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2.9
Solving Linear
Inequalities
Linear Inequalities
An inequality is a statement that contains of
the symbols: < , >, ≤ or ≥.
Equations
x=3
Inequalities
x>3
12 = 7 – 3y
12 ≤ 7 – 3y
Graphing Solutions
Graphing solutions to linear inequalities in
one variable
• Use a number line
• Use a bracket at the endpoint of a interval
if you want to include the point
• Use a parenthesis at the endpoint if you
DO NOT want to include the point
Represents the set {xx  7}
Represents the set {xx > – 4}
Solutions to Linear Inequalities
x<3
Interval notation: (–∞, 3)
NOT included
–2 < x < 0
Interval notation: (–2, 0)
Included
–1.5  x  3
Interval notation: [–1.5, 3)
Example
Graph: x  1
Addition Property of Inequality
If a, b, and c are real numbers, then
a < b and a + c < b + c
are equivalent inequalities.
Multiplication Property of
Inequality
1. If a, b, and c are real numbers, and c is positive,
then
a < b and ac < bc
are equivalent inequalities.
2. If a, b, and c are real numbers, and c is
negative, then
a < b and ac > bc
are equivalent inequalities.
Solving Linear Inequalities
Step 1: Clear the inequality of fractions by multiplying both
sides of the inequality by the LCD of all fractions in the
inequality.
Step 2: Remove grouping symbols such as parentheses
by using the distributive property.
Step 3: Simplify each side of inequality by combining like
terms.
Step 4: Write the inequality with variable terms on one side
and numbers on the other side by using the addition
property of inequality.
Step 5: Get the variable alone by using the multiplication
property of inequality.
Example
Solve: 3x + 8 ≥ 5. Graph the solution set.
3x  8  5
3x  8  8  5  8
3 x  3
3x 3

3
3
x  1
Example
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
The graph of solution
set is{x|x ≤ 7}.
Apply the distributive property.
Subtract 3x from both sides.
Simplify.
Add 5 to both sides.
Simplify.
Divide both sides by 2.
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
The graph of solution
set is {x|x > –9/2}.
Apply the distributive property.
Combine like terms.
Subtract 4x from both sides.
Simplify.
Add 14 to both sides.
Simplify.
Divide both sides by 4.
Compound Inequalities
A compound inequality contains two
inequality symbols.
0  4(5 – x) < 8
This means 0  4(5 – x) and 4(5 – x) < 8.
To solve the compound inequality, perform
operations simultaneously to all three parts of the
inequality (left, middle and right).
Example
Graph: 2  x  5
Example
Solve the inequality. Graph the solution set and
write it in interval notation.
9 < z + 5 < 13
9 < z + 5 < 13
9 – 5 < z + 5 – 5 < 13 – 5
4<
(4, 8)
z
<
8
Subtract 5 from all three parts.
Example
Solve the inequality. Graph the solution set and
write it in interval notation.
–7 < 2p – 3 ≤ 5
–7 < 2p – 3 ≤ 5
–4 < 2p ≤ 8
–2 < p ≤ 4
(–2, 4]
Add 3 to all three parts.
Divide all three parts by 2.
Example
Solve the inequality. Graph the solution set and
write it in interval notation.
0  20 – 4x < 8
0  20 – 4x < 8
Use the distributive property.
0 – 20  20 – 20 – 4x < 8 – 20
Subtract 20 from each part.
– 20  – 4x < – 12
Simplify each part.
5x>3
Divide each part by –4.
Remember that the sign changes direction when you divide by a
negative number.
The solution is (3,5].
Inequality Applications
Example: Six times a number, decreased by 2, is at
least 10. Find the number.
1.) UNDERSTAND
Let x = the unknown number.
“Six times a number” translates to 6x,
“decreased by 2” translates to 6x – 2,
“is at least 10” translates ≥ 10.
Continued
Finding an Unknown Number
Example continued:
2.) TRANSLATE
Six times
a number
decreased
by 2
is at least
10
6x
–
2
≥
10
Continued
Finding an Unknown Number
Example continued:
3.) SOLVE
6x – 2 ≥ 10
6x ≥ 12
x≥2
Add 2 to both sides.
Divide both sides by 6.
4.) INTERPRET
Check: Replace “number” in the original statement of the
problem with a number that is 2 or greater.
Six times 2, decreased by 2, is at least 10
6(2) – 2 ≥ 10
10 ≥ 10
State: The number is 2.
Example
You are having a catered event. You can spend at
most $1200. The set up fee is $250 plus $15 per
person, find the greatest number of people that can
be invited and still stay within the budget.
Let x represent the number of people
Set up fee + cost per person × number of people ≤ 1200
250 +
15x
≤ 1200
continued
continued
You are having a catered event. You can spend at most $1200. The
set up fee is $250 plus $15 per person, find the greatest number of
people that can be invited and still stay within the budget.
250  15x  1200
15 x  950
15 x 950

15
15
x  63.3
The number of people who can be invited must be 63
or less to stay within the budget.