In this explainer, we will learn how to solve simple and compound linear inequalities and how to express their solutions using interval notation.

Before we begin discussing solving inequalities over the real numbers, we should recall the properties and methods of solving inequalities over the rational numbers.

First, we recall that we can add or subtract rational constants from both sides of an inequality to get an equivalent inequality.

We can also multiply or divide both sides of an inequality by a nonzero rational number to get an equivalent inequality, where we switch the direction of the inequality if the number is negative.

Second, we recall that we can sketch inequalities on a number line by using an open circle to represent that an endpoint is not included in the set and a closed circle to represent that an endpoint is included in the set.

For example, we can represent on a number line by using an open circle at and an arrow pointing left to represent all numbers less than .

This indicates that can take any value less than and it satisfies the inequality.

has the same representation as on a number line with the only difference being that we use a closed circle at to show that this value is included.

This indicates that can take any value less than or equal to and it satisfies the inequality.

All of these results also hold for real numbers. For example, we can add or subtract any real number from both sides of an inequality to find an equivalent inequality. We can also multiply or divide by any nonzero real number to find an equivalent inequality where we switch the direction of the inequality if the real number is negative. Finally, we can also represent solutions over the real numbers on a number line in the same way.

We can write the properties for rewriting inequalities over the real numbers formally as follows.

### Properties: Equivalent Inequalities over the Real Numbers

- Adding or subtracting a real constant from either side of an inequality gives an equivalent inequality. For example, if , then .
- Multiplying or dividing both sides of an inequality by a positive real number gives an equivalent inequality. For example, if and , then .
- Multiplying or dividing both sides of an inequality by a negative real number reverses the direction of the inequality. For example, if and , then .

These results allow us to solve any inequality over the real numbers where both sides of the inequality have linear expressions in a single variable. We do this by isolating the variable on one side of the inequality.

Before we see our first example, it is worth recalling that we can also represent inequalities using interval notation. We use an open parenthesis, ) or (, to represent that the endpoint is not included and a closed bracket, ] or [, to show that the endpoint is included.

For example, we can represent the set of real numbers greater than 2 but less than or equal to 3 as the compound inequality or as the interval . We can also sketch the set on a number line as shown.

We also note that we can use the infinity symbol, , to represent an unbounded interval. For example, we know that the inequality has solutions of that can take infinitely large values. To represent this in interval notation, we use a left open interval at 4, since 4 is not included in the interval, and a right open interval at infinity: . On a number line, this is given by the following.

In our first example, we will rewrite the solution set of an inequality in interval notation.

### Example 1: Representing an Inequality in Interval Notation

Find the solution set of the inequality in . Give your answer in interval notation.

### Answer

We first note that the solution set of the inequality includes all real numbers less than or equal to 20. We can sketch this solution set on a number line using a closed circle at 20 to show it is included in the set. This gives us the following.

To write this set in interval notation, we recall that since the left side of the interval is unbounded, we will need to use the symbol . We also note that the right endpoint is closed at 20. We use an open left parenthesis at and a closed right bracket at 20 to get the solution set .

In our next example, we will solve an inequality over the real numbers by using the properties of equivalent inequalities and we will represent the solution set on a number line.

### Example 2: Representing the Solution Set of an Inequality on a Number Line

Which of the following diagrams represents the inequality ?

### Answer

To find the solution set of this inequality, we will start by isolating the variable on the left-hand side of the inequality. To do this, we want to divide the inequality through by . We can do this by recalling that we can divide both sides of an inequality through by a positive real number to find an equivalent inequality. This gives us

We now want to rewrite by noting that , and we recall that we can take the square root of each factor separately. Thus,

Therefore, the inequality simplifies to .

We can sketch this on a number line by drawing an open circle at 2 to show that this value is not included and then including all numbers greater than 2. This gives us the following diagram.

This is given as option B.

In our next example, we will solve a compound inequality over the real numbers by using the properties of equivalent inequalities and we will represent the solution set on a number line.

### Example 3: Representing the Solution Set of a Multistep Inequality on a Number Line

Which of the following shows the solution set of the inequality ?

### Answer

To sketch the solution set of this inequality, we first want to isolate the variable on the left-hand side of the inequality. We can do this in two ways.

