zero, we’ll have negative π‘₯. So our equation would be negative π‘₯ is equal to 94. We can solve this second equation by multiplying through by negative one, which gives us that π‘₯ must be equal to negative 94.

So let’s check our solutions are valid. We get that π‘₯ is equal to 94, which occurs when π‘₯ is greater than or equal to zero. And 94 is greater than or equal to zero, so that would work. And then we got the solution that π‘₯ must be equal to negative 94. And that does occur when π‘₯ is less than zero. Therefore, both of our solutions are valid. So let’s write those as a solution set. We can therefore give the answer that the solution set of the equation the absolute value of π‘₯ equals 94 is the set containing negative 94 and 94.

Next, we’ll see how we can solve an equation involving an absolute value that also has an unknown on both sides.

Find the solution set of the equation the absolute value of π‘₯ plus three equals negative three π‘₯ plus seven.

It can be helpful to recall the definition for the absolute value. That is, if we’re finding the absolute value of 𝑝, it’s equal to 𝑝 if 𝑝 is greater than or equal to zero or negative 𝑝 if 𝑝 is less than zero. In order to solve this equation, we’ll have to consider the two possible options for the absolute value of π‘₯ plus three.

We’ll either have π‘₯ plus three or we’ll have negative π‘₯ plus three. We travel along this first path, if you like, whenever π‘₯ plus three is greater than or equal to zero. We can, of course, alternatively think of this by subtracting three from both sides to say that π‘₯ must be greater than or equal to negative three. We travel along this path on the right when π‘₯ plus three is less than zero. By subtracting three from both sides of this inequality, that occurs when π‘₯ is less than negative three.

Now that we found both options for this expression on the left, let’s write what that means in terms of the equation. When we have π‘₯ plus three, the equation will be π‘₯ plus three equals negative three π‘₯ plus seven. So let’s solve this to find the value of π‘₯. Firstly, adding three π‘₯ to both sides, we have four π‘₯ plus three equals seven. Next, we can subtract three from both sides. And finally, we can divide both sides by four to give us π‘₯ equals one.

Now that we found a possible solution, let’s see what happens when we have negative π‘₯ plus three. This time our equation will be negative π‘₯ plus three equals negative three π‘₯ plus seven. Simplifying this, on the left-hand side we’d have negative π‘₯ minus three equals minus three π‘₯ plus seven. We can then add three π‘₯ to both sides. And adding three would give us two π‘₯ equals 10. And dividing through would give us that π‘₯ is equal to five.

Now that we have two possible solutions for this equation, let’s check our results. For the first solution of π‘₯ is equal to one, remember that this will only happen when π‘₯ is greater than or equal to negative three. One is greater than or equal to negative three, so this solution would be valid. For the second solution then, π‘₯ is equal to five. Remember that we got here by traveling along the path for π‘₯ is less than negative three. However, five is not less than negative three. So π‘₯ equals five is not a valid solution.

In order to see just how we ended up with a solution that isn’t valid, let’s have a look at this problem graphically. Let’s begin by drawing the graph of 𝑦 equals the absolute value of π‘₯ plus three. If we consider the graph of 𝑦 equals π‘₯ plus three, that’s going to have a slope of one and a 𝑦-intercept of three. We can start with the coordinate zero, three and then the slope of one means that for every one unit across, it’s going to go one unit up.

We can continue this line, but let’s pause for a second at the coordinate negative three, zero. If we were just drawing the line 𝑦 equals π‘₯ plus three, then it would continue in this direction. However, the absolute value of π‘₯ plus three is a little different at this point. Every negative 𝑦-coordinate would instead be the positive 𝑦-coordinate.

Now let’s consider the graph of 𝑦 equals negative three π‘₯ plus seven. For this line, the 𝑦-intercept will be seven and the slope will be negative three. That slope means that for every positive increase on the π‘₯-axis of one, the 𝑦-value decreases by three. We can draw a line through these points to indicate the graph of 𝑦 equals negative three π‘₯ plus seven.

So let’s think about the solutions then. We have this point here where they cross. Notice that this is the point where π‘₯ is equal to one. So where did this value of π‘₯ equals five come from? Well, it’s really an incorrect extension of this portion of the graph. However, we know that down here when π‘₯ is equal to five, well, that portion of the line isn’t on the graph 𝑦 equals the absolute value of π‘₯ plus three.

