Transcript Algebra 2
Algebra 2
Lesson 1-6 Part 2
Absolute Value Inequalities
Goals
Goal
• To write and solve
inequalities involving
absolute values.
Rubric
Level 1 – Know the goals.
Level 2 – Fully understand the
goals.
Level 3 – Use the goals to
solve simple problems.
Level 4 – Use the goals to
solve more advanced problems.
Level 5 – Adapts and applies
the goals to different and more
complex problems.
Essential Question
Big Idea: Absolute Value Inequalities
• How can you solve an absolute value inequality?
• Students will understand that an absolute value
quantity is nonnegative.
Vocabulary
• None
Absolute Value
Inequalities
• When an inequality contains an absolute-value
expression, it can be written as a compound
inequality.
• The inequality |x| < 5 describes all real numbers
whose distance from 0 is less than 5 units.
• The solutions are all numbers between –5 and 5, so
|x|< 5 can be rewritten as –5 < x < 5, or as x > –5
AND x < 5.
“Less Than” - Conjunction
The solutions of |x| < 3 are the points that are less than 3 units
from zero. The solution is a conjunction: –3 < x < 3.
Helpful Hint
Think: Less thand inequalities involving < or ≤
symbols are conjunctions.
“Less Than” Absolute
Value Inequalities
Example:
Solve the inequality and graph the solutions.
|x|– 3 < –1
Since 3 is subtracted from |x|, add 3 to both
sides to undo the subtraction.
|x|– 3 < –1
+3 +3
|x| < 2
x > –2 AND x < 2
2 units
–2
–1
Write as a compound inequality.
The solution set is {x: –2 < x < 2}.
2 units
0
1
2
Example:
Solve the inequality and graph the solutions.
|x – 1| ≤ 2
x – 1 ≥ –2 AND x – 1 ≤ 2
+1 +1
+1 +1
x ≥ –1 AND
Write as a compound inequality.
Solve each inequality.
x≤3
Write as a compound inequality.
The solution set is
{x: –1 ≤ x ≤ 3}.
–3
–2
–1
0
1
2
3
Procedure Solving “Less
Than” Inequalities
Solving Absolute Inequalities:
1. Isolate the absolute value.
2. Break it into 2 inequalities (and statement) –
one positive and the other negative
reversing the sign.
3. Solve both inequalities.
4. Check both answers.
Helpful Hint
Just as you do when solving absolute-value equations, you
first isolate the absolute-value expression when solving
absolute-value inequalities.
Your Turn:
Solve the inequality and graph the solutions.
2|x| ≤ 6
2|x| ≤ 6
2
2
|x| ≤ 3
Since x is multiplied by 2, divide both sides
by 2 to undo the multiplication.
Write as a compound inequality.
The solution set is {x: –3 ≤ x ≤ 3}.
x ≥ –3 AND x ≤ 3
3 units
–3
–2
–1
3 units
0
1
2
3
Your Turn:
Solve each inequality and graph the solutions.
|x + 3|– 4.5 ≤ 7.5
|x + 3|– 4.5 ≤ 7.5
+ 4.5 +4.5
Since 4.5 is subtracted from |x + 3|,
add 4.5 to both sides to undo
the subtraction.
|x + 3| ≤ 12
x + 3 ≥ –12 AND x + 3 ≤ 12
–3 –3
–3 –3
Write as a compound inequality.
The solution set is
{x: –15 ≤ x ≤ 9}.
x ≥ –15 AND x ≤ 9
–20
–15
–10
–5
0
5
10
15
“Greater Than” Absolute
Value Inequalities
• The inequality |x| > 5 describes all real numbers
whose distance from 0 is greater than 5 units.
• The solutions are all numbers less than –5 or
greater than 5.
• The inequality |x| > 5 can be rewritten as the
compound inequality x < –5 OR x > 5.
“Greater Than” - Disjunction
The solutions of |x| > 3 are the points that are more than 3 units from
zero. The solution is a disjunction:
x < –3 or x > 3.
Helpful Hint
Think: Greator inequalities involving > or ≥
symbols are disjunctions.
“Greater Than”
Absolute Value
Inequalities
Example:
Solve the inequality and graph the solutions.
|x| + 14 ≥ 19
|x| + 14 ≥ 19
– 14 –14
|x| ≥ 5
x ≤ –5 OR x ≥ 5
Since 14 is added to |x|, subtract 14 from both
sides to undo the addition.
Write as a compound inequality. The solution
set is {x: x ≤ –5 OR x ≥ 5}.
5 units
–10
–8 –6 –4
–2
5 units
0
2
4
6
8 10
Example:
Solve the inequality and graph the solutions.
3 + |x + 2| > 5
Since 3 is added to |x + 2|, subtract 3
from both sides to undo the addition.
3 + |x + 2| > 5
–3
–3
|x + 2| > 2
x + 2 < –2 OR x + 2 > 2
–2 –2
–2 –2
Write as a compound inequality. Solve
each inequality.
