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Algebra 2
Solving Inequalities
Lesson 1-5 Part 2
Goals
Goal
• To write and solve
compound 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.
Essential Question
Big Idea: Compound Inequalities
• In Part 1 of the lesson, you learned how to solve
and graph inequalities.
• In this lesson you will use what you learned to
write and solve compound inequalities.
Vocabulary
• Compound Inequality
Definition
• The inequalities you have seen so far are simple
inequalities. When two simple inequalities are
combined into one statement by the words AND
or OR, the result is called a compound inequality.
• Compound Inequality – the result of combining
two inequalities. The words and and or are used to
describe how the two parts are related.
What is the Difference
Between and and or?
A
B
AND means intersection
-what do the two items
have in common?
OR means union
-if it is in one item, it is in
the solution
A
B
Disjunction
A disjunction is a compound statement that uses the
word or.
Disjunction: x ≤ –3 OR x > 2
x ≤ –3 U x > 2
A disjunction is true if and only if at least one of its parts is true.
Number Line and
Compound Inequalities
You can graph the solutions of a compound inequality
involving OR by using the idea of combining regions.
The combined regions are called the union and show
the numbers that are solutions of either inequality.
>
Conjunction
A conjunction is a compound statement that uses the word and.
Conjunction: x ≥ –3 AND x < 2
x ≥ –3 ⋂ x < 2
A conjunction is true if and only if all of its parts are true.
Conjunctions can be written as a single statement as shown.
x ≥ –3 and x< 2
–3 ≤ x < 2
Number Line and
Compound Inequalities
You can graph the solutions of a compound inequality
involving AND by using the idea of an overlapping region.
The overlapping region is called the intersection and shows
the numbers that are solutions of both inequalities.
Reading Math
Dis- means “apart.” Disjunctions have two separate pieces.
Con- means “together” Conjunctions represent one piece.
Writing Math
The “and” compound inequality y < –2 and y < 4 can be written
as –2 < y < 4.
The “or” compound inequality y < 1 or y > 9 must be written
with the word “or.”
Compound Inequalities
Example: Writing
Compound Inequalities
Write the compound inequality shown by the graph.
The shaded portion of the graph is not between two values, so the
compound inequality involves OR.
On the left, the graph shows an arrow pointing left, so use
either < or ≤. The solid circle at –8 means –8 is a solution so use ≤.
x ≤ –8
On the right, the graph shows an arrow pointing right, so use
either > or ≥. The empty circle at 0 means that 0 is not a
solution, so use >.
x>0
The compound inequality is x ≤ –8 OR x > 0.
Example: Writing
Compound Inequalities
Write the compound inequality shown by the graph.
The shaded portion of the graph is between the values –2 and 5, so the
compound inequality involves AND.
The shaded values are on the right of –2, so use > or ≥. The empty circle
at –2 means –2 is not a solution, so use >.
m > –2
The shaded values are to the left of 5, so use < or ≤. The
empty circle at 5 means that 5 is not a solution so use <.
m<5
The compound inequality is m > –2 AND m < 5 (or –2 < m < 5).
Your Turn:
Write the compound inequality shown by the graph.
The shaded portion of the graph is between the values –9 and –2, so
the compound inequality involves AND.
The shaded values are on the right of –9, so use > or . The empty circle
at –9 means –9 is not a solution, so use >.
x > –9
The shaded values are to the left of –2, so use < or ≤. The
empty circle at –2 means that –2 is not a solution so use <.
x < –2
The compound inequality is –9 < x AND x < –2 (or –9 < x < –2).
Your Turn:
Write the compound inequality shown by the graph.
The shaded portion of the graph is not between two values, so the
compound inequality involves OR.
On the left, the graph shows an arrow pointing left, so use
either < or ≤. The solid circle at –3 means –3 is a solution, so use ≤.
x ≤ –3
On the right, the graph shows an arrow pointing right, so use
either > or ≥. The solid circle at 2 means that 2 is a solution,
so use ≥.
x≥2
The compound inequality is x ≤ –3 OR x ≥ 2.
Example:
Solve the compound inequality. Then graph the solution set.
6y < –24 OR y +5 ≥ 3
Solve both inequalities for y.
6y < –24
y + 5 ≥3
or
y < –4
y ≥ –2
The solution set is all points that satisfy
y < –4 or y ≥ –2.
