1: Equations and Inequalities Involving Absolute Value
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11.1 • Equations and Inequalities Involving Absolute Value
453
2
The solution set is e- , 2 f , and its graph is shown in Figure 11.3.
3
−2
3
−5 − 4 − 3 −2 −1
0
1
2
3
4
5
Figure 11.3
The “distance interpretation” for absolute value also provides a good basis for solving
inequalities involving absolute value. Consider the following examples.
Classroom Example
Solve and graph the solutions for
ƒ x ƒ 6 5.
EXAMPLE 4
Solve and graph the solutions for ƒ x ƒ 6 2.
Solution
The number x must be less than two units away from zero. Thus ƒ x ƒ 6 2 implies
x 7 -2
x 6 2
and
The solution set is 5x ƒ x 7 - 2 and x 6 26, and its graph is shown in Figure 11.4. The
solution set is (Ϫ2, 2) written in interval notation.
−5 − 4 − 3 −2 −1
0
1
2
3
4
5
Figure 11.4
Classroom Example
Solve and graph the solutions for
ƒ x ϩ 2 ƒ 6 3.
Solve and graph the solutions for ƒ x Ϫ 1 ƒ 6 2.
EXAMPLE 5
Solution
The number x Ϫ 1 must be less than two units away from zero. Thus ƒ x Ϫ 1 ƒ 6 2 implies
x Ϫ 1 7 -2
x 7 -1
and
xϪ1 6 2
x 6 3
and
The solution set is 5x ƒ x 7 - 1 and x 6 36, and its graph is shown in Figure 11.5. The
solution set is (Ϫ1, 3) written in interval notation.
−5 − 4 − 3 −2 −1
0
1
2
3
4
5
Figure 11.5
Classroom Example
Solve and graph the solutions for
ƒ 4x ϩ 2 ƒ … 10.
EXAMPLE 6
Solve and graph the solutions for ƒ 2x ϩ 5 ƒ … 1.
Solution
The number 2x ϩ 5 must be equal to or less than one unit away from zero. Therefore
ƒ 2x ϩ 5 ƒ … 1 implies
2x ϩ 5 Ú - 1
2x Ú - 6
x Ú -3
and
and
and
2x ϩ 5 … 1
2x … - 4
x … -2
The solution set is 5x ƒ x Ú - 3 and x … - 26, and its graph is shown in Figure 11.6. The solution set is [Ϫ3, Ϫ2] written in interval notation.
454
Chapter 11 • Additional Topics
−5 −4 −3 −2 −1
0
1
2
3
4
5
Figure 11.6
Classroom Example
Solve and graph the solutions for
ƒ x ƒ 7 4.
EXAMPLE 7
Solve and graph the solutions for ƒ x ƒ 7 2.
Solution
The number x must be more than two units away from zero. Thus ƒ x ƒ 7 2 implies
x 6 - 2 or x 7 2
The solution set is 5x ƒ x 6 - 2 or x 7 26, and its graph is shown in Figure 11.7. The
solution set is 1- q , -22 ʜ12, q 2 written in interval notation.
− 5 − 4 − 3 −2 −1
0
1
2
3
4
5
Figure 11.7
Classroom Example
Solve and graph the solutions for
ƒ 3x ϩ 4 ƒ 7 5.
Solve and graph the solutions for ƒ 3x Ϫ 1 ƒ 7 4.
EXAMPLE 8
Solution
The number 3x Ϫ 1 must be more than four units away from zero. Thus ƒ 3x Ϫ 1 ƒ 7 4 implies
3x Ϫ 1 6 - 4
3x 6 - 3
or
or
x 6 -1
or
3x Ϫ 1 7 4
3x 7 5
5
x 7
3
The solution set is ex ƒ x 6 - 1 or x 7
5
f , and its graph is shown in Figure 11.8.
3
5
The solution set is (- q , - 1) ഫ a , qb written in interval notation.
3
5
3
−5 − 4 −3 − 2 −1
0
1
2
3
4
5
Figure 11.8
The solutions for equations and inequalities such as ƒ 3x Ϫ 7 ƒ ϭ -4, ƒ x ϩ 5 ƒ Ͻ -3, and
ƒ 2x Ϫ 3 ƒ 7 - 7 can be found by inspection. Notice that in each of these
examples the right side is a negative number. Therefore, using the fact that the
absolute value of any number is nonnegative, we can reason as follows:
ƒ 3x Ϫ 7 ƒ ϭ - 4 has no solutions because the absolute value of a number cannot be negative.
ƒ x ϩ 5 ƒ 6 - 3 has no solutions because we cannot obtain an absolute value less than Ϫ3.
