Section 8.3: Absolute Value Equations And Inequalities
What is the definition of the absolute value? It is the distance from zero on the number line. Because the
number line includes the positive and negative numbers, we conclude that the absolute value has two solutions.
For example, x = 4 means that the distance from x and 0 on the number line is equal to 4 units. Hence we can
conclude that x can be 4 or x can also be negative 4!
What about something like x > 4 ? This is asking for all values of x on the number line whose distance from
zero is greater than 4. You must be careful here. For example, we know that any number greater than 4 works
here. We also know that any number LESS THAN negative 4 also works!
Lastly, what about x < 4 ? This is asking for all values of x on the number line whose distance from zero is less
than 4. Hence we know that any number less than 4 works AND any number greater than -4 works.
These are the three cases we will be dealing with in this section. You must always isolate the absolute value
expression. So make sure you move any terms that are not inside the absolute value to the other side of the
equation or inequality. Also, always remember that there are two solutions to deal with – the positive solution
and the negative solution. For an inequality, whenever we deal with the negative solution you must remember
to switch the direction of the inequality!
k
§1 Solve Absolute Value Equations Of The Form ax + b =
6 . We are looking for the value of x that makes this absolute value equation true.
Look at the example x − 4 =
Remember, there are two cases – the positive case and the negative case. So we have to set up two equations.
6. The other equation deals with the negative case – we simply make the other
One of them is simply x − 4 =
side negative. So the other equation becomes x − 4 =−6 . We can solve both of these to get that x = 10 and x =
-2. Note that if you plug these back into the original equation, they both work!
Keep in mind that we must isolate the absolute value expression before we can make our two equations. For
13 . Before we can write our two equations we must make sure
example, say we want to solve 2 3 x − 4 − 1 =
that the absolute value expression is isolated, so there should be no other terms. First we need to add the 1 to
14 . Then we divide both sides by 2 to get 3 x − 4 =
7 . Now we notice that the
both sides to get 2 3 x − 4 =
absolute value expression is isolated. Now we can go ahead and make our two equations. The first one is
3x − 4 =
7 . The second one is simply 3 x − 4 =−7 . Solve both of these to get that x = 11/ 3 or x = −1
PRACTICE
1) Solve
1
2x + 3 =
7
2
§2 Solve Absolute Value Inequalities Of The Form ax + b > k
Let’s try an example first. Solve 2 x − 1 > 3 . We know that there are two cases here - the positive case and the
negative case. The positive case simply means that the left side is positive. So we have 2 x − 1 > 3 . Here’s where
it gets a little tricky. The negative case means that the left side becomes negative. So we end up with
− ( 2 x − 1) > 3 . Note what happen when we divide both sides by the negative 1. Because we divide by a
negative 1, the direction of the inequality changes! So we end up with 2 x − 1 < −3. You should be able to solve
these inequalities. The first one becomes x > 2 . The second one becomes x < −1 . What does this solution
look like on the number line?
In interval notation, the answer is ( −∞, −1) ∪ ( 2, ∞ )
YOU MUST REMEMBER TO CHANGE THE DIRECTION OF THE INEQUALITY FOR THE NEGATIVE CASE. You must
also remember to isolate the absolute value expression before you write the two inequalities. For example, say
you want to solve 2 x − 3 − 1 > 7 . We must add the one to both sides first to get 2 x − 3 > 8 . Then we divide
both sides by 2 to get x − 3 > 4 . Now that the absolute value expression is isolated, we can go ahead and
write our two inequalities. They are x − 3 > 4 or x − 3 < −4 . Solve these to get x > 7 or x < −1 . On the
number line the solution looks like the following:
In interval notation, the solution is ( −∞, −1) ∪ ( 7, ∞ )
PRACTICE
2) Solve 2 x − 3 > 7 and graph the solution on the number line and express the solution in interval notation.
3) Solve x + 4 + 2 > 10 and graph the solution on the number line and express the solution in interval notation.
§3 Solve Absolute Value Inequalities Of The Form ax + b < k
First of all, note that the direction of the inequality is opposite the previous examples. How does this change the
solution? Well, as we saw in the beginning of this section, in the case the solution should consist of a single
interval. For example, solve x − 2 < 5 . Keeping mind the same principles, we write our two inequalities as
x − 2 < 5 AND x − 2 > −5 . Solve these to get that x < 7 and x > −3 . On the number line the solution looks
like the following:
In interval notation, the solution is ( −3, 7 ) . Make sure you understand what this interval means. Any number
between -3 and 7 is a solution to the original absolute value inequality. Hence a number like -4 will not work, nor
will a number like 8.
PRACTICE
4) Solve x + 4 + 2 < 10 and graph the solution on the number line and express the solution in interval notation.
§4 Solve Absolute Value Equations With Two Inequalities
Let’s look at an example: Solve 4 x − 1 = 3 x + 5 . How do we go about solving this?
Remember, this means that the distance from zero of the expression on the left side must be the same as the
distance from zero of the expression on the right side. Hence we actually have only two possibilities. Either the
sign of both sides is the same OR the sign of both sides is opposite. Hence we have to set up two equations. In
the first one, we simply remove the absolute value symbols to get 4 x − 1 = 3 x + 5 . For the other equation, we
pick one side to be positive and one side to be negative – it doesn’t matter which one is negative, but most
people pick the right side to be negative. Then the equation to solve becomes 4 x − 1 =− ( 3 x + 5 ) . Make sure
you distribute this negative sign all the way through! We can solve both of these. The first one becomes x = 6
while the second one becomes 4 x − 1 =−3 x − 5 , so we get that x = −4 / 7.
PRACTICE
5) Solve 2 x − 6 = x + 9