1. College Algebra Fifth Edition
James Stewart  Lothar Redlin  Saleem Watson

2. Equations and Inequalities1

3. Absolute Value Equations and Inequalities
1.7

4. Absolute Value of a Number
Recall from Section P.3 that the absolute value of a number a is given by

5. Absolute Value of a Number
It represents the distance from a to the origin on the real number line

6. Absolute Value of a Number
More generally, |x – a| is the distance between x and a on the real number line.
The figure illustrates the fact that the distance between 2 and 5 is 3.

7. Absolute Value Equations

8. E.g. 1—Solving an Absolute Value Equation
Solve the inequality |2x – 5| = 3
The equation |2x – 5| = 3 is equivalent to two equations 2x – 5 = 3 or 2x – 5 = – x = 8 or x = x = 4 or x = 1
The solutions are 1 and 4.

9. E.g. 2—Solving an Absolute Value Equation
Solve the inequality 3|x – 7| + 5 = 14
First, we isolate the absolute value on one side of the equal sign. 3|x – 7| + 5 = |x – 7| = |x – 7| = 3 x – 7 = 3 or x – 7 = – x = 10 or x = 4
The solutions are 4 and 10.

10. Absolute Value Inequalities

11. Properties of Absolute Value Inequalities
We use these properties to solve inequalities that involve absolute value.

12. Properties of Absolute Value Inequalities
These properties can be proved using the definition of absolute value.
For example, to prove Property 1, the inequality |x| < c says that the distance from x to 0 is less than c.

13. Properties of Absolute Value Inequalities
From the figure, you can see that this is true if and only if x is between -c and c.

14. E.g. 3—Absolute Value Inequality
Solution 1
Solve the inequality |x – 5| < 2
The inequality |x – 5| < 2 is equivalent to –2 < x – 5 < 2 (Property 1) < x < 7 (Add 5)
The solution set is the open interval (3, 7).

15. E.g. 3—Absolute Value Inequality
Solution 2
Geometrically, the solution set consists of:
All numbers x whose distance from 5 is less than 2.

16. E.g. 3—Absolute Value Inequality
Solution 2
From the figure, we see that this is the interval (3, 7).

17. E.g. 4—Solving an Absolute Value Inequality
Solve the inequality |3x + 2| ≥ 4
By Property 4, the inequality |3x + 2| ≥ 4 is equivalent to: x + 2 ≥ 4 or 3x + 2 ≤ – x ≥ x ≤ –6 (Subtract 2) x ≥ 2/ x ≤ –2 (Divide by 3)

18. E.g. 4—Solving an Absolute Value Inequality
So, the solution set is: {x | x ≤ –2 or x ≥ 2/3} = (-∞, -2] [2/3, ∞)

19. E.g. 5—Piston Tolerances
The specifications for a car engine indicate that the pistons have diameter in. with a tolerance of in.
This means that the diameters can vary from the indicated specification by as much as in. and still be acceptable.

20. E.g. 5—Piston Tolerances
Find an inequality involving absolute values that describes the range of possible diameters for the pistons.
Solve the inequality.

21. Let d represent the actual diameter of a piston.
E.g. 5—Piston Tolerances
Example (a)
Let d represent the actual diameter of a piston.
The difference between the actual diameter (d) and the specified diameter (3.8745) is less than
So, we have |d – | ≤

22. The inequality is equivalent to –0.0015 ≤ d – 3.8745 ≤ 0.0015
E.g. 5—Piston Tolerances
Example (b)
The inequality is equivalent to
– ≤ d – ≤
≤ d ≤
So, acceptable piston diameters may vary between in. and in.