Therefore, the solution set of this equation is the set containing one.

In the next question, we’ll see how we can solve an equation that has an absolute value on both sides.

Find the solution set of the absolute value of π‘₯ plus three equals the absolute value of two π‘₯ minus six.

In this equation, we have two expressions with an absolute value on both sides of the equation. So let’s consider each one in turn.

When we’re finding something like the absolute value of π‘₯ plus three, we really have two different options. We can say that it’s equal to π‘₯ plus three if π‘₯ plus three is greater than or equal to zero, i.e., it’s positive. Or the absolute value of π‘₯ plus three is equal to negative π‘₯ plus three if π‘₯ plus three is less than zero.

Sometimes it’s nice to visualize these as though they were two different paths we could take, something a little like this. Notice that this inequality π‘₯ plus three is greater than or equal to zero can also be written as π‘₯ is greater than or equal to negative three simply by subtracting three from both sides of the inequality. In the same way, if π‘₯ plus three is less than zero, then π‘₯ must be less than negative three.

Now let’s do the same thing for the absolute value of two π‘₯ minus six. It will either be equal to two π‘₯ minus six if two π‘₯ minus six is greater than or equal to zero or negative two π‘₯ minus six if two π‘₯ minus six is less than zero.

We can simplify these inequalities a little just like last time. The two different pathways will occur when π‘₯ is greater than or equal to three or π‘₯ is less than three. It can be useful to create a number line with the different alternatives for the absolute value expressions. For example, we can see that we would have π‘₯ plus three when π‘₯ is greater than or equal to negative three.

Now let’s consider the different values that we can get for this equation. We can start with π‘₯ plus three is equal to two π‘₯ minus six. This would occur in this region on our number line diagram. Then we could have π‘₯ plus three equals negative two π‘₯ minus six. This would occur in the central region of our number line diagram. Then a third possibility would be negative π‘₯ plus three equals two π‘₯ minus six.

However, where exactly would this occur on the diagram? Well, it can’t really occur because minus π‘₯ plus three is only obtained when π‘₯ is less than negative three and two π‘₯ minus six is obtained when π‘₯ is greater than or equal to three. The value of π‘₯ can’t both be less than negative three and greater than or equal to three. So this equation isn’t actually a possibility.

We can, however, solve negative π‘₯ plus three equals minus two π‘₯ minus six, as this is what occurs in the region to the left on our number line diagram. Now we take these equations and solve them to find the values of π‘₯. When π‘₯ plus three equals two π‘₯ minus six, we’re going to start by subtracting π‘₯ from both sides. Doing this and then adding six would give us that nine equals π‘₯ or, of course, π‘₯ is equal to nine.

For the second equation, we can start by simplifying this expression on the right-hand side. So π‘₯ plus three is equal to negative two π‘₯ plus six. Adding two π‘₯ to both sides and then subtracting three gives us that three π‘₯ equals three. Therefore, π‘₯ is equal to one.

In the third equation, we might notice that we could begin by dividing through by negative one. Observe that since this would give us π‘₯ plus three equals two π‘₯ minus six, we notice that that’s the same as the first equation that we solved. Both equations would therefore give the solution that π‘₯ is equal to nine. We can therefore give the answer that the solution set to this equation is the set containing nine and one.

But before we finish with this question, let’s check using a graph. We can get some squared paper. And let’s start with the graph of 𝑦 equals the absolute value of π‘₯ plus three. The graph of 𝑦 equals π‘₯ plus three has a slope of one and a 𝑦-intercept of three. However, instead of continuing this line downwards when we’re drawing the graph of 𝑦 equals the absolute value of π‘₯ plus three, every negative 𝑦-coordinate will instead be the positive 𝑦-coordinate.

Next, we can consider the graph of 𝑦 equals the absolute value of two π‘₯ minus six. If we were just drawing the graph of 𝑦 equals two π‘₯ minus six, it would have a 𝑦-intercept of negative six. However, every negative 𝑦-coordinate is going to be reflected, so this graph will actually contain the coordinate zero, six. The graph of 𝑦 equals the absolute value of two π‘₯ minus six will look like this in pink. We can see that there are two points of intersection: here when π‘₯ is equal to one and here when π‘₯ is equal to nine, thus confirming our original answer.

