Math 091
Handout D
Solving Compound Inequalities (Section 2.8)
4 Types of Compound Inequality Situations:
1. The graphs of the inequalities overlap and go in opposite directions.
Example : Look at the inequalities
4x  6  10
 5x  3  13
Step 1: Solve both inequalities separately
4 x  6  10
 5 x  3  13
+6 +6
3
+3
5 x 10

Note:the inequality sign changed because of division by a negative number.
5
5
x2
4 x 16

4
4
x4
Step 2: Graph
)
-1
0
1
2
[
3
the graph of x < 4
4
the graph of x  2
Step 3: Interval Notation
For union,  , (indicated by “or”) include everything. Thus the solution in interval notation is (  ,  ). Notice that
there is no end to  or  , so parentheses are used to indicate that the endpoints are not included.
For intersection,  , (indicated by “and”) include what the inequalities have in common. Thus the solution is [2, 4).
Notice that the endpoint of 2 is included: shown by “ [ “ and that the endpoint of 4 is excluded: shown by “ ) ”
2. The graphs of the inequalities overlap and go in the same direction.
Example: Look at the inequalities
x  6  9x  30
8  6x  4x  12
Step 1: Solve both inequalities separately
x  6  9 x  30
9 x  9 x
8 x  6  30
6 6
8  6 x  4 x  12
 4x  4x
8  10 x  12
8
8
8 x 24

8 8
x  3
10 x 20

10 10
x 2
Step 2: Graph
)
-4
-3
the graph of
-2
-1
x  3
0
1
2
]
the graph of
x2
Step 3: Interval Notation
For union,  , (indicated by ”or”) include everything. Thus the solution in interval notation is (, 2] .
For intersection,  , (indicated by “and”) include what the inequalities have in common. Thus the solution is (  , 3) .
3. The graphs of the inequalities don’t overlap.
3x  4  10
Example: Look at the inequalities
2x  5  x  9
Step 1: Solve both inequalities separately
3x  4  10
4 4
3x 6

3 3
x 2
2x  5  x  9
x
x
x  5  9
+5 +5
x  4
Step 2: Graph
[
-4
]
-3
-2
the graph of
-1
0
1
the graph of
x2
2
x  4
Step 3: Interval Notation
For union,  , (indicated by “or”) include everything. Thus the solution in interval notation is (, 4]  [2,  ) .
For intersection,  , (indicated by “and”) include what the inequalities have in common. Thus the solution is
.
4. You start with the union or intersection of two intervals already in interval notation.
Example: Look at the intervals
[2,5)
[ 1 ,4)
Step 1: There is nothing to solve, so go to step 2, graph.
Step 2: Graph
[
-1
[
0
1
2
) the graph of [2,5)
3
4
)
5
the graph of [ 1 ,4)
Step 3: Interval Notation
For union,  , (indicated by “or”) include everything. Thus the solution in interval notation is [1, 5) .
For intersection,  , (indicated by “and”) include what the inequalities have in common. Thus the solution is [ 2, 4) .
EXERCISES
A. Solve each inequality, graph, and write the solution in interval notation.
1. x  3 and x  7
2. x  2 and x  1
3. x  4 and x  0
4. x  2 or x  6
5. x  8 or x  12
6. x  3 and x  5
7. x  3 and x  5
8. x  3 or x  5
9. x  3 or x  5
10. x  3 or x  5
11. x  3  5 and
13. 3x  3 and x  2  0
14. 3x  3 and x  2  2
15. 2x 1  3 and 2x  8  4
16. 5x  2  2 and 5x  2  7
17. 6x  8  16 and 4x 1  15
2
18.  x  4 and  2x  6
3
3
x  2  6 or  x  3
4
20. x  3  2 or x  4  3
21. 2x  3  7 or 4x 1  3
22. 3x  x 12 or x 1  5
23. 2x  5  10 or 4x  4
24. 3x  6  15 or 2x >15
25. 5x  2  17 or 2x  1  9
26. 3x  4  9 or 4x  5  13
19.
1
x2
2
B.
Graph each interval and write the solution in interval notation.
27.
  , 1  4, 
28.  1,     ,9
30.
  , 6  9,  
31.
33.
5,11  6, 
34.   9,1    , 3
36. 3,6   4,9 
37.
  ,3    , 2
1, 2   0,5
12. x  5  9 and x  3  2
29.  4,      ,12
32.
  ,5   0, 
35.
 1, 4   2,7 
38. 3,8   5,11
Answers To Selected Exercises.
1. (3,7) <---l----l----l----(----l----l----l----)----->
0
3
7
5.
  , 8 <----l----]-----l-----l-----l-----l---->
8 6 4 2
9.
13. (  2,
0
1
0
11. [4,8] <---l----l----l----l----[----l----l----l----]--->
15.

27. [  4,
0
<----l-----(-----l------l---->
 1]
  ,3
17.
8
  , 4 <----l------l------l------l-------]---->
21.
4
  ,1   2,  <-----l----l----)----(----l---->
4
1
25.
2
  ,   <----l------l------l------l------l---->
2 1
1
0
1
2
<----[------l------l------]------l---->
4
1
29. [4, 12]
<----[------l------l------l------]----->
4
6
8
10 12
<----l-----l-----l-----)---->
33. [6, 11]
<----[----l----l----l----l----]----->
3
35. (2, 4)
4
0
  ,  4   4,  <--l--)--l--l--l--l--l--l--l--(--l-->
1, 
5
0
 1) <----l-----(-----)-----l----->
2 1 0
0
31.
0
2
4
23.
5, <----l----l----l----l----l----[----l---- >
7.
  ,   <----l------l------l------l------l---->
2 1
19.
3. (0,4) <----l----(----l----l----l----)----->
0
4
<----l----(----l----)----l----->
2
4
6
37. [  1, 5)
11
<----[----l----l----l----l----l----)----->
5
1