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Solving
Inequalities
Using Addition or
Subtraction
Section 3-2
Goals
Goal
• To use addition or
subtraction to solve
inequalities.
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.
Vocabulary
• Equivalent Inequalities
Solving Inequalities
• Solving one-step inequalities is much
like solving one-step equations.
• To solve an inequality, you need to
isolate the variable using the properties
of inequality and inverse operations.
• At each step, you will create an
inequality that is equivalent to the
original inequality.
• Equivalent inequalities have the same
solution set.
Addition & Subtraction Properties
of Inequalities
Addition & Subtraction Properties
of Inequalities
When you add or subtract the same number on
both sides of an inequality, the resulting statement
will still be true.
–2 < 5
+7 +7
5 < 12
You can find solution sets of inequalities
the same way you find solutions of
equations, by isolating the variable.
Inequality Solutions
In Lesson 3-1, you saw that one way to show the solution
set of an inequality is by using a graph. Another way is to
use set-builder notation.
The set of all numbers x such that x has the given property.
{x : x < 6}
Read the above as “the set of all numbers x
such that x is less than 6.”
Example:
Solve and graph the inequality.
x + 3 > –5
x + 3 > –5
–3 –3
> –8
x
–9
Since 3 is added x, subtract 3 from
both sides.
-8 –7 -6 -5 -4 -3 -2 -1
0
1
2
3
4
5
Example: Continued
Check
According to the graph –4 should be a solution and –
9 should not be a solution.
x + 3 > –5
–4 + 3 > –5
x + 3 > –5
Substitute –4
for x.
–1 > –5
So –4 is a
solution.
–9 + 3 > –5
–6 > –5
Substitute –9
for x.

So –9 is not a solution.
Checking Inequality Solutions
Since there can be an infinite number of solutions to an
inequality, it is not possible to check all the solutions. You can
check the endpoint and the direction of the inequality symbol.
The solutions of x + 9 < 15 are given by x < 6.
Helpful Hint
Use an inverse operation to “undo” the
operation in an inequality. If the inequality
contains addition, use subtraction to undo
the addition.
Example:
Solve the inequality and graph the solutions.
x + 12 < 20
x + 12 < 20
–12 –12
Since 12 is added to x, subtract 12
from both sides to undo the
addition.
x+0 < 8
x < 8
–10 –8 –6 –4 –2
The solution set is {x: x < 8}.
0
2
4
6
8 10
Example:
Solve the inequality and graph the solutions.
d – 5 > –7
d – 5 > –7
+5 +5
d + 0 > –2
d > –2
–10 –8 –6 –4 –2
Since 5 is subtracted from d, add 5
to both sides to undo the
subtraction.
The solution set is {d: d > –2}.
0
2
4
6
8 10
Example:
Solve the inequality and graph the solutions.
0.9 ≥ n – 0.3
0.9 ≥ n – 0.3
+0.3
+0.3
Since 0.3 is subtracted from n,
add 0.3 to both sides to undo
the subtraction.
1.2 ≥ n – 0
1.2 ≥ n or n ≤1.2
0
The solution set is {n: n ≤ 1.2}.
1
1.2

2
Your Turn:
Solve each inequality and graph the solutions.
a. s + 1 ≤ 10
Since 1 is added to s, subtract 1 from both
sides to undo the addition.
s + 1 ≤ 10
–1 –1
9
s+0≤ 9
s ≤ 9
–10 –8 –6 –4 –2
0
2
4
The solution set is {s: s ≤ 9}.
6
8 10
Your Turn:
Solve each inequality and graph the solutions.
5
b.
> –3 + t
25
2
+3
> –3 + t
Since –3 is added to t, add 3 to both sides.
+3
7
> 0+t
2
7
t<
2
–10 –8 –6 –4 –2
0
2
4
6
8 10
Solve each inequality and graph the solutions.
c. q – 3.5 < 7.5
q – 3.5 < 7.5
+ 3.5 +3.5
q – 0 < 11
q < 11
Since 3.5 is subtracted from q, add 3.5 to
both sides to undo the subtraction.
–7
–5
–3 –1
1
3
5
7
9 11 13
Example: Application
While training for a race, Ann’s goal is to run at least 3.5
miles each day. She has already run 1.8 miles today. Write
and solve an inequality to find out how many more miles
she must run today.
Let m = the number of additional miles.
1.8 miles plus additional miles is at least 3.5 miles.
1.8
+
m
≥
3.5
1.8 + m ≥ 3.5
Since 1.8 is added to m, subtract
–1.8
–1.8
1.8 from both sides.
m ≥ 1.7
Ann should run at least 1.7 more miles.
Your Turn:
Tim’s company produces recycled paper. They produce
60.5 lb of paper each day. They have already produced at
least 20.2 lb today. Write and solve an inequality to find out
how many more pounds Tim’s company must produce.
Let p = the number of additional pounds of paper.
20.2 lbs
20.2
plus
additional pounds is at least 60.5 lb.
+
p
20.2 + p ≥ 60.5
–20.2
– 20.2
p ≥ 40.3
≥
60.5
Since 20.2 is added to p,
subtract 20.2 from both sides.
Tim’s company should produce at least 40.3 lb more
of paper.
Joke Time
• What flower grows between your nose and
your chin?
• Tulips
• How many sides are there to a circle?
• 2 – inside and outside.
• What do you get when you cross an elephant
and Darth Vader?
• An elevader.
Assignment
3.2 Exercises Pg. 187 – 190: #10 – 66 even