Write as a compound inequality.
x < –4 OR x > 0
–10
–8 –6 –4
–2
The solution set is
{x: x < –4 or x > 0}.
0
2
4
6
8
10
Procedure Solving
“Greater Than”
Inequalities
Solving Absolute Inequalities:
1. Isolate the absolute value.
2. Break it into 2 inequalities (or statement) – one
positive and the other negative reversing the
sign.
3. Solve both inequalities.
4. Check both answers.
Your Turn:
Solve each inequality and graph the solutions.
|x| + 10 ≥ 12
|x| + 10 ≥ 12
– 10 –10
|x|
Since 10 is added to |x|, subtract 10 from
both sides to undo the addition.
≥ 2
x ≤ –2 OR x ≥ 2
Write as a compound inequality. The solution set
is {x: x ≤ –2 or x ≥ 2}.
2 units 2 units
–5 –4 –3 –2 –1
0
1
2
3
4
5
Your Turn:
Solve the inequality and graph the solutions.
Since is added to |x + 2 |, subtract
from both sides to undo the addition.
Write as a compound inequality.
Solve each inequality.
OR
x ≤ –6
x≥1
–7 –6 –5 –4 –3 –2 –1
Write as a compound
inequality. The solution
set is {x: x ≤ –6 or x ≥ 1}
0
1
2
3
Example: Application
A pediatrician recommends that a baby’s bath water be 95°F,
but it is acceptable for the temperature to vary from this amount
by as much as 3°F. Write and solve an absolute-value inequality
to find the range of acceptable temperatures. Graph the
solutions.
Let t represent the actual water temperature.
The difference between t and the ideal temperature is at
most 3°F.
t – 95
≤
3
Example: Continued
t – 95
≤
3
|t – 95| ≤ 3
t – 95 ≥ –3 AND t – 95 ≤ 3
+95 +95
+95 +95
t
≥ 92 AND t
90
Solve the two
inequalities.
≤ 98
92
94
96
98
100
The range of acceptable temperature is 92 ≤ t ≤ 98.
Your Turn:
A dry-chemical fire extinguisher should be pressurized to 125
psi, but it is acceptable for the pressure to differ from this value
by at most 75 psi. Write and solve an absolute-value inequality
to find the range of acceptable pressures. Graph the solution.
Let p represent the desired pressure.
The difference between p and the ideal pressure is at
most 75 psi.
p – 125
≤
75
Your Turn: Continued
p – 125
≤
75
|p – 125| ≤ 75
p – 125 ≥ –75 AND p – 125 ≤ 75
+125 +125
+125 +125
p
≥
50 AND p
25
50
Solve the two
inequalities.
≤ 200
75
100
125
150
175
200
The range of pressure is 50 ≤ p ≤ 200.
225
Solutions to Absolute Value
Inequalities
• When solving an absolute-value inequality, you may
get a statement that is true for all values of the
variable.
• In this case, all real numbers are solutions of the
original inequality.
• If you get a false statement when solving an absolutevalue inequality, the original inequality has no
solutions. Its solution set is ø.
Example:
Solve the inequality.
|x + 4|– 5 > – 8
|x + 4|– 5 > – 8
+5 +5
|x + 4| > –3
Add 5 to both sides.
Absolute-value expressions are
always nonnegative. Therefore,
the statement is true for all real
numbers.
The solution set is all real numbers.
Example:
Solve the inequality.
|x – 2| + 9 < 7
|x – 2| + 9 < 7
–9 –9
|x – 2| < –2
Subtract 9 from both sides.
Absolute-value expressions are
always nonnegative. Therefore,
the statement is false for all values
of x.
The inequality has no solutions. The solution set is ø.
Remember!
An absolute value represents a distance, and distance
cannot be less than 0.
Your Turn:
Solve the inequality.
|x| – 9 ≥ –11
|x| – 9 ≥ –11
+9 ≥ +9
|x| ≥ –2
Add 9 to both sides.
Absolute-value expressions are
always nonnegative. Therefore,
the statement is true for all real
numbers.
The solution set is all real numbers.
Your Turn:
Solve the inequality.
4|x – 3.5| ≤ –8
4|x – 3.5| ≤ –8
4
4
|x – 3.5| ≤ –2
Divide both sides by 4.
Absolute-value expressions are
always nonnegative. Therefore,
the statement is false for all values
of x.
The inequality has no solutions. The solution set is ø.
SUMMARY
Absolute Value Inequalities (< , )
Inequalities of the Form < or Involving Absolute Value
If a is a positive real number and if u is an algebraic expression,
then
|u| < a is equivalent to u < a and u > -a
|u| a is equivalent to u a and u > -a
Note: If a = 0, |u| < 0 has no real solution, |u| 0 is equivalent
to u = 0. If a < 0, the inequality has no real solution.
