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Linear and Compound Inequalities

In the past we have solved inequalities that have a solution set.

For example, let’s solve:

8 − (4𝑥 − 2) ≥ −5(𝑥 + 1) − 5

At this point, we have a choice to either graph our solution, write the solution using interval notation, or we can

write the solution using set-builder notation.

Graph:

Interval Notation:

Set-builder Notation:

This solution means that the variable x can be any real value that is 20 or larger to make the inequality

TRUE!!

Notes about interval notation:

When you write the interval notation that does NOT include the endpoint, use parenthesis.

Always use a parenthesis when you use a ∞ or −∞.

When using interval notation, is always listed first (e.g. , 4 ) and is always listed second (e.g.

4, ).

Remember when you divide or multiply by a negative number, you need to the inequality.


Objective 1: Solving Linear Inequalities with No Solution and All Real Numbers

Sometimes you may run across an inequality that has no solution or all real numbers. Let’s see what this may

look like.

Example: Solve the inequality, graph and write the solution using interval and set-builder notation.

−29 − 42𝑥 > 6(1 − 7𝑥)

As you may notice, 29 is NOT greater than 6. This inequality is false and so we would say that it has “No

Solution”.

The graph of a “No Solution” answer would be just a plain number line.

To write the solution using interval or set-builder notation we use the “empty set” symbol:

What will happen if we switch the inequality around in the same exact problem from before?

−29 − 42𝑥 < 6(1 − 7𝑥)

We end up with a TRUE inequality. 29 is less than 6 which means no matter what value we give the variable

x, the inequality will always be true.

Here is how we would write the solution in all 3 forms:

Graph:

Interval Notation: (−∞, ∞)

Set-builder Notation: {𝑥|𝑥 ∈ ℝ}



6 4 6 24x x


6 3 3 6x x


Pause the video and try these problems.

Solve the inequality and write the solution using set-builder and interval notation.

1. 5 8 3 9 2

2. 2 4 3 8 5 7

3. 4 10 7 3 2

4. 6 3 2 4 5 17

x x x

x x x

x x x

x x


Objective 2: Solving Compound Inequalities with “and”

Compound inequalities involve a connection between more than one inequality. First let's look at two

inequalities connected with the word "and". When we solve for a conjunction (“and”) we look to see where the

two sets intersect or overlap.

It is often helpful to use a graph to determine the solution.

Example: Solve, graph the solution set, and write the solution in interval and set-builder notation.

5 6 2 3 2 4x x and x

Let’s now graph both of the solutions on the same number line.

The solution to this problem is where the two graphs overlap.

Graph:

Interval Notation:




3 8 6 7 9 2 4 10x x and x x x

Graph both of the solutions on the same number line.

Graph:

Interval Notation:




7𝑥 + 3(2𝑥 − 4) < 5(3𝑥 + 1) − 3 𝑎𝑛𝑑 5𝑥 − 10 > 2(2𝑥 − 4)

Graph both of the solutions on the same number line.

Graph:

Interval Notation:



At times, compound inequalities can be written like:

−6 < 6𝑥 + 12 ≤ 9

There are a couple of methods to solving compound inequalities like this.

Method 1:

The goal is to isolate the variable in the middle of the inequality. You need to remember that whichever

inverse operation you use to isolate the variable, you need to do that same operation to all sides of the

inequality.

Graph:

Interval Notation:


Method 2:

Break up the inequality by re-writing it as 2 different inequalities.

−6 < 6𝑥 + 12 ≤ 9

−6 < 6𝑥 + 12 𝑎𝑛𝑑 6𝑥 + 12 ≤ 9

From here you would solve each inequality and the solution would still be the overlapped portion of the

two graphs together.


Example: Solve, graph the solution set, and write the solution in interval and set-builder notation

−10 < 4 − 2𝑥 ≤ −6

Graph:

Interval Notation:



For each problem, graph the solution and write the solution using interval and set-builder notation.

1. 8 4 5 16x


2 1

2. 4 5 6 and 1 83 4

x x

3. 2 2 4 14x

1 4

4. 3 4 6 and 7 72 5

x x


Objective 3: Solving Compound Inequalities with “or”

Next let's look at two inequalities combined with the word "or". Think of a family reUNION, where all are

invited from any of the family components. Anyone that is in any of the sets is part of the answer.


𝑥 < 3 𝑜𝑟 𝑥 > 7

Since 𝑥 is already by itself in both inequalities, let’s graph both of them on the same number line.

When two inequalities are combined with the word "or", it is the union of the two solution sets. So here

we want the set of numbers that are either less than 3 OR greater than 7.

Graph:

Interval Notation:



6 + 7𝑥 < 6𝑥 − 5 𝑜𝑟 3𝑥 − 7 ≤ 5 + 5𝑥


Let’s graph both of them on the same number line.

Graph:

Interval Notation:



2 5 7 8 5 07

xx x or



Graph:

Interval Notation:



3 5 5 8x or x


Graph:

Interval Notation:




For each problem, graph the solution and write the solution using interval and set-builder notation.

1. 2 10 or 4 1 14x x x x

2. 3 7 1 4 2 or 8 2 3 6x x x

1

3. 6 2 2 3 1 4 or 8 2 3 3 14

x x x x x

1

4. 2 4 3 3 4 or 2 64

x x x x


Objective 4: Solving Compound Inequalities with All Real Numbers and No Solution

Let us look at the following compound inequalities.

I. Solve: 5 and 7x x

Graph:

Interval Notation:


II. Solve: 2 or 4x x

Graph:

Interval Notation:



Summary of Compound Inequalities:

Inequalities AND vs. OR

𝑥 > 2 𝑥 > 5

Graph:

𝑥 > 2 𝑎𝑛𝑑 𝑥 > 5

Graph:

Interval Notation:


𝑥 > 2 𝑜𝑟 𝑥 > 5

Graph:

Interval Notation:


𝑥 ≤ 2 𝑥 > 5

Graph:

𝑥 ≤ 2 𝑎𝑛𝑑 𝑥 > 5

Graph:

Interval Notation:


𝑥 ≤ 2 𝑜𝑟 𝑥 > 5

Graph:

Interval Notation:


𝑥 > 2 𝑥 ≤ 5

Graph:

𝑥 > 2 𝑎𝑛𝑑 𝑥 ≤ 5

Graph:

Interval Notation:


𝑥 > 2 𝑜𝑟 𝑥 ≤ 5

Graph:

Interval Notation:



For each problem, write the solution using interval and set-builder notation.

1. 6 8 or 5 3 2 9x x x


2. 9 12 3 and 2 8 10 2x x x

1 3

3. 4 3 5 and 2 65 4

x x

4. 4 1 3 7 and 2 4 3 8 1x x x x