Download ► Absolute Value Equations and Inequalities

 Absolute Value
Inequalities
1.7
Equations and
Absolute Value of a Number
 
Recall from Section P.3 that the absolute value of a
number a is given by
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 = 8
or
x=4
or
The solutions are 1 and 4.
2x – 5 = –3
2x = 2
x=1
E.g. 2—Solving an Absolute Value Equation
  Solve
 
the equation
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.
E.g. 3—Absolute Value Inequality
  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)
Or if you want to have two equations
–2 < x – 5
x–5<2
 
The solution set is the open interval (3, 7).
E.g. 3—Absolute Value Inequality
  Geometrically, the
solution set
consists of:
  All
numbers x whose distance from 5
is less than 2.
Solution 2
E.g. 3—Absolute Value Inequality
  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.
 
A tolerance - a concept which means acceptable
variance, in this case 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
  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
  The
 
 
 
Example (b)
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.
Exercise 1.7 - 84
  A
company manufactures industrial laminates
of thickness 0.020 in with a tolerance of
0.003
  What is the inequality?
  Let x=0.020, then |x-0.020|
−0.003 ≤ x − 0.020 ≤ 0.003
Exercise
 
Now solve for x to find out the sizes of this laminate
-0.003<= x-0.020 <= 0.003
+0.020
+0.020
0.017 <= x <= 0.023