Compound Inequalities Lesson

In our last lesson, we learned how to solve a linear inequality in one variable. In this lesson, we will learn how to solve a compound inequality. A compound inequality consists of two inequalities linked by the connective word "and" or "or". Before we solve compound inequalities, we have to refresh our memory on the intersection and union of two sets.

Intersection of Two Sets

The intersection of two sets, A and B, will be a set that contains all elements that are common to both set A and set B. We denote the intersection of two sets by placing the intersection symbol "∩" between the two sets. Let's suppose set A contains the numbers: 1, 3, 5, and 7, whereas set B contains the numbers 5, 7, 9, and 11.
A = {1, 3, 5, 7}
B = {5, 7, 9, 11}
The intersection of set A and set B is a set that contains the elements 5 and 7 only. These two numbers are the only two elements that are common to both sets:
A ∩ B = {5, 7}
In many cases, a Venn diagram will help us to visualize the intersection of two sets:

Solving Compound Inequalities with "and"

• Solve each inequality separately
• The solution for the compound inequality contains all numbers that satisfy both inequalities
  • This is the intersection of the two solution sets

Let's look at an example.
Example 1: Solve each compound inequality, write the solution in interval notation, graph the interval.
4x + 10 > 6x - 10
and
5x + 10 ≤ 4 + 8x
Let's begin by solving the top inequality:
4x + 10 > 6x - 10
4x - 6x > -10 - 10
-2x > -20
-2/-2 x > -20/-2
x < 10
Now let's solve the bottom inequality:
5x + 10 ≤ 4 + 8x
5x - 8x ≤ 4 - 10
-3x ≤ -6
-3/-3 x ≤ -6/-3
x ≥ 2
Now we think about the intersection of the two solution sets.
For the top inequality, x can be any number that is less than 10. For the bottom inequality, x can be any number that is greater than or equal to 2. The intersection of the two solution sets is a set that contains all numbers from 2 up to but not including 10. This means x can be 2 or any number that is larger, up to but not including 10.
2 ≤ x < 10
[2, 10) Example 2: Solve each compound inequality, write the solution in interval notation, graph the interval.
3x + 2 > -4 + 4x
and
10 + x ≥ -10 - 3x
Let's begin by solving the top inequality:
3x + 2 > -4 + 4x
3x - 4x > -4 - 2
-x > -6
-1/-1 x > -6/-1
x < 6
Now, let's solve the bottom inequality:
10 + x ≥ -10 - 3x
x + 3x ≥ -10 - 10
4x ≥ -20
4/4 x ≥ -20/4
x ≥ -5
For the top inequality, x can be any number that is less than 6. For the bottom inequality, x can be any number that is greater than or equal to -5. The intersection of the two solution sets is a set that contains all numbers from -5 up to but not including 6. This means x can be -5 or any number that is larger, up to but not including 6.
-5 ≤ x < 6
[-5, 6)

Solving Compound Inequalities with No Solution

In some cases, we will see a compound inequality with "and" that has no solution. This will occur when there are no common elements between the two solution sets. In other words, we will not have a solution for a compound inequality with "and" when there isn't a number that satisfies both inequalities. Let's look at an example.
Example 3: Solve each compound inequality, write the solution in interval notation, graph the interval.
10x - 9 ≤ 9x - 6
and
-5 + 3(1 + 5x) > 103
Let's begin by solving the top inequality:
10x - 9 ≤ 9x - 6
10x - 9x ≤ -6 + 9
x ≤ 3
Now, let's solve the bottom inequality:
-5 + 3(1 + 5x) > 103
-5 + 3 + 15x > 103
-2 + 15x > 103
15x > 103 + 2
15x > 105
15/15 x > 105/15
x > 7
When we look at our two solutions, we can see that there are no values that will satisfy both inequalities. For our top inequality, x can be 3 or any number that is smaller. For our bottom inequality, x can be any number that is larger than 7. Since there aren't any numbers that are both less than or equal to 3 and greater than 7, we can say there is "no solution" for this compound inequality.

Union of Two Sets

The union of two sets, A and B, will be a set that contains all elements of both sets. We denote the union of two sets by placing the union symbol "∪" between the two sets. Let's revisit our example with set A and set B.
A = {1, 3, 5, 7}
B = {5, 7, 9 , 11}
The union of set A and set B is a set that contains all elements of both sets.
A ∪ B = {1, 3, 5, 7, 9, 11}
Again, we can use a Venn diagram to visualize the union of two sets:

Solving Compound Inequalities with "or"

• Solve each inequality separately
• The solution for the compound inequality contains all numbers that satisfy either inequality
  • This is the union of the two solution sets

Let's look at an example.
Example 4: Solve each compound inequality, write the solution in interval notation, graph the interval.
-9x - 9 > 6 - 10x
or
7 + 7x < 5x - 3
Let's begin by solving the top inequality:
-9x - 9 > 6 - 10x
-9x + 10x > 6 + 9
x > 15
Now, let's solve the bottom inequality:
7 + 7x < 5x - 3
7x - 5x < -3 - 7
2x < -10
2/2 x < -10/2
x < -5
For our top inequality, x can be any number that is greater than 15. For the bottom inequality, x can be any number that is less than -5. The union of the two solution sets is a set that contains all numbers less than -5 and all numbers greater than 15.
We write this in interval notation as the union of the two solution sets:
(-∞, -5) ∪ (15, ∞) Example 5: Solve each compound inequality, write the solution in interval notation, graph the interval.
-5(8x - 3) ≥ 60 + 5x
or
12x - 5(-11x - 1) > 5 - 3x
Let's begin by solving the top inequality:
-5(8x - 3) ≥ 60 + 5x
-40x + 15 ≥ 60 + 5x
-40x - 5x ≥ 60 - 15
-45x ≥ 45
-45/-45 x ≥ 45/-45
x ≤ -1
Now let's solve the bottom inequality:
12x - 5(-11x - 1) > 5 - 3x
12x + 55x + 5 > 5 - 3x
67x + 5 > 5 - 3x
67x + 3x > 5 - 5
70x > 0
70/70 x > 0/70
x > 0
For our top inequality, x can be any number that is less than or equal to -1. For the bottom inequality, x can be any number that is greater than 0. The union of the two solution sets is a set that contains all numbers less than or equal to -1 and all numbers greater than 0. We write this in interval notation as the union of the two solution sets:
(-∞, -1] ∪ (0, ∞)

---

Skills Check: