1 Understand absolute value on the number line.
2 Solve equations of the form x = a, a 7 0.
3 Solve inequalities of the form x 6 a, a 7 0.
4 Solve inequalities of the form x 7 a, a 7 0.
5 Solve inequalities of the form x 6 a or x 7 a, a 6 0.
6 Solve inequalities of the form x 6 0, x … 0, x 7 0, or x Ú 0.
7 Solve equations of the form
0x0 = 0y0.
1 Understand Absolute Value on the Number Line
Absolute value
The absolute value of a number x, symbolized x, is the distance x is from 0 on the number line.
3 = 3 because the number 3 is 3 units from 0 on the number line.
-3 = 3 because the number -3 is 3 units from 0 on the number line.
Now consider the equation x = 3. We are looking for values of x that are exactly 3 units from 0 on the number line. Thus, the solutions to x = 3 are x = 3 and x = -3 (see Fig. 2.15a on page 128).
Next, consider the inequality x 6 3. We are looking for values of x that are less than 3 units from 0 on the number line. Thus, the solutions to x 6 3 are the numbers that are between -3 and 3 on the number line (see Fig. 2.15b on page 128).
Finally, consider the inequality x 7 3. We are looking for values of x that are greater than 3 units from 0 on the number line. Thus, the solutions to x 7 3 are the numbers that are less than -3 or greater than 3 on the number line (see Fig. 2.15c on page 128).
a) counting numbers b) whole numbers c) rational numbers d) real numbers
We will use these examples and their illustrations on the number line to develop methods for solving equations and inequalities involving absolute value.
2 Solve Equations of the Form 0 x 0 = a, a + 0
When solving an equation of the form 0x0 = a, a 7 0, we are determining the values that are exactly a units from 0 on the number line.
3 2 1 0 23 22 21
3 units 3 units uxu 3
(a) FIGURE 2.15
3 2 1 0 23 22 21
uxu , 3 less than
3 units
less than 3 units
(b)
3 2 1 0 23 22 21
uxu . 3 greater than
3 units greater than
3 units
(c)
To Solve Equations of the Form 0x0 = a
If 0x0 = a and a 7 0, then x = a or x = -a.
EXAMPLE 1 Solve each equation.
a) 0x0 = 6 b) 0y0 = 0 c) 0z0 = -6
Solution
a) We need to determine the values of x that are exactly 6 units from 0 on the number line. Thus, x = 6 or x = -6. The solution set is 5-6, 66.
b) The only real number whose absolute value equals 0 is 0. The solution set is {0}.
c) The absolute value of a number is never negative. Thus, there are no solutions to this equation. The solution set is ∅.
Now Try Exercise 13 Understanding Algebra
The absolute value of a number is never negative.
EXAMPLE 2 Solve the equation 02w - 10 = 5.
Solution We are looking for the values of w such that 2w - 1 is exactly 5 units from 0 on a number line. Thus, the quantity 2w - 1 must be equal to 5 or -5.
2w - 1 = 5 or 2w - 1 = -5 2w = 6 2w = -4 w = 3 w = -2 Check w = 3 02w - 10 = 5
02132 - 10 ≟5
06 - 10 ≟5
050 ≟5
5 = 5 True
w = -2 02w - 10 = 5 021-22 - 10 ≟ 5 0-4 - 10 ≟ 5 0-50 ≟ 5 5 = 5 True The solutions 3 and -2 each result in 2w - 1 being 5 units from 0 on the number line.
The solution set is 5-2, 36.
Now Try Exercise 19 Consider the equation 02w - 10 - 3 = 2. The first step in solving this equation is to isolate the absolute value term. We do this by adding 3 to both sides of the equation. This results in the equation 2w - 1 = 5, which we solved in Example 2.
3 Solve Inequalities of the Form x 6 a, a 7 0
Recall our discussion earlier regarding x 6 3. The solutions to x 6 3 are the numbers that are between -3 and 3 on the number line (See Fig. 2.15b on page 128). Similarly, the solutions to x 6 a are the numbers that are between -a and a on the number line.
To solve inequalities of the form 0x0 6 a, we can use the following procedure.
To Solve Inequalities of the Form 0x0 * a
If 0x0 6 a and a 7 0, then -a 6 x 6 a.
EXAMPLE 3 Solve the inequality 02x - 30 6 5.
Solution The solution to this inequality will be the set of values such that the dis- tance between 2x - 3 and 0 on a number line will be less than 5 units (see Fig. 2.16).
Using Figure 2.16, we can see that -5 6 2x - 3 6 5.
Solving, we get -5 6 2x - 3 6 5 -2 6 2x 6 8 -1 6 x 6 4 The solution set is 5x-1 6 x 6 46.
