SUPPLEMENTAL SHEET 12
TOPIC: SOLVING MODULUS INEQUALITIES
There are many opportunities for mistakes with absolute-value inequalities, so we're going to take
this slowly, and look at some pictures along the way. When we're done, I hope you will have a
good picture in your head of what is going on, so you won't make some of the more common
errors. Once you catch on to how these inequalities work, this stuff really isn't so bad.
Let's first return to the original definition of absolute value: "| x | is the distance of x from zero."
For instance, since both –2 and 2 are two units from zero, we have | –2 | = | 2 | = 2:
With this definition and picture in mind, let's look at some absolute value inequalities.
Suppose you're asked to graph the solution to | x | < 3. The solution is going to be all the points
that are less than three units away from zero. Look at the number line:
The number 1 will work, as will –1; the number 2 will work, as will –2. But 4 will not work, and
neither will –4, because they are too far away. Even 3 and –3 won't work (though they're right on
the edge). But 2.99 will work, as will –2.99. In other words, all the points between –3 and 3, but
not actually including –3 or 3, will work in this inequality. Then the solution looks like this:
Translating this into symbols, you find that the solution is
–3 < x < 3.
This pattern always holds: Given the inequality | x | < a, the solution is always of the form
< a. Even when the problems get more complicated, the pattern still holds.
–a < x
We are ready for our first inequality. Find the set of solutions for
|x|<5.
Translate into English: we are looking for those real numbers x whose distance from the origin is
less than 5 units.
Obviously we are talking about the interval (-5,5):

Solve
| 2x + 3 | < 6.
Since this is a "less than" absolute value inequality, the first step is to clear the absolute
value according to the pattern: Copyright © Elizabeth Stapel 2000-2007 All Rights Reserved
| 2x + 3 | < 6
–6 < 2x + 3 < 6 [this is the pattern for "less than"]
–6 – 3 < 2x + 3 – 3 < 6 – 3
–9 < 2x < 3
–9
/2 < x < 3/2
Then the solution to | 2x
+ 3 | < 6 is the interval –9/2 < x < 3/2.
There really isn’t much to do other than plug into the formula. As with equations p simply
represents whatever is inside the absolute value bars. So, with this first one we have,
Now, this is nothing more than a fairly simply double inequality to solve so let’s do that.
The interval notation for this solution is
.
Not much to do here.
The interval notation is
.
We’ll need to be a little careful with solving the double inequality with this one, but other than that
it is pretty much identical to the previous two parts.
In the final step don’t forget to switch the direction of the inequalities since we divided everything
by a negative number. The interval notation for this solution is
.
The other case for absolute value inequalities is the "greater than" case. Let's first return to the
number line, and consider the inequality | x | > 2.
The solution will be all points that are more than two units away from zero. For instance, –3 will
work, as will 3; –4 will work, as will 4. But –1 will not work, and neither will 1, because they're too
close; –2 will not work, and neither will 2 (although they're right on the edge). In other words, the
solution will be two separate sections: one being all the points more than two units from zero off
to the left, and the other being all the points more than two units from zero off to the right. The
solution looks like this:
Translating this solution into symbols, you get "x < –2 or x > 2". That is, the solution is TWO
inequalities, not one. DO NOT try to write this as one inequality. If you try to write this solution as
"–2 > x > 2", take out the x in the middle, and you'll see that you would be saying "–2 > 2",
which certainly isn't true. Take the extra half a second, and write the solution correctly!
The pattern for "greater than" inequalities always holds: the solution is always in two parts. Given
the inequality | x | > a, the solution always starts by splitting the inequality into two pieces: x < –a
or x > a. For instance:

Solve
| 2x – 3 | > 5.
The first thing to do is clear the absolute value bars by splitting the inequality into two:
| 2x – 3 | > 5
2x – 3 < –5 or 2x – 3 > 5
2x < –2 or 2x > 8
x < –1 or x > 4
[this is the pattern for "greater than"]
This PAIR of inequalities is the solution;
the solution to | 2x – 3 | > 5 consists of the two intervals x
< –1 and x > 4.
Again, p represents the quantity inside the absolute value bars so all we need to do here is plug
into the formula and then solve the two linear inequalities.
The interval notation for these are
or
.
Let’s just plug into the formulas and go here,
The interval notation for these are
Again, not much to do here.
or
.
Notice that we had to switch the direction of the inequalities when we divided by the negative
number! The interval notation for these solutions is
or
.
Okay, we next need to take a quick look at what happens if b is zero or negative. We’ll do these
with a set of examples and let’s start with zero.
Example 3 Solve each of the following.
(a)
(b)
(c)
(d)
]
Solution
These four examples seem to cover all our bases.
(a) Now we know that
and so can’t ever be less than zero. Therefore, in this case there is
no solution since it is impossible for an absolute value to be strictly less than zero (i.e. negative).
(b) This is almost the same as the previous part. We still can’t have absolute value be less than zero,
however it can be equal to zero. So, this will have a solution only if
and we know how to solve this from the previous section.
(c) In this case let’s again recall that no matter what p is we are guaranteed to have
means that no matter what x is we can be assured that
values will always be positive or zero.
. This
will be true since absolute
The solution in this case is all real numbers, or all possible values of x. In inequality notation this would
be
.
(d) This one is nearly identical to the previous part except this time note that we don’t want the absolute
value to ever be zero. So, we don’t care what value the absolute value takes as long as it isn’t zero.
This means that we just need to avoid value(s) of x for which we get,
The solution in this case is all real numbers except
Taken from
www.sosmath.com
www.purplemath.com
.
http://tutorial.math.lamar.edu/Classes/Alg/SolveAbsValueIneq.aspx