Let’s take a look at one final question.

Find algebraically the solution set of the equation the absolute value of π‘₯ plus three times the absolute value of π‘₯ minus three equals 39.

In this question, we need to work out the solution to two absolute value expressions multiplied together. Let’s consider the different possible values for each expression in turn. The absolute value of π‘₯ plus three will either be π‘₯ plus three or negative π‘₯ plus three. It will be π‘₯ plus three when π‘₯ plus three is greater than or equal to zero. We can simplify this by subtracting three from both sides so that the inequality is π‘₯ is greater than or equal to negative three. And we get a negative result when π‘₯ plus three is less than zero or π‘₯ is less than negative three.

Next, for the absolute value of π‘₯ minus three, we’ll either have π‘₯ minus three or negative π‘₯ minus three. These will occur when π‘₯ minus three is greater than or equal to zero or when it’s less than zero. These inequalities can alternatively be written as when π‘₯ is greater than or equal to three or π‘₯ is less than three.

Now we can consider the different possibilities of expressions that will multiply together. To begin with, we could have π‘₯ plus three multiplied by π‘₯ minus three equals 39. Or we could have π‘₯ plus three multiplied by negative π‘₯ minus three equals 39. The third option is that we have negative π‘₯ plus three multiplied by π‘₯ minus three equals 39. Or finally, we could have negative π‘₯ plus three multiplied by negative π‘₯ minus three equals 39.

Before we rush into solving all four equations, there is something we can observe. Looking at the second and third equation, because of the commutative property of multiplication, this negative here could equivalently be put here and the product would not change. That means that the equations two and three would produce the same result.

And then what do we notice about the first and final equations. Observe that this negative π‘₯ plus three multiplied by a negative π‘₯ minus three would produce a positive. In other words, it’s the same as π‘₯ plus three multiplied by π‘₯ minus three.

So let’s have a look at solving the first equation, and we know that it would give the same result as the fourth equation. We could use a method such as the FOIL method to expand the parentheses on the left-hand side, observing that the terms plus three π‘₯ minus three π‘₯ would simplify to zero. Thus, π‘₯ squared minus nine equals 39. We can then add nine to both sides. We could then take the square root of both sides, so we’d have that π‘₯ is equal to the square root of 48. However, as we want to consider both the positive and negative values of the square root, then we can use the plus or minus symbol. As 48 can be written as 16 multiplied by three, it means that we can write this more simply as π‘₯ equals plus or minus four root three.

Now let’s take a look at solving one of the other equations, either equation two or three. Choosing negative π‘₯ plus three multiplied by π‘₯ minus three equals 39, we can begin by expanding the parentheses. Using our previous result that π‘₯ plus three multiplied by π‘₯ minus three would give us π‘₯ squared minus nine, we have that negative π‘₯ squared minus nine equals 39. Multiplying both sides of this equation by negative one would give us π‘₯ squared minus nine equals negative 39. Then adding nine would give us π‘₯ squared equals negative 30.

At this point, we may notice that we have a problem. For as we begin to take the square root of both sides, we see that we’re trying to take the square root of a negative value. In this case, there would be no real solution for π‘₯, so this solution would not be valid.

Before we finalize our answer, let’s just think about whether or not the two solutions to the equation π‘₯ plus three times π‘₯ minus three equals 39 are valid for the absolute value equation. This particular equation only applies when π‘₯ is less than negative three or when π‘₯ is greater than or equal to three. So if the solution for π‘₯ is in the region negative three is less than or equal to π‘₯ is less than three, then it won’t be valid. As it turns out, negative four root three is less than negative three; it’s about negative 6.93. And four root three is greater than or equal to three; it’s about 6.93. Both of these solutions are valid. We therefore just have one set of values for the solution to this equation.

The solution set will be negative four root three and four root three.

We can now summarize the key points of this video. We saw that the absolute value is the distance of a number from zero on a number line. Finally, when solving absolute value equations, we must always check our solutions are valid, either by using an analytic or a graphical method.

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