Now Try Exercise 33
5 6 4 3 2 1 0 24
25
26 232221
2x 2 3
FIGURE 2.16
EXAMPLE 5 Solve the inequality 07.8 - 4x0 - 5.3 6 14.1 and graph the solution on a number line.
Solution First isolate the absolute value by adding 5.3 to both sides of the inequality.
Then solve as in the previous examples.
07.8 - 4x0 - 5.3 6 14.1
07.8 - 4x0 6 19.4
-19.4 6 7.8 - 4x 6 19.4 -27.2 6 -4x 6 11.6 -27.2
-4 7 -4x -4 7 11.6
-4
6.8 7 x 7 -2.9 or -2.9 6 x 6 6.8
5
4 6 7 8
3 2 1 0 24
6.8 22.9
232221
The solution set is 5x-2.9 6 x 6 6.86.
Now Try Exercise 39
EXAMPLE 4 Solve the inequality 02x + 10 … 9 and graph the solution on a number line.
Solution Since this inequality is of the form 0x0 … a, we write -9 … 2x + 1 … 9
-10 … 2x … 8 -5 … x … 4
5 4 3 2 1 0 24
25
4 25
26 232221
Now Try Exercise 75 The solution set is 5x-5 … x … 46.
4 Solve Inequalities of the Form x 7 a, a 7 0
Recall our discussion earlier regarding x 7 3. The solutions to x 7 3 are the numbers that are less than -3 or greater than 3 on the number line (See Fig. 2.15c on page 128).
Similarly, the solutions to x 7 a are the numbers that are less than -a or greater than a on the number line.
To solve inequalities of the form 0x0 7 a, we can use the following procedure.
To Solve Inequalities of the Form 0x0 + a
If 0x0 7 a and a7 0, then x 6 -a or x 7 a.
EXAMPLE 6 Solve the inequality 02x - 30 7 5 and graph the solution on a number line.
Solution The solution to 02x - 30 7 5 is the set of values such that the distance be- tween 2x - 3 and 0 on a number line will be greater than 5. The quantity 2x - 3 must be either less than -5 or greater than 5 (see Fig. 2.17).
5 6 7 8 4
3 2 1 0 24
25 26 27
28 232221
2x 2 3 2x 2 3
FIGURE 2.17
Since 2x - 3 must be either less than -5 or greater than 5, we set up and solve the fol- lowing compound inequality:
2x - 3 6 -5 or 2x - 3 7 5
2x 6 -2 2x 7 8
x 6 -1 x 7 4
5 6 7 4 3 2 1 0 24
25 232221
The solution set to 02x - 30 7 5 is 5xx 6 -1 or x 7 46.
Now Try Exercise 51
EXAMPLE 7 Solve the inequality 02x + 10 Ú 7 and graph the solution on the number line.
Solution The quantity 2x + 1 must be less than or equal to -7 or greater than or equal to 7.
2x + 1 … -7 or 2x + 1 Ú 7 2x … -8 2x Ú 6 x … -4 x Ú 3
5 6 4 3 2 1 0 4
5
6 321
Now Try Exercise 53
EXAMPLE 8 Solve the inequality `3x - 4
2 ` Ú 9 and graph the solution on a number line.
Solution Since this inequality is of the form 0x0 Ú a, we write 3x - 4
2 … -9 or 3x - 4
2 Ú 9
Now multiply both sides of each inequality by the least common denominator, 2. Then solve each inequality.
2 a3x - 4
2 b … -9#2 or 2 a3x - 4
2 b Ú 9#2
3x - 4 … -18 3x - 4 Ú 18
3x … -14 3x Ú 22
x … -14
3 x Ú 22
3
5 6 7 8 9 10 4
3 2 1 0 24
25
2726 232221
22 3 14
23
Now Try Exercise 57
Understanding Algebra
Any inequality of the form x 6a, where a is a negative number, will have solution set ∅.
EXAMPLE 9 Solve the inequality 05x - 110 + 7 6 3.
Solution First, isolate the absolute value by subtracting 7 from both sides of the inequality.
05x - 110 + 7 6 3
05x - 110 6 -4
Since 05x - 110 is always greater than or equal to 0 for any real number x, there are no solutions to this inequality. The solution set is ∅.
Now Try Exercise 41 Now consider the inequality 0x0 7 -3. Since 0x0 will always have a value greater than or equal to 0 for any real number x, this inequality will always be true and the solu- tion is the set of all real numbers, ℝ.
EXAMPLE 10 Solve the inequality 05x + 30 + 4 Ú -9.
Solution Begin by subtracting 4 from both sides of the inequality.
05x + 30 + 4 Ú -9
05x + 30 Ú -13
Since 05x + 30 will always be greater than or equal to 0 for any real number x, this in- equality is true for all real numbers. Thus, the solution is the set of all real numbers, ℝ.
