### Video Transcript

In this video, we will learn how to
solve equations involving the absolute value. Weβll recap how to solve absolute
value expressions, for example, the absolute value of negative four. Weβll then look at how we solve
absolute value equations. For example, weβll 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, letβs recap absolute
value. The absolute value is the distance
of a number from zero on a number line. So letβs say, for example, weβre
considering the value four. The value four would be at a
distance of four units from zero. Next, letβs 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. Itβs got the expression within two
vertical lines. So for the absolute value of four,
well, thatβs 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 π equals π if π is greater than or equal to zero and the absolute
value of π equals negative π if π 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. Letβs 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, weβll have π₯. So the equation would be π₯ is
equal to 94. When π₯ is less than zero, weβll
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 letβs 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 letβs 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, weβll 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 weβre finding the
absolute value of π, itβs equal to π if π is greater than or equal to zero or
negative π if π is less than zero. In order to solve this equation,
weβll have to consider the two possible options for the absolute value of π₯ plus
three.

Weβll either have π₯ plus three or
weβll 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, letβs 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 letβs 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, letβs 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 weβd 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, letβs 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 isnβt valid, letβs have a look at this problem
graphically. Letβs begin by drawing the graph of
π¦ equals the absolute value of π₯ plus three. If we consider the graph of π¦
equals π₯ plus three, thatβs going to have a slope of one and a π¦-intercept of
three. We can start with the coordinate
zero, three and then the slope of one means that for every one unit across, itβs
going to go one unit up.

We can continue this line, but
letβs pause for a second at the coordinate negative three, zero. If we were just drawing the line π¦
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 π¦-coordinate would
instead be the positive π¦-coordinate.

Now letβs consider the graph of π¦
equals negative three π₯ plus seven. For this line, the π¦-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 π¦-value decreases by three. We can draw a line through these
points to indicate the graph of π¦ equals negative three π₯ plus seven.

So letβs 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, itβs 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 isnβt on the graph π¦
equals the absolute value of π₯ plus three.

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

In the next question, weβll 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 letβs consider each one in
turn.

When weβre finding something like
the absolute value of π₯ plus three, we really have two different options. We can say that itβs equal to π₯
plus three if π₯ plus three is greater than or equal to zero, i.e., itβs
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 itβs 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 letβs 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 letβs 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 canβt 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₯ canβt both be less
than negative three and greater than or equal to three. So this equation isnβt 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, weβre 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 thatβs 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, letβs check using a graph. We can get some squared paper. And letβs start with the graph of
π¦ equals the absolute value of π₯ plus three. The graph of π¦ equals π₯ plus
three has a slope of one and a π¦-intercept of three. However, instead of continuing this
line downwards when weβre drawing the graph of π¦ equals the absolute value of π₯
plus three, every negative π¦-coordinate will instead be the positive
π¦-coordinate.

Next, we can consider the graph of
π¦ equals the absolute value of two π₯ minus six. If we were just drawing the graph
of π¦ equals two π₯ minus six, it would have a π¦-intercept of negative six. However, every negative
π¦-coordinate is going to be reflected, so this graph will actually contain the
coordinate zero, six. The graph of π¦ 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.

Letβs 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. Letβs 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, weβll 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 itβs 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, itβs the same as π₯
plus three multiplied by π₯ minus three.

So letβs 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 weβd 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 letβs 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 weβre 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,
letβs 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 wonβt
be valid. As it turns out, negative four root
three is less than negative three; itβs about negative 6.93. And four root three is greater than
or equal to three; itβs 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.