INEQUALITIES
INVOLVING ABSOLUTE VALUES
Objective: To learn how to solve inequalities that involve absolute values.
Recall 1: The absolute value of a non-zero number a is defined to be the positive value of a. It is denoted as |a|. For example,
|4| = 4 and |-4| = 4.
But what happens in the
case of |x - 4| ?
Is |x - 4| = x - 4 or -(x - 4)
?
The answer is that it depends on the
value of
x. We have two possibilities:
1. If x >= 4, then x - 4 is not negative and |x - 4| = x - 4
2. If x < 4, then x - 4 is negative and |x - 4| = -(x - 4).
Note 1
Take
x to be 5. Then the first possibility applies
and you
have
|x - 4| = |5 - 4| = |1| = 1.
Take x to be 2. Then the second possibility applies
and you
have
|x - 4| = |2 - 4| = |-2| =
2.
( Notice that 2 = -(x - 4),
where x = 2.)
Note 2
The absolute value is a shorthand notation
that represents
the magnitude of the number or variable.
Graphically, if we represent
the real number a on the number line,
then the absolute value of
a represents a's distance from 0.
This is illustrated below
for the case where a = -7/2
.
Recall 2: An inequality may involve an absolute value, as in the case of |x - 3| < 5.
Such an inequality restricts the possibilities for x. In other words, only certain values of x will satisfy such an inequality. We find those values by recalling that |x - 3| is shorthand for two possibilities:
1. If x - 3 >= 0, then |x - 3| = x -3, in which case x - 3 < 5 and x < 8. Remember that x >= 3, so we get 3 <= x < 8.
2. If x - 3 < 0, then |x - 3| = -(x - 3), in which case -(x - 3) < 5.
This last result simplifies to -x + 3 < 5 and further to -x < 2. Multiplying the last inequality by -1 gives x > -2. Remember that x < 3, which means that the solution in this case is -2 < x < 3.
Putting the two cases together we conclude that x must be less than 8 and greater than -2.
Symbolically, this is represented as
-2 < x < 8.
Graphically, this is represented as
Recall 3: The values of x that satisfy the inequality |x - 3| < 5 lie in a single interval, as represented by the interval above. What happens if the inequality is reversed as in the example
|x - 3| > 5?
Again, we split the symbol |x - 3| into two possibilities:
1. If x - 3 >= 0, then |x - 3| = x -3--in which case x - 3 > 5 and x > 8. We are in the case x >= 3, so the answer is still x > 8.
2. If x - 3 < 0, then |x - 3| = -(x - 3)--in which case -(x - 3) > 5. This last result simplifies to -x + 3 > 5 and further to -x > 2. Multiplying the last inequality by -1 gives x < -2. We are in the case x < 3, so the answer is still x < -2.
We conclude that x must be less than -2 orgreater than 8.
Symbolically, this is written as
x < -2 or x > 8.
This is exactly the reverse of the result in Recall 2. You can see this graphically.
Notice that the interval split into two disjoint parts.
Recall 4: If you try to find the interval of values of
x that satisfy the inequality |x| < 3 you will find that -3 <
x < 3. This is the pattern for all absolute value inequalities
of this form.
|x|
< 4 implies that -4 < x < 4,
|x| < 5 implies that -5 < x <
5,
|x| < 2.718 implies that
-2.718 <
x < 2.718.
This pattern permits us to translate problems about inequalities of absolute values into problems about inequalities that do not involve absolute values.
More generally, to find the interval of real numbers that satisfy |3x - 4| < 2, we use the pattern:
|3x - 4| < 2 implies that -2 < 3x - 4 < 2,
and solve for x.
Examples
Example 1. Find the interval of values of x that satisfy
|3x -4| < 2
and represent it graphically.
Solution: Use Recall 4 to find that
|3x - 4| < 2 implies that -2 < 3x - 4 < 2.
Solve the two
inequalities
3x - 4 <
2
and
-2 < 3x - 4.
The first tells us that 3x < 6 or that x < 2, while the second tells us that 2< 3x or that 2/3 < x. Putting this together we have 2/3 < x < 2. This is graphically represented below.
Example 2. Find the interval of values of x that satisfy
|2x + 5| > 2
and represent it graphically.
Solution: The values of x that
satisfy |2x + 5|
> 2 are precisely the values that do not
satisfy
|2x + 5| <= 2.
We shall find the values of x that do satisfy |2x + 5|
<=
2. The remaining real numbers will be precisely
the ones
that we want.
|2x + 5| <= 2 implies that -2 <= 2x + 5 <= 2.
Solve the two
inequalities
2x + 5 <=
2
and
-2 <= 2x + 5.
The first tells us that 2x <= -3 or that x <= -3/2, while the second tells us that -7<= 2x or that -7/2 <= x. Putting this together we have -7/2<= x <= -3/2 . This is graphically represented below.
The intervals that we want are precisely the opposite of this. Symbolically, they are x < -7/2 and x > -3/2 .
EXERCISES
Find the interval(s) of real numbers that satisfy the following inequalities:
1. |x - 4| < 0
2.
|x + 3| > 0
3. |3x - 2| <
0
4. |6x + 3| <
4
5. |2x + 5| *
3
6 0 < |3 -
2x|
7 | [(2-3y) / 5]|
< 2
8 |1 - x| + 2 <
3
9 Use the number line to
graphically represent the interval
of real numbers x which
satisfy
|3x + 4| <=
2.
10 Use the number line to
graphically represent the interval
of real numbers x which
satisfy |2x - 3| <=
5.
11 Use the number line to graphically represent the interval
of
real numbers x which satisfy |2x + 3| >
2.
12 Use the line below to graphically
represent the interval
of real numbers y which
satisfy
| [(2-3y) / 5]| <=
2.
13 Use the
number line to graphically represent the interval
of real numbers x
which satisfy
| [(2-3y) / 5]| >
2.