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2.4 – Linear Inequalities and Problem Solving
An inequality is a statement that contains one of the
symbols: < , >, ≤ or ≥.
Equations
Inequalities
x=3
12 = 7 – 3y
x>3
12 ≤ 7 – 3y
A solution of an inequality is a value of the variable that
makes the inequality a true statement.
The solution set of an inequality is the set of all
solutions.
2.4 – Linear Inequalities and Problem Solving
2.4 – Linear Inequalities and Problem Solving
2.4 – Linear Inequalities and Problem Solving
Example
Graph each set on a number line and then write it in
interval notation.
a. {x | x  2}
b. {x | x  1}
c. {x | 0.5  x  3}
a.
[2,  )
b.
c.
(0.5, 3]
2.4 – Linear Inequalities and Problem Solving
Addition Property of Inequality
If a, b, and c are real numbers, then
a < b and a + c < b + c
are equivalent inequalities.
2.4 – Linear Inequalities and Problem Solving
Example
Solve: 3x  4  2 x  6 Graph the solution set.
3x  4  2 x  6
3x  4  2 x  2 x  6  2 x
x  4  6
x  4  4  6  4
x  10
{x | x  10} or  10, 
[
2.4 – Linear Inequalities and Problem Solving
Multiplication Property of Inequality
If a, b, and c are real numbers, and c is positive, then
a < b and ac < bc
are equivalent inequalities.
If a, b, and c are real numbers, and c is negative, then
a < b and ac > bc
are equivalent inequalities.
2.4 – Linear Inequalities and Problem Solving
Example
Solve: 2.3x  6.9 Graph the solution set.
2.3x  6.9
2.3x 6.9

2.3 2.3
x  3
The inequality symbol is reversed since we divided by a
negative number.
{x | x  3} or
(
 3, 
2.4 – Linear Inequalities and Problem Solving
Solve: 3x + 9 ≥ 5(x – 1). Graph the solution set.
3x + 9 ≥ 5(x – 1)
3x + 9 ≥ 5x – 5
3x – 3x + 9 ≥ 5x – 3x – 5
9 ≥ 2x – 5
9 + 5 ≥ 2x – 5 + 5
14 ≥ 2x
7≥x
x≤7
[
2.4 – Linear Inequalities and Problem Solving
Example
Solve: 7(x – 2) + x > –4(5 – x) – 12. Graph the solution set.
7(x – 2) + x > –4(5 – x) – 12
7x – 14 + x > –20 + 4x – 12
8x – 14 > 4x – 32
8x – 4x – 14 > 4x – 4x – 32
4x – 14 > –32
4x – 14 + 14 > –32 + 14
4x > –18
9
x
2
x > –4.5
(
2.5 – Compound Inequalities
Intersection of Sets
The solution set of a compound inequality formed with
and is the intersection of the individual solution sets.
2.5 – Compound Inequalities
Example
Find the intersection of: {2, 4,6,8}  {3, 4,5,6}
The numbers 4 and 6 are in both sets.
The intersection is {4, 6}.
2.5 – Compound Inequalities
Example
Solve and graph the solution for x + 4 > 0 and 4x > 0.
First, solve each inequality separately.
x+4>0
x>–4
and
4x > 0
x>0
(
-4
(
0
(
(0, )
2.5 – Compound Inequalities
Example
0  4(5 – x) < 8
0  20 – 4x < 8
0 – 20  20 – 20 – 4x < 8 – 20
– 20  – 4x < – 12 Remember that the sign direction
changes when you divide by a
5x>3
number < 0!
(
[
(3,5]
2.5 – Compound Inequalities
Union of Sets
The solution set of a compound inequality formed with or is
the union of the individual solution sets.
2.5 – Compound Inequalities
Example
Find the union of:
{2, 4,6,8}  {3, 4,5,6}
The numbers that are in either set are
{2, 3, 4, 5, 6, 8}.
This set is the union.
2.5 – Compound Inequalities
Example
Solve and graph the solution for
5(x – 1)  –5 or 5 – x < 11
5(x – 1)  –5
or
5 – x < 11
5x – 5  –5
–x < 6
5x  0
x>–6
x0
[
0
(
-6
(
-6
(–6, )
2.5 – Compound Inequalities
Example
or
 ,  
2.6 – Absolute Value Equations
Absolute Value Definition
If a is a positive number, then X= a is equivalent to
x = a or x = a.
Solving Equations of the Form |X| = a
Solve 6 + 2n = 4.
6 + 2n = 4
6 + 2n = 4
2n = 2
2n = 10
n = 1
n = 5
The solutions are 1 and 5.
{1, 5}
2.6 – Absolute Value Equations
Example
Solve  2x- 6 = 4.
 2x= 10
2x = 10
2x = 10
x = 5
x=5
The solutions are 5 and 5.
{5, 5}
2.6 – Absolute Value Equations
Example
Solve
7x  0
The solution is the set of all numbers whose distance
from 0 is 0 units.
The only number is 0.
The solution set is 0.
{0}
2.6 – Absolute Value Equations
Example
Solve 3z  2 + 8 = 1
3z  2 = 7
No solution
An absolute value can NEVER be equal to a negative
number.
2.6 – Absolute Value Equations
Example
Solve x  3  5  x
x–3=5-x
2x - 3 = 5
2x = 8
x=4
x – 3 = (5 – x)
x - 3 = 5 + x
– 3 = 5
False
The only solution for the original absolute value equation
is 4.
The solution set is {4}.