Unit 3 – Linear Inequalites and Absolute Value
Unit 3.1 – Linear Inequalities in One Variable
Inequalities are algebraic expressions related by “is less than”, “is less than or equal to”, etc.
Graphing inequalities on a number line: graph the interval on the number line that represents all
the numbers that satisfy the inequality and use interval notation to write the interval.
For example:
Examples of open intervals:
Set
Interval Notation
Graph
{x | a < x}
(a, ∞)
(
a
{x | a < x < b}
(a, b)
{x | x < b}
(-∞, a)
{x | x is a real #}
(-∞, ∞)
(
a
)
a
Examples of half-open intervals:
Set
Interval Notation
{x | a ≤ x}
[a, ∞)
{x | a ≤ x < b}
{x | x ≤ a}
)
b
Graph
[
a
[a, b)
(-∞, a]
[
a
)
b
]
a
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Unit 3 – Linear Inequalites and Absolute Value
Examples of closed interval:
Set
{x | a ≤ x ≤ b}
Interval Notation
[a, b]
Graph
[
a
]
b
A linear inequality in one variable:
Ax  B  C where A, B, and C are real numbers and A ≠ 0.
Solving linear inequalities requires the same properties as solving linear equations.
Solving linear inequalities with the addition property:
For all real numbers A, B, and C, the inequalities A < B and A + C < B + C are equivalent. In
other words, you can add the same number to both sides of an inequality and not change the
solution set.
Examples:
x  7  12
Solve and show solution set as interval notation and a graph.
x  5
14  2m  3m
Solve and show solution set as interval notation and a graph.
14  m
Note: Rewrite inequalities as necessary so that the variable is on the left. In other words, rewrite
14 ≤ m as m ≥ 14. This will help avoid errors.
Solving linear inequalities using the multiplication property:
What happens when you multiply both sides of the true statement
-2 < 5 by 8? You get another true statement -16 < 40.
What happens when you multiply both sides of the true statement
-2 < 5 by -8? You get a false statement 16 < -40! To make it a true statement you must reverse
the inequality symbol.
Multiplication Property of Inequality
For all real numbers A, B, and C, with C ≠ 0
a) the inequalities A < B and AC < BC are equivalent if C > 0 ;
b) the inequalities A < B and AC > BC are equivalent if C < 0 .
In other words, each side of an inequality may be multiplied (or divided) by a positive number
without changing the direction of the inequality symbol, but multiplying (or dividing) by a
negative number requires we reverse the inequality symbol.
Examples:
5m  30 Solve and show solution set as interval notation and a graph.
 4k  32 Solve and show solution set as interval notation and a graph.
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Unit 3 – Linear Inequalites and Absolute Value
Solving a linear inequality.
1. Simplify each side separately. Use the distributive property if necessary and combine like
terms.
2. Isolate the variable terms on one side of the inequality. Use the addition property.
3. Isolate the variable. Use the multiplication property to isolate the variable. Remember to
reverse the inequality symbol is multiplying (or dividing) by a negative number.
Linear inequalities with three parts
3 x 8
3 x28
In this case x+2 is between 3 and 8 (and doesn’t include 3 or 8).
Note: Be sure to write three part inequalities in the correct order. It is incorrect to write
8  x  2  3 as this implies that 8 is less than 3.
To solve this three part inequality, we would subtract 2 from all three parts.
3  2  x  2  2  8  2 resulting in this 1  x  6 . The solution set is (1, 6).
Solving applied problems using linear inequalities.
Be able to interpret terms such as:
Word Expression
a is less than b
a is greater than b
a is at least b
a is no less than b
a is at most b
a is no more than b
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Interpretation
a<b
a>b
a≥b
a≥b
a≤b
a≤b
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Unit 3 – Linear Inequalites and Absolute Value
Unit 3.2 – Set Operations and Compound Inequalities
The intersection of two sets is defined using the word and. We are looking for elements that are
members of both sets simultaneously.
A
B
Here we are looking for those elements that are ‘in the green’, i.e. members of set A and of set B.
A B  {x | x is an element of A and x is an element of B}.
Example: A = {1, 2, 3, 4} and B = {2, 4, 6} then A
B  2, 4
A compound inequality consists of two inequalities linked by a connective word such as and or
or.
x  1  9 and x  2  3
2x  4 or 3x  6  5
Solving a compound inequality with the word and.
1. Solve each inequality separately.
2. Since the inequalities are joined with and, the solution set includes all numbers that satisfy
both inequalities.
Union of Sets
The union of two sets is defined using the word or. We are looking for elements that are
members of either set (or both).
A
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B
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Unit 3 – Linear Inequalites and Absolute Value
A
B = {x | x is an element of A or x is an element of B}.
Example: A = {1, 2, 3, 4} and B = {2, 4, 6} then A
B  1,2,3, 4,6
Solving a compound inequality with the word or.
3. Solve each inequality separately.
4. Since the inequalities are joined with or, the solution set includes all numbers that satisfy
either inequality.
Unit 3.3 – Absolute Value Equations and Inequalities
Use the distance definition of absolute value. The absolute value of a number, x, represents the
distance from x to 0 on the number line.
What are the solutions to the absolute value equation x  4 ? 4 & -4
In set notation this is {-4, 4}
-4
4
What about x  4 ? Use the distance definition of absolute value to describe the answer. All
numbers that are more than 4 units from 0.
-4
4
In interval notation  ,4  4, 
Another way to look at this is x < -4 or x > 4.
What about x  4 ? Use the distance definition of absolute value to describe the answer. All
numbers that are less than 4 units from 0.
-4
4
In interval notation (-4, 4).
Another way to look at this is -4 < x < 4 (which means x > -4 and x < 4).
Absolute value equations and inequalities take the form:
ax  b  k , ax  b  k , or ax  b  k . (k is a positive number and a is not 0)
The above examples show solution sets of:
x  4 has the same solution set as x = -4 or x =4.
x  4 has the same solution set as x < -4 or x > 4.
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Unit 3 – Linear Inequalites and Absolute Value
x  4 has the same solution set as x > -4 and x < 4 (-4 < x < 4).
Solving Absolute Value Equations and Inequalities
(k is a positive real number)
To solve ax  b  k , solve the compound equation
ax  b  k or ax  b  k . The solution set is usually of the form {p, q}.
p
q
To solve ax  b  k , solve the compound inequality
ax  b  k or ax  b  k . The solution set is of the form  , p   q,  .
p
q
To solve ax  b  k , solve the three part inequality  k  ax  b  k . The solution set is of the
form  p, q  .
p
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q
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