Now Try Exercise 59
5 Solve Inequalities of the Form x 6 a or x 7 a, a 6 0
Consider the inequality 0x0 6 -3. Since 0x0 will always have a value greater than or equal to 0 for any real number x, this inequality can never be true, and the solution is the empty set, ∅.
HELPFUL HINT
Some general information about equations and inequalities containing absolute value follows.
For real numbers a, b, and c, where a ≠0 and c 7 0:
Form of Equation
or Inequality The Solution Will Be:
Solution on a Number Line:
0ax +b0 = c Two distinct numbers, p and q p q
0ax +b0 6 c The set of numbers between two
numbers, p 6 x 6 q p q
0ax +b0 7 c The set of numbers less than one number or greater than a second number, x6 p or x 7 q
q p
Understanding Algebra
Any inequality of the form x 7a, where a is a negative number, will have solution set ℝ.
6 Solve Inequalities of the Form x 6 0, x … 0, x 7 0, or x Ú 0
Each of the following examples is based on the fact that the absolute value of a number can never be negative.
Inequality Solution Set Explanation
x - 5 6 0 ∅ The absolute value of a number can never be 6 0.
x - 5 … 0 {5} The absolute value of a number can never be 6 0 but it can equal 0. When x = 5, we get
x -5 … 0 5 -5 … 0
0 … 0 True
x - 5 7 0 5xx ≠ 56 Substituting any real number for x except for 5 makes x - 5 positive. When x = 5, we get
x -5 7 0 5 -5 7 0
0 7 0 False
x - 5 Ú 0 ℝ The absolute value of any number is always Ú 0.
EXAMPLE 11 Solve each inequality.
a) 0x + 20 7 0 b) 03x - 80 … 0 c) 02x - 30 Ú 0 d) 05x + 40 6 0
Solution
a) The inequality will be true for every value of x except -2. The solution set is 5xx 6 -2 or x 7 -26.
b) Determine the number that makes the absolute value equal to 0 by setting the ex- pression within the absolute value sign equal to 0 and solving for x.
3x - 8 = 0 3x = 8 x = 8 3 The inequality will be true only when x = 8
3. The solution set is e8 3f.
c) Since the absolute value of any number is always greater than or equal to 0, the solution set is all real numbers, or ℝ.
d) Since the absolute value of any number can never be less than zero, there is no solution and the solution set is ∅.
Now Try Exercise 61
7 Solve Equations of the Form 0 x 0 = 0 y 0
Now we will discuss absolute value equations where an absolute value appears on both sides of the equation.
When solving an absolute value equation with an absolute value expression on each side of the equals sign, the two expressions must have the same absolute value. Therefore, the expressions must be equal to each other or be opposites of each other.
To Solve Equations of the Form 0x0 = 0y0
If 0x0 = 0y0, then x = y or x = -y.
Understanding Algebra
If x = y, then x = y or x = -y.
The solution set is e4 3, 10f.
Now Try Exercise 63
EXAMPLE 12 Solve the equation 0z + 30 = 02z - 70.
Solution If we let z + 3 be x and 2z - 7 be y, this equation is of the form 0x0 = 0y0. Using the procedure given above, we obtain the two equations
z + 3 = 2z - 7 or z + 3 = -12z - 72 Now solve each equation.
z + 3 = 2z - 7 or z + 3 = -12z - 72 3 = z - 7 z + 3 = -2z + 7 10 = z 3z + 3 = 7
3z = 4 z = 4 3 Check z = 10 0z + 30 = 02z - 70
010 + 30 ≟ 021102 - 70
0130 ≟ 020 - 70
0130 ≟ 0130
13 = 13 True
z = 4
3 0z + 30 = 02z - 70
`4
3 + 3` ≟ `2a4 3b - 7`
`13 3 ` ≟ `8
3 - 21 3 `
`13
3 ` ≟ `-13 3 ` 13
3 = 13 3 True
EXAMPLE 13 Solve the equation 05x - 30 = 017 - 5x0. Solution 5x - 3 = 17 - 5x or 5x - 3 = -117 - 5x2
10x - 3 = 17 5x - 3 = -17 + 5x
10x = 20 -3 = -17 False
x = 2
Since the equation 5x - 3 = -117 - 5x2 results in a contradiction, the absolute value equation has only one solution. The solution set is {2}.
Now Try Exercise 65
Summary of Procedures for Solving Equations and Inequalities Containing Absolute Value
For a 7 0,
If 0x0 = a, then x = a or x = -a.
If 0x0 6 a, then -a 6 x 6 a.
If 0x0 7 a, then x 6 -a or x 7 a.
If 0x0 = 0y0, then x = y or x = -y.