Pre-Calculus 6.0
Two different fundamental approaches that were used for solving absolute value equations will now be adapted to the solution of absolute value inequalities. Again, our intention this year is to move beyond memorization of steps to a better understanding of the technical “behind the scene” reasons for what we do.
The only difference between the distance approach for inequalities versus that of equations is that we must now find the complete set of numbers that satisfy the inequality instead of a limited number of discrete solutions. Keep in mind that this method may be used only if the variable terms are all contained within the absolute value expression.
Example 1: 2 x 1 2 5
1) Factor out coefficient of linear term:
2

 x
−
1
2
2) Use properties of absolute value to “extract the coefficient:


2 5
2

 x
−
1
2


2 5
2 x 2 5
2
3) Isolate the absolute value block:
2 x 2 5
2
+
2
+
2
1
 
2 x
2
7
1 7 x
2 2
4) Identify central value and “Critical Point” values (i.e. Solution Values for an analogous absolute value equation):
Central value: h
=
1
2

1 7 1
Critical Points: x = ,
+
7
2 2 2 2
=
{ − }
Note that the solution will be exclusive of the critical points in this case given the nature of the given inequality
(i.e. less than type)
5) Graph solution and note regions of number line that meet the distance criteria given by the isolated absolute value
inequality (in this case, the distance away from the central value of 1/2 must be less than 7/2 units. SHADE solution
region and state solution in Set Builder and Interval Notation:
Solution
-3
Set Builder Notation:
{ x 4
}
4
; Interval Notation:
( −
3, 4
)

Pre-Calculus 6.0
When the variable is located both inside and outside of the absolute value expression, then the case analysis approach is appropriate. As previously demonstrated with absolute value equations, the formal case analysis approach requires that we develop solutions for the two scenarios, or cases, where the absolute value argument is either non-negative or negative.
Example 2:
1
3
1) Isolate the absolute value “block”:
1
2 x 1 1 4
2 x 1 1 4 x x
3
+
1
+
1


1
3
2 x
+
1


≥ (
5
− x
)( )
2 x 1 15
−
3 x
2) Create two cases:
Case 1: Argument is positive or 0
Case Domain: 2 x 1 0; x
Case Inequality: 2 x 1 15
−
3 x
1
2
3) Solve independently:
2 x 1 15
−
3 x
−
1
−
1
OR
Case 2: Argument is negative
Case Domain: 2 x 1 0; x
Case Inequality:
− (
1
2 x 1
)
15
−
3
2 x
− (
2 x
+
1
) ≥
15
−
3 x
1
 
2 x
≥
14
−
3 x
+
3 x
+
3 x
−
2 x 1 15
−
3 x
+
1
+
1
−
2 x
≥
16
−
3 x
+
3 x
+
3 x x
≥
14
5
Subject to intersection with case domain restriction: x
≥
16
Subject to intersection with case domain restriction: x
≥
14
5
 x
≥ −
1
2 occurs where x
≥
14
5 x
≥
16  x
< −
1
2 which has no solution
4) Determine solution from union of case solutions: x
≥
14
5

(
No Solution
Final Answer: 
 x x
≥
14
5 
)
occurs where x
≥
14
5
or more simply


14
5
,
∞


Graphical Solution:
Graphing Calculator Check:
14
5
Enter y
1
=
1
2 x 1 1 and y
2 x into the equation editor and then graph. Range values
3
for the absolute value function are greater than or equal to those of the linear function for
domain values that are greater than or equal to 14/5 thereby confirming the algebraic solution.

Pre-Calculus 6.0
Example 3:
2
5
1) Isolate the absolute value “block”:
2
5
3 x 2 3
1
2 x
+
1
3 x 2 3
1
2 x
+
+
3
+
3
1
5 2
  5
3 x
−
2
 
 
1
2 x
+
4
2) Create two cases:
3 x 2
5
4 x
+
OR
Case 1: Argument is positive or 0
10
Case 2: Argument is negative
Case Domain: 3 x 2 0; x
2
3
Case Domain: 3 x 2 0; x
2
3
Case Inequality: 3 x 2
5
4 x
+
10 Case Inequality:
− (
3 x
−
2
) <
5
4 x
+
10
3) Solve independently:
3 x 2
+
2
3 x
<
−
5
4 x
4
 
7
4 x


<
5
4 x
+
10
+
2
5
4 x
+
12
−
5
4 x
− (
3 x
−
2
−
3 x 2
−
2
) <
5
4 x
+
10
5
4 x
+
10
−
2


−
4
17


−
17
4
−
3 x
<
−
5
4 x

 x
<
5
4 x
+
8
−
5
4 x


−
4
17

 x
<
48
7
Subject to intersection with case domain restriction: x
> −
32
17
Subject to intersection with case domain restriction: x
<
48
7
 x
≥
2
3 occurs where
2
3
48
7 x
> −
32
17
 x
<
2
3 occurs where
−
32
< <
17 x
4) Determine solution from union of case solutions:
2
3
48
7
Final Answer:
 −
32
< <
17 x x
−
32
< <
17 x
7
2
3
occurs where
−
32
< <
17 x
48
or more simply


48
7
−
32 48
,
17 7


2
3
Graphical Solution:
Graphing Calculator Check:
−
32
17
48
7
Enter y
1
=
2
5
3 x 2 3 and y
2
=
1
2 x
+
1 into the equation editor and then graph. Range values
for the absolute value function are less than those of the linear function for domain values that
are between -32/17 and 48/7 exclusive thereby confirming the algebraic solution.