–6 –5 –4 –3 –2 –1
0
1
2
3
Example:
Solve the compound inequality. Then graph the solution set.
Solve both inequalities for c.
and
2c + 1 < 1
c ≥ –4
c<0
The solution set is the set of points that satisfy both
c ≥ –4 and c < 0 (-4 ≤ c < 0).
–6 –5 –4 –3 –2 –1
0
1
2
3
Example:
Solve the compound inequality. Then graph the solution set.
x – 5 < –2 OR –2x ≤ –10
Solve both inequalities for x.
x – 5 < –2
–2x ≤ –10
or
x<3
x≥5
The solution set is the set of all points that satisfy
x < 3 or x ≥ 5.
–3 –2 –1
0
1
2
3
4
5
6
Your Turn:
Solve the compound inequality. Then graph the solution set.
x – 2 < 1 OR 5x ≥ 30
Solve both inequalities for x.
x–2<1
5x ≥ 30
x≥6
or
x<3
The solution set is all points that satisfy
x < 3 U x ≥ 6.
–1
0
1
2
3
4
5
6
7
8
Your Turn:
Solve the compound inequality. Then graph the solution set.
2x ≥ –6 AND –x > –4
Solve both inequalities for x.
2x ≥ –6
–x > –4
and
x ≥ –3
x<4
The solution set is the set of points that satisfy both x ≥ –3 ⋂ x < 4.
–4 –3 –2 –1 0
1
2
3
4
5
Your Turn:
Solve the compound inequality. Then graph the solution set.
x –5 < 12 OR 6x ≤ 12
Solve both inequalities for x.
x –5 < 12
or
6x ≤ 12
x≤2
x < 17
Because every point that satisfies x < 17 also satisfies x ≤ 2, the
solution set is x < 17.
2
4
6
8
10 12 14 16 18 20
Your Turn:
Solve the compound inequality. Then graph the solution set.
–3x < –12 AND x + 4 ≤ 12
Solve both inequalities for x.
–3x < –12
x + 4 ≤ 12
and
x < –4
x≤8
The solution set is the set of points that satisfy both 4 < x ≤ 8.
2
3
4
5
6
7
8
9 10 11
Example:
Solve the compound inequality and graph the
solutions.
Since 1 is added to x, subtract 1
–5 < x + 1 < 2
from each part of the inequality.
–5 < x + 1 AND x + 1 < 2
–1
–1
–1 –1
–6 < x
AND x
The solution set is
{x:–6 < x AND x < 1}.
<1
-6 < x < 1
Graph -6 < x < 1.
1
–10
–8 –6 –4 –2
0
2
4
6
8 10
Example:
Solve the compound inequality and graph the solutions.
8 < 3x – 1 ≤ 11
Since 1 is subtracted from 3x, add 1 to
each part of the inequality.
8 < 3x – 1 ≤ 11
+1
+1 +1
9 < 3x ≤ 12
Since x is multiplied by 3, divide each
part of the inequality by 3 to undo
the multiplication.
3<x≤4
–5 –4 –3 –2 –1
The solution set is 3 < x ≤ 4.
0
1
2
3
4
5
Your Turn:
Solve the compound inequality and graph the
solutions.
Since 10 is subtracted from x, add 10
to each part of the inequality.
–9 < x – 10 < –5
–9 < x – 10 < –5
+10
+10 +10
1<x
The solution set is {x:1 < x < 5}.
< 5
1<x<5
Graph 1 < x < 5.
–5 –4 –3 –2 –1
0
1
2
3
4
5
Your Turn:
Solve the compound inequality and graph the
solutions.
–4 ≤ 3n + 5 < 11
Since 5 is added to 3n, subtract 5 from
each part of the inequality.
–4 ≤ 3n + 5 < 11
–5
–5 –5
Since n is multiplied by 3, divide each part
–9 ≤ 3n
< 6
of the inequality by 3 to undo the
multiplication.
The solution set is {n:–3 ≤ n < 2}.
–3 ≤ n < 2
Graph -3 ≤ x < 2.
–5 –4 –3 –2 –1
0
1
2
3
4
5
Assignment
Section 1-5 part 2, Pg 40 – 42; #1 – 6 all, 8 –
34 even.