ƒ 2x Ϫ 3 ƒ 7 - 7 is satisfied by all real numbers because the absolute value of 2x Ϫ 3,
regardless of what number is substituted for x, will always be greater than Ϫ7.
11.1 • Equations and Inequalities Involving Absolute Value
455
Concept Quiz 11.1
For Problems 1–10, answer true or false.
1.
2.
3.
4.
5.
6.
7.
8.
9.
The absolute value of a negative number is the opposite of the number.
The absolute value of a number is always positive or zero.
The absolute value of a number is equal to the absolute value of its opposite.
The solution set for ƒ x Ϫ 1 ƒ ϭ 4 is {- 3, 3}.
The solution set for the equation |x ϩ 5 ƒ ϭ 0 is the null set, л.
The solution set for ƒ x Ϫ 2 ƒ Ú - 6 is all real numbers.
The solution set for ƒ x ϩ 1 ƒ 6 - 3 is all real numbers.
The solution set for ƒ x Ϫ 4 ƒ … 0 is {4}.
If a solution in interval notation is (-4, - 2), then using set builder notation, it can be
expressed as {x|x 7 - 4 and x 6 - 2}.
10. If a solution in interval notation is (- q , - 2)ഫ (4, q ), then using set builder notation, it can be expressed as {x |x 6 - 2 or x 7 4}.
Problem Set 11.1
For Problems 1– 26, solve the equation or inequality. Graph
the solutions. (Objectives 1 and 2)
For Problems 27– 42, solve each of the following. (Objectives
1 and 2)
1. ƒ x ƒ ϭ 4
2. ƒ x ƒ ϭ 3
27. ƒ 3x Ϫ 1 ƒ ϭ 17
28. ƒ 4x ϩ 3 ƒ ϭ 27
3. ƒ x ƒ 6 1
4. ƒ x ƒ 6 4
29. ƒ 2x ϩ 1 ƒ 7 9
30. ƒ 3x Ϫ 4 ƒ 7 20
5. ƒ x ƒ Ú 2
6. ƒ x ƒ Ú 1
31. ƒ 3x Ϫ 5 ƒ 6 19
32. ƒ 5x ϩ 3 ƒ 6 14
7. ƒ x ϩ 2 ƒ ϭ 1
8. ƒ x ϩ 3 ƒ ϭ 2
33. ƒ -3x Ϫ 1 ƒ ϭ 17
34. ƒ -4x ϩ 7 ƒ ϭ 26
9. ƒ x Ϫ 1 ƒ ϭ 2
10. ƒ x Ϫ 2 ƒ ϭ 1
35. ƒ 4x Ϫ 7 ƒ … 31
36. ƒ 5x Ϫ 2 ƒ … 21
11. ƒ x Ϫ 2 ƒ … 2
12. ƒ x ϩ 1 ƒ … 3
37. ƒ 5x ϩ 3 ƒ Ú 18
38. ƒ 2x Ϫ 11 ƒ Ú 4
13. ƒ x ϩ 1 ƒ 7 3
14. ƒ x Ϫ 3 ƒ 7 1
39. ƒ -x Ϫ 2 ƒ 6 4
40. ƒ -x Ϫ 5 ƒ 6 7
15. ƒ 2x ϩ 1 ƒ ϭ 3
16. ƒ 3x Ϫ 1 ƒ ϭ 5
41. ƒ -2x ϩ 1 ƒ 7 6
42. ƒ -3x ϩ 2 ƒ 7 8
17. ƒ 5x Ϫ 2 ƒ ϭ 4
18. ƒ 4x ϩ 3 ƒ ϭ 8
For Problems 43–50, solve each equation or inequality by
inspection. (Objectives 1 and 2)
19. ƒ 2x Ϫ 3 ƒ Ú 1
20. ƒ 2x ϩ 1 ƒ Ú 3
43. ƒ 7x ƒ ϭ 0
44. ƒ 3x Ϫ 1 ƒ ϭ -4
21. ƒ 4x ϩ 3 ƒ 6 2
22. ƒ 5x Ϫ 2 ƒ 6 8
45. ƒ x Ϫ 6 ƒ 7 - 4
46. ƒ 3x ϩ 1 ƒ 7 - 3
23. ƒ 3x ϩ 6 ƒ ϭ 0
24. ƒ 4x Ϫ 3 ƒ ϭ 0
47. ƒ x ϩ 4 ƒ 6 - 7
48. ƒ 5x Ϫ 2 ƒ 6 - 2
25. ƒ 3x Ϫ 2 ƒ 7 0
26. ƒ 2x ϩ 7 ƒ 6 0
49. ƒ x ϩ 6 ƒ … 0
50. ƒ x ϩ 7 ƒ 7 0
Thoughts Into Words
51. Explain why the equation ƒ 3x ϩ 2 ƒ ϭ -6 has no real
number solutions.