Absolute Value Inequalities (> , )
Inequalities of the Form > or Involving Absolute Value
If a is a positive real number and u is an algebraic expression, then
|u| > a is equivalent to u < – a or u > a
|u| a is equivalent to u – a or u a.
Procedure
Solving Absolute Inequalities:
1. Isolate the absolute value.
2. Break it into 2 inequalites (“< , ≤ and
statement” – “> , ≥ or statement”) – one
positive and the other negative reversing the
sign.
3. Solve both inequalities.
4. Check both answers.
Solving an Absolute-Value Equation
Recall that |xx is
| isthe
thedistance
distancebetween
betweenxxand
and0.0.IfIf x
| x
| 8,8,then
then
any number between 8 and 8 is a solution of the inequality.
8 7 6 5 4 3 2 1
0
1
2
3
4
5
6
You can use the following properties to solve
absolute-value inequalities and equations.
7
8
SOLVING ABSOLUTE-VALUE INEQUALITIES
SOLVING ABSOLUTE-VALUE INEQUALITIES
| ax b | c
means
ax b c
and
a x b c.
a number,
means valueais
| a xWhen
b | an
c absolute
x less
b than
c
and
a x the
b c.
inequalities are connected by and. When an absolute
value is greater than a number, the inequalities are
connected by or.
| ax b | c
means
ax b c
or
a x b c.
| ax b | c
means
ax b c
or
a x b c.
Solving an Absolute-Value Inequality
Solve | x 4 | < 3
x 4 IS POSITIVE
x 4 IS NEGATIVE
|x4|3
|x4|3
x 4 3
x 4 3
x7
x1
Reverse
inequality
symbol.
The solution is all real numbers greater than 1 and
less than
7.
This can be written as 1 x 7.
–10
–8 –6 –4
–2
0
2
4
6
8
10
Solving an Absolute-Value Inequality
Solve
1 | 3 6 and graph
2x + 1| 2x
IS POSITIVE
2x +the
1 ISsolution.
NEGATIVE
| 2x 1 | 3 6
| 2x 1 | 3 6
2x + 1 IS POSITIVE
| 2x|2x1 | 13| 69
2x + 1 IS NEGATIVE
1 | 69
| 2x|2x1 | 3
2x 1 9
| 2x 1 | 9
2x 10
2x 8
2x 1 9
2x 1 +9
x4
x 5
2x 10
2x 8
The solution is all real numbers greater than or equal
x4
x 5
to 4 or less than or equal to 5. This can be written as
the compound inequality
x 5 or x 4.
Reverse
2x11| +9
| 2x
9
inequality symbol.
6 5 4 3 2 1
0
1
2
3
4
5
6
MORE PROBLEMS FOR YOU
Your Turn:
Solve the compound inequality. Then graph the solution set.
|p – 2| ≤ –6
|p – 2| ≤ –6 and p – 2 ≥ 6
p ≤ –4 and p ≥ 8
Multiply both sides by –2, and
reverse the inequality symbol.
Rewrite the absolute value
as a conjunction.
Add 2 to both sides of
each inequality.
Because no real number satisfies both p ≤ –4 and
p ≥ 8, there is no solution. The solution set is ø.
Your Turn:
Solve the compound inequality. Then graph the solution set.
|x – 5| ≤ 8
Multiply both sides by 2.
x – 5 ≤ 8 and x – 5 ≥ –8
Rewrite the absolute
value as a conjunction.
x ≤ 13 and x ≥ –3
Add 5 to both sides
of each inequality.
Continued
The solution set is {x|–3 ≤ x ≤ 13}.
–10
–5
0
5
10
15
20
25
Your Turn:
Solve the compound inequality. Then graph the solution set.
|2x +7| ≤ 3
Multiply both sides by 3.
2x + 7 ≤ 3 and 2x + 7 ≥ –3
Rewrite the absolute
value as a conjunction.
2x ≤ –4 and 2x ≥ –10
Subtract 7 from both
sides of each inequality.
x ≤ –2 and x ≥ –5
Divide both sides of
each inequality by 2.
Continued
The solution set is {x|–5 ≤ x ≤ 2}.
–6 –5 –3 –2 –1
0
1
2
3 4
Your Turn:
Solve the compound inequality. Then graph the solution set.
–2|x +5| > 10
|x + 5| < –5
x + 5 < –5 and x + 5 > 5
x < –10 and x > 0
Divide both sides by –2, and
reverse the inequality symbol.
Rewrite the absolute value
as a conjunction.
Subtract 5 from both
sides of each inequality.
Because no real number satisfies both x < –10 and x > 0, there is
no solution. The solution set is ø.
Your Turn:
Solve:
1. |3k| + 11 > 8
2. –2|u + 7| ≥ 16
3. |1 – 2x| > 7
All Real Numbers
No Solution
x < –3 or x > 4
Assignment
• Section 1-6 Part 2, Pg 49 – 52; #1 – 5 all, 6
– 52 even