In the previous section we
solved equations that contained absolute values. In this section we want
to look at inequalities that contain absolute values. We will need to examine
two separate cases.
Inequalities Involving < and
≤
As we did with equations
let’s start off by looking at a fairly simple case.
|p| ≤ 4
This says that no matter what
p is it must have a distance of no more than 4 from the origin. This means
that p must be somewhere in the range,
– 4
≤ p ≤ 4
We could so a similar
inequality with the < and get a similar result.
In general we have the
following formulas to use here,
If
|p| ≤ b, b >
0 then – b ≤ p ≤ b
If
|p| < b, b
> 0 then – b < p < b
|
Notice that this does require
b to be positive just as we did with equations.
Let’s take a look at a couple
of examples.
Example 1 Solve each of the following.
(a) |2x – 4|
< 10
(b) |9m + 2| ≤ 1
(c) |3 – 2z| ≤ 5
Solution
(a) |2x – 4| < 10
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,
– 10 < 2x – 4 < 10
Now, this is nothing more
than a fairly simply double inequality to solve so let’s do that.
– 6 < 2x < 14
– 3 < x < 7
The interval notation for
this solution is (– 3, 7).
(b) |9m + 2| ≤ 1
Not
much to do here.
– 1 ≤ 9m + 2 ≤
2
– 3 ≤ 9m + 2 ≤
– 2
The
interval notation is
.
.(c) |3 – 2z| ≤ 5
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.
– 5 ≤ 3 – 2z ≤
5
– 5 ≤ 3 – 2z ≤
5
4 ≥ z ≥ – 1
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 [– 1, 4].
Inequalities
Involving > and ≥
Once
again let’s start off with a simple number example.
| p | ≥ 4
This
says that whatever p is it must be at least a distance of 4 from the origin and
so p must be in one of the following two ranges,
p ≤ – 4 or p ≥ 4
Before
giving the general solution we need to address a common mistake that students
make with these types of problems. Many students try to combine these into a
single double inequality as follows,
– 4 ≥ p ≥ 4
While
this may seem to make sense we can’t stress enough that THIS IS NOT CORRECT!! Recall
what a double inequality says. In a double inequality we require that both of
the inequalities be satisfied simultaneously. The double inequality above would
then mean that p is a number that is simultaneously smaller than -4 and larger
than 4. This just doesn’t make sense. There is no number that satisfies this.
These
solutions must be written as two inequalities.
Here
is the general formula for these.
If
|p| ≥ b, b > 0 then p ≤ – b or
p ≥ b
If |p|
> b, b > 0 then p < – b or
p > b
|
Again, we will require that b
be a positive number here. Let’s work a couple of examples.
Example 2 Solve each of the following.
(a) |2x – 3| > 7
(b) |6t + 10| ≥ 3
(c) |2 – 6y| > 10
Solution
(a) |2x – 3| > 7
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.
2x < – 4 or
2x >
10
x <
– 2 or x > 5
The interval notation for these
are (– ∞, – 2) or (5, ∞).
(b) |6t + 10| ≥ 3
Let’s just plug into the
formulas and go here,
6t ≤ – 13 or 6t ≥ – 7
or 
The interval notation for
these are
.
.(c) |2 – 6y| > 10
Again, not much to do here.
2 – 6y < – 10 or 2 – 6y
> 10
– 6y < – 12 or – 6y
> 8
y > 2 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 (2, ∞) 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) |3x
+ 2| > 0
(b) |x
– 9| ≤ 0
(c) |2x
– 4| ≥ 0
(d) |3x
– 9| > 0
Solution
These
four examples seem to cover all our bases.
(a)
Now we know that | p | ≥ 0 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,
|x – 9| = 0
and
we know how to solve this from the previous section.
x – 9 = 0 ⇒ x = 9
(c)
In this case let’s again recall that no matter what p is we are guaranteed
to have |p| ≥ 0. This means that no matter what x is we can be
assured that |2x – 4| ≥ 0 will be true since absolute values will always
be positive or zero.
The
solution in this case is all real numbers, or all possible values of x. In inequality
notation this would be – ∞ < x < ∞.
(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,
|3x – 9| = 0 ⇒ 3x
– 9 = 0 ⇒ x = 3
The
solution in this case is all real numbers except x = 3.
Now,
let’s do a quick set of examples with negative numbers.
Example
4 Solve each of the following.
(a) |4x
+ 15| < – 2 and |4x + 15| ≤ – 2
(b) |2x
– 9| ≥ – 8 and |2x – 9| > – 8
Solution
Notice
that we’re working these in pairs, because this time, unlike the previous set
of examples the solutions will be the same for each.
Both
(all four?) of these will make use of the fact that no matter what p is we are
guaranteed to have |p| ≥ 0 . In other words, absolute values are always
positive or zero.
(a)
Okay, if absolute values are always positive or zero there is no way they can
be less than or equal to a negative number.
Therefore,
there is no solution for either of these.
(b)
In this case if the absolute value is positive or zero then it will always be
greater than or equal to a negative number.
The
solution for each of these is then all real numbers.
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