We can divide the inequality through by to get

Then, we can distribute over the parentheses to get

Next, we recall that we can add any real number to both sides of the inequality to get an equivalent inequality. We then add to both sides to get

Then, we divide through by 3 to get

Finally, we want on the left-hand side of the inequality. So, we switch the sides of the inequality and its direction to get

We can also start by distributing on the left-hand side to get

Thus, we can add to both sides of the inequality to get

Similarly, we can subtract from both sides to get

We can now isolate on the left-hand side by dividing both sides of the inequality through by . We note that since this is negative, we need to switch the direction of the inequality. This gives

We can sketch this on a number line by drawing an open circle at to show that this value is not included and then including all numbers greater than . This gives us the following diagram.

This is given as option E.

In our next example, we will solve a compound inequality over the set of real numbers, and we will give our answer as an interval.

### Example 4: Solving a Compound Inequality and Giving the Solution as an Interval

Find all the values of that satisfy . Give your answer in interval form.

### Answer

Since the variable only appears in the middle of the compound inequality, we can solve this inequality by isolating in the middle. First, we subtract 4 from all parts of the inequality to get the equivalent inequality

Next, we divide all parts of the inequality through by 7 to get

Therefore, the values of that satisfy this compound inequality are all of the real values between 1 and 4, excluding 1 but including 4. We can sketch this set on a number line by connecting an open circle at 1 to show that this value is not included and a closed circle at 4 to show that this value is included with a line.

Finally, we need to write this in interval notation. We use an open left parenthesis at 1 to show this value is not included and a closed right bracket at 4 to show this value is included in the solution set and satisfies the inequality.

This gives us that the values of that satisfy the inequality are any of the members of the set .

In our next example, we will once again solve a compound inequality over the set of real numbers and we will give our answer in interval notation.

### Example 5: Solving a Compound Inequality Where the Variable Appears on All Sides of the Inequality

Find the solution set of the inequality in . Give your answer in interval notation.

### Answer

We first recall that we can solve compound inequalities by treating them as two separate inequalities. We have and . We can solve each of these inequalities by isolating on the left-hand side of the inequality.

Letβs start with . We can multiply the inequality through by 10 to write the coefficients of as integers. This gives

We can then add to both sides of the inequality to get

Next, we add 1 to both sides of the inequality, giving us

Now, to isolate , we divide both sides of the inequality through by 25. This gives us

We also need the second inequality to be satisfied by solutions to the original compound inequality. We can rewrite the second inequality by first multiplying through by 2 to get

Next, we subtract from both sides of the inequality. This gives

Then, we subtract 10 from both sides of the inequality. We get

We can now divide the inequality through by ; we recall that since this value is negative, we need to switch the direction of the inequality to get

Therefore, for to be a solution to the original compound inequality, we must have that and . It is worth noting that and . So, this is equivalent to saying that must be greater than 1.4 and less than 2.04. We can write this as an interval by using open parentheses to show that the endpoints are not included.

Hence, the solution set of the inequality is .

In our final example, we will solve a compound inequality involving radical coefficients.

### Example 6: Solving a Compound Inequality Involving Radical Coefficients

Find the solution set of the inequality in . Give your answer in interval notation.

### Answer

We begin by recalling that we can solve compound inequalities by treating them as two separate inequalities. So, for to be a solution of the inequality, it must be a solution of both and . We can solve each of these inequalities by isolating on either side of the inequality.

Letβs start with . We can simplify the inequality by multiplying through by . This gives us

We then distribute over the parentheses in the right-hand side of the inequality and simplify to get

We can then add to both sides of the inequality. This gives us

Now, we divide both sides of the inequality through by 3. This yields

We also need the second inequality to be satisfied by solutions to the original compound inequality. We can rewrite the second inequality by multiplying through by to get

We then distribute over the parentheses in the left-hand side of the inequality and simplify to get

Next, we add to both sides of the inequality. This gives

We can now divide the inequality through by 8 to get

We can combine these two inequalities to get . We can write this inequality as an interval by using open parentheses to show that the endpoints are not included.

Hence, the solution set of the given inequality is .

Letβs finish by recapping some of the important points from this explainer.

### Key Points

- Adding or subtracting a real constant from either side of an inequality gives an equivalent inequality.
- We can also add or subtract constant real multiples of the variable from either side of an inequality to get an equivalent inequality.
- Multiplying or dividing both sides of an inequality by a positive real number gives an equivalent inequality.
- Multiplying or dividing both sides of an inequality by a negative real number reverses the direction of the inequality.
- We can represent inequalities over in set-builder notation, on number lines, or in interval notation.
- We can solve compound inequalities by treating them as two separate inequalities. So, for to be a solution of the inequality, it must be a solution of both.