Transcript 4.3
§ 4.3
Equations and Inequalities Involving Absolute
Value
Absolute Value Equations
Rewriting an Absolute Value Equation Without
Absolute Value Bars
If c is a positive real number and X represents any algebraic expression, then
|X| = c is equivalent to X = c or X = -c.
Rewriting an Absolute Value Equation with Two
Absolute Values, Without Absolute Value Bars
If |X| = |Y|, then X = Y or X = -Y.
Blitzer, Intermediate Algebra, 4e – Slide #42
Absolute Value Equations
EXAMPLE
Solve: 3|y + 5| = 12.
SOLUTION
We must first isolate the absolute value expression, |y + 5|.
3|y + 5| = 12
|y + 5| = 4
y+5=4
or
y + 5 = -4
y = -1
y = -9
Divide both sides by 3
Rewrite equation without
absolute value bars
Subtract 5 from both sides
Now we will check the two solutions using the original equations.
Blitzer, Intermediate Algebra, 4e – Slide #43
Absolute Value Equations
CONTINUED
Check -1:
Check -9:
3|y + 5| = 12
3|y + 5| = 12
3|(-1) + 5| =? 12
3|(-9) + 5| =? 12
Substitute the proposed
solutions
Simplify
?
3|-9 + 5| = 12
3|4| = 12
?
3|-4| = 12
Add
3(4) =? 12
3(4) =? 12
Simplify
3|-1 + 5| = 12
12 = 12 true
?
Original equation
?
12 = 12 true
Multiply
The solutions are -1 and -9. We can also say that the solution set is
{-1,-9}.
Blitzer, Intermediate Algebra, 4e – Slide #44
Absolute Value Equations
EXAMPLE
Solve: |3x - 5| = |3x + 5|.
SOLUTION
We rewrite the equation without absolute value bars.
3x - 5 = 3x + 5
or
3x - 5 = -(3x + 5)
We now solve the two equations that do not contain absolute
value bars.
3x - 5 = 3x + 5
-5 = 5
or
3x - 5 = -(3x + 5)
3x - 5 = -3x - 5
6x - 5 = - 5
6x = 0
x=0
Blitzer, Intermediate Algebra, 4e – Slide #45
Absolute Value Equations
CONTINUED
Since the first equation yielded -5 = 5, which is clearly a false
statement, there is no solution for the first equation. However,
the second equation yielded a potentially legitimate solution,
x = 0. We now check it.
Check 0:
|3x - 5| = |3x + 5|
Original equation
?
Replace x with 0
|3(0) – 5| = |3(0) + 5|
Multiply
|0 – 5| =? |0 + 5|
|-5| =? |5|
Simplify
5 = 5 true
Simplify
The solution is 0. We can also say that the solution set is {0}.
Blitzer, Intermediate Algebra, 4e – Slide #46
Absolute Value Inequalities
Solving an Absolute Value Inequality
If X is an algebraic expression and c is a positive number,
1) The solutions of |X| < c are the numbers that satisfy –c < X < c.
2) The solutions of |X| > c are the numbers that satisfy X < -c or X > c.
These rules are valid if < is replaced by and > is replaced by .
Absolute Value Inequalities with Unusual
Solution Sets
If X is an algebraic expression and c is a negative number,
1) The inequality |X| < c has no solution.
2) The inequality |X| > c is true for all real numbers for which X is
defined.
Blitzer, Intermediate Algebra, 4e – Slide #47
Absolute Value Inequalities
EXAMPLE
Solve and graph the solution set on a number line: |x - 2| > 5.
SOLUTION
We rewrite the inequality without absolute value bars.
x - 2 < -5
or
x-2>5
We now solve the compound inequality.
x - 2 < -5
or
x-2>5
x < -3
x>7
Add 2 to both sides
Blitzer, Intermediate Algebra, 4e – Slide #48
Absolute Value Inequalities
CONTINUED
The solution set consists of all numbers that are less than -3 or
greater than 7. The solution set is {x|x < -3 or x > 7}, or, in interval
notation, ,3 7, . The graph of the solution set is shown as
follows:
x < -3
or
x>7
)
-6
-5
-4
-3
(
-2
-1
0
1
2
3
4
5
Blitzer, Intermediate Algebra, 4e – Slide #49
6
7
Absolute Value Inequalities
EXAMPLE
Solve and graph the solution set on a number line: 3x 1 2 20 .
SOLUTION
We rewrite the inequality without absolute value bars.
20 3x 1 2 20
We now solve the compound inequality.
20 3x 3 2 20
Multiply
20 3x 1 20
Simplify
19 3x 21
Add 1 to all three sides
19
x7
3
Divide all three sides by 3
Blitzer, Intermediate Algebra, 4e – Slide #50
Absolute Value Inequalities
CONTINUED
19
3
The solution set is all real numbers greater than or equal to
19
and less than or equal to 7, denoted by x | 319 x 7 or [ 3 ,7].
The graph of the solution set is as follows.
19
x7
3
[
-7
-6
]
-5
-4
-3
-2
-1
0
1
2
3
4
5
19
3
Blitzer, Intermediate Algebra, 4e – Slide #51
6
7
Absolute Value Inequalities
EXAMPLE
The specifications for machine parts are given with tolerance
limits that describe a range of measurements for which the part
is acceptable. In this example, x represents the length of a
machine part, in centimeters. The tolerance limit is 0.01
centimeters. Solve: x 9.4 0.01. If the length of the machine
part is supposed to be 9.4 centimeters, interpret the solution.
SOLUTION
First we solve the given inequality.
x 9.4 0.01
0.01 x 9.4 0.01
9.39 x 9.41
Original inequality
Rewrite inequality
Add 9.4 to all three sides
Blitzer, Intermediate Algebra, 4e – Slide #52
Absolute Value Inequalities
CONTINUED
Therefore, the solution set is x | 9.39 x 9.41 or 9.39,9.41. So,
a machine part that is supposed to be 9.4 centimeters is
acceptable between a low of 9.39 centimeters and a high of
9.41 centimeters.
Blitzer, Intermediate Algebra, 4e – Slide #53