Download Absolute Value of a Number

College Algebra
Sixth Edition
James Stewart  Lothar Redlin

Saleem Watson
1
Equations and
Inequalities
1.9
Solving Absolute Value
Equations and Inequalities
Absolute Value of a Number
Recall from Section P.2 that the absolute
value of a number a is given by
a if a  0
a 
a if a  0
Absolute Value of a Number
It represents the distance from a to the origin
on the real number line
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.
Absolute Value Equations
E.g. 1—Solving an Absolute Value Equation
Solve the equation
|2x – 5| = 3
• The equation |2x – 5| = 3 is equivalent
to two equations:
2x – 5 = 3 or
2x – 5 = –3
2x = 8 or
2x = 2
x = 4 or
x=1
• The solutions are 1 and 4.
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 = 14
3|x – 7| = 9
|x – 7| = 3
x–7=3
or
x – 7 = –3
x = 10 or
x=4
• The solutions are 4 and 10.
Absolute Value Inequalities
Properties of Absolute Value Inequalities
We use these properties to solve
inequalities that involve absolute value.
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.
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.
E.g. 3—Solving an Absolute Value Equation
Solution 1
Solve the inequality
|x – 5| < 2
• The inequality |x – 5| < 2 is equivalent
to
–2 < x – 5 < 2
(Property 1)
3<x<7
(Add 5)
• The solution set is the open interval (3, 7).
E.g. 3—Solving an Absolute Value Equation Solution 2
Geometrically, the solution set
consists of:
• All numbers x whose distance from 5
is less than 2.
E.g. 3—Solving an Absolute Value Equation
Solution 2
From the figure, we see that this is
the interval (3, 7).
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
3x + 2 ≥ 4
3x ≥ 2
x ≥ 2/3
or
3x + 2 ≤ –4
3x ≤ –6
x ≤ –2
(Subtract 2)
(Divide by 3)
E.g. 4—Solving an Absolute Value Inequality
So, the solution set is
{x | x ≤ –2 or x ≥ 2/3}
= (– ∞, – 2]
[2/3, ∞)
E.g. 5—Piston Tolerances
The specifications for a car engine indicate
that the pistons have diameter 3.8745 in.
with a tolerance of 0.0015 in.
• This means that the diameters can vary from
the indicated specification by as much as
0.0015 in. and still be acceptable.
E.g. 5—Piston Tolerances
a) Find an inequality involving absolute
values that describes the range of
possible diameters for the pistons.
b) Solve the inequality.
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 0.0015.
• So, we have
|d – 3.8745| ≤ 0.0015
E.g. 5—Piston Tolerances
Example (b)
The inequality is equivalent to
–0.0015 ≤ d – 3.8745 ≤ 0.0015
3.8730 ≤ d ≤ 3.8760
• So, acceptable piston diameters may vary
between 3.8730 in. and 3.8760 in.