52. Explain why the inequality ƒ x ϩ 6 ƒ 6 - 4 has no real
number solutions.
456
Chapter 11 • Additional Topics
Further Investigations
A conjunction such as x Ͼ Ϫ2 and x Ͻ 4 can be written in
a more compact form Ϫ2 Ͻ x Ͻ 4, which is read as “Ϫ2 is
less than x, and x is less than 4.” In other words, x is clamped
between Ϫ2 and 4. The compact form is very convenient for
solving conjunctions as follows:
-3 6 2x Ϫ 1 6 5
For Problems 53– 62, solve the compound inequalities
using the compact form.
53. -2 6 x Ϫ 6 6 8
54. -1 6 x ϩ 3 6 9
55. 1 … 2x ϩ 3 … 11
56. -2 … 3x Ϫ 1 … 14
57. -4 6
-2 6 2x 6 6
Add 1 to the left side,
middle, and right side
-1 6 x 6 3
Divide through by 2
xϪ1
6 2
3
58. 2 6
xϩ1
6 5
4
59. ƒ x ϩ 4 ƒ 6 3
[Hint: ƒ x ϩ 4 ƒ 6 3 implies Ϫ3 Ͻ x ϩ 4 Ͻ 3]
Thus the solution set can be expressed as 5x ƒ -1 Ͻ x Ͻ 36.
60. ƒ x Ϫ 6 ƒ 6 5
61. ƒ 2x Ϫ 5 ƒ 6 7
62. ƒ 3x ϩ 2 ƒ 6 14
Answers to the Concept Quiz
1. True
11.2
2. True
3. True
4. False
5. False
6. True
7. False
8. True
9. True
10. True
3 ϫ 3 Systems of Equations
OBJECTIVES
1
Solve systems of three linear equations in three variables
2
Use a system of three linear equations to solve word problems
When we find the solution set of an equation in two variables, such as 2x ϩ y ϭ 9, we are
finding the ordered pairs that make the equation a true statement. Plotted in two dimensions,
the graph of the solution set is a line.
Now consider an equation with three variables, such as 2x Ϫ y ϩ 4z ϭ 8. A solution set of
this equation is an ordered triple, (x, y, z), which makes the equation a true statement. For
example, the ordered triple (3, 2, 1) is a solution of 2x Ϫ y ϩ 4z ϭ 8 because 2(3) Ϫ 2 ϩ 4(1)
ϭ 8. The graph of the solution set of an equation in three variables is a plane, not a line. In fact,
graphing equations in three variables requires the use of a three-dimensional coordinate system.
A 3 ϫ 3 (read “3 by 3”) system of equations is a system of three linear equations in three
variables. To solve a 3 ϫ 3 system such as
2x Ϫ y ϩ 4z ϭ 5
3x ϩ 2y ϩ 5z ϭ 4
P
Q
4x Ϫ 3y Ϫ z ϭ 11
means to find all the ordered triples that satisfy all three equations. In other words, the solution set of the system is the intersection of the solution sets of all three equations in the system. Using a graphing approach to solve systems of three linear equations in three variables
is not at all practical. However, a graphic analysis will provide insight into the types of possible solutions.
In general, each linear equation in three variables produces a plane. A system of three
such equations produces three planes. There are various ways that the planes can intersect.
For our purposes at this time, however, you need to realize that a system of three linear equations in three variables produces one of the following possible solution sets.
11.2 • 3 ϫ 3 Systems of Equations 457
1. There is one ordered triple that satisfies all three equations. The three planes have a
common point of intersection as indicated in Figure 11.9.
Figure 11.9
2. There are infinitely many ordered triples in the solution set, all of which are coordinates of points on a line common to the planes. This can happen when three planes
have a common line of intersection, as in Figure 11.10(a), or when two of the planes
coincide, and the third plane intersects them, as in Figure 11.10(b).
(a)
(b)
Figure 11.10
3. There are infinitely many ordered triples in the solution set, all of which are coordinates of points on a plane. This happens when the three planes coincide, as illustrated in Figure 11.11.
Figure 11.11
4. The solution set is empty; it is л. This can happen in various ways, as you can see in
Figure 11.12. Notice that in each situation there are no points common to all three planes.
(a) Three parallel planes
(b) Two planes coincide
and the third one is
parallel to the
coinciding planes.
(c) Two planes are
parallel and the third
intersects them in
parallel lines.
(d) No two planes are
parallel, but two of
them intersect in a
line that is parallel
to the third plane.
Figure 11.12