# Solving Inequalities Using Addition and Subtraction

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When learning how to solve inequalities, algebra 1 students usually start with solving linear inequalities with one variable. They thus become proficient in solving inequalities using addition and subtraction.

Early lessons on inequalities can be challenging for some students, which is why math teachers need to have specific strategies in place to address potential difficulties that students encounter while learning this topic.

## Strategies for Teaching How to Solve Inequalities

### Review Inequalities

Children should already have an understanding of inequalities, and they should be familiar with checking if a given value is the solution to an inequality, but it’s good to briefly review this to identify any gaps. So you can start your lesson by defining what inequalities are.

You can define an inequality as a mathematical statement comparing values of two expressions with the help of an inequality symbol and thus showing whether one value is less than, greater than, or equal to the other.

Remind students that there are five inequality symbols that we use for this purpose, including < (less than), > (greater than), ≤ (less than or equal to), ≥ (greater than or equal to), ≠ (not equal to). For more guidelines, you can also refer to our introductory article on inequalities.

### What Is Solving Inequalities?

Explain to students that you’ll be solving linear inequalities in one variable, that is, inequalities involving a linear function. Provide an example of a linear inequality on the whiteboard. For instance:

x – 6 ≥ 4

Explain that solving this inequality means having the x (or n, y, or whatever variable is being used in the inequality) on the left of the inequality sign and on its own. You can provide the solution to the above inequality to illustrate what it should look like:

x ≥ 10

Once x is on its own on the left, we say that we have solved the inequality. After you’ve provided a brief introduction on what solving inequalities implies, you can move on to providing step-by-step guidelines on how we can actually solve the inequality.

Point out to children that we solve inequalities with the help of the addition property of inequalities, as well as the subtraction property of inequalities. You can start by explaining the addition property of inequalities.

According to this property, if we add any number to each side of a true inequality, the resulting inequality will also be true. You can present this on the whiteboard by stating that for any number a, b and c, the following are true:

• If a > b, then a + c > b + c
• If a < b, then a + c < b + c
• If a ≥ b, then a + c ≥ b + c
• If a ≤ b, then a + c ≤ b + c

#### Examples:

Provide examples of how this property works in practice, that is:

11 > 8
11 + 2 > 8 + 2
13 > 10

Point out that the initial inequality was 11 is greater than 8. By adding 2 on the left and on the right side, we can observe that the resulting inequality (13 is greater than 10) remains true.

9 ≥ 8
9 + 4 ≥ 8 + 4
13 ≥ 12

So the initial inequality was 9 is greater than or equal to 8. By adding 4 on the left and on the right side, we can observe that the inequality remains true nevertheless. That is, the resulting inequality 13 is greater than or equal to 12 is also true.

#### Example of Solving Inequalities Using the Addition Property:

Now you can provide an example of how we use this property when solving inequalities. This time, you can write a simple linear inequality on the whiteboard, such as:

x – 3 > 1

Remind children that the aim of solving an inequality is to get the variable alone on the left side. Once x is on its own on the left, we say that we have solved the inequality. How can we isolate x with the help of the previously demonstrated addition property of inequalities?

Point out that to get x alone on the left, we’ll try to get rid of -3. And we can get rid of -3 by adding + 3, as this will mean that -3 and + 3 will cancel each other out. But make sure to point out to students that they must add + 3 to both sides so that the inequality remains true:

x – 3 + 3 > 1 + 3

x > 4

So the solution is x is greater than 4. You can also show students how to check this solution. Take a number that’s greater than 4, such as 5. Then substitute this number for the variable in the inequality:

x – 3 > 1

5 – 3 > 1

2 > 1

We see that 2 is greater than 1, which means the solution is correct. You can also try it out with other values that are greater than 4.

### Subtraction Property of Inequalities

As mentioned earlier, we also solve inequalities by using the subtraction property of inequalities. This property follows a similar logic with the addition property of inequality, the difference being that we’re subtracting the same value from both sides.

You can simply explain to students that according to this property, if we subtract any number on each side of a true inequality, the resulting inequality will also be true. You can present this on the whiteboard by stating that for any number a, b and c, the following are true:

• If a > b, then a – c > b – c
• If a < b, then a – c < b – c
• If a ≥ b, then a – c ≥ b – c
• If a ≤ b, then a – c ≤ b – c

#### Examples:

Provide a few examples of how this property works in practice:

12 ≤ 15

12 – 4 ≤ 15 – 4

8 ≤ 11

So the initial inequality was 12 is less than or equal to 15. By subtracting 4 on the left and on the right side, we can observe that the inequality remains true nevertheless. That is, the resulting inequality 8 is less than or equal to 11 is also true.

16 < 20

16 – 3 < 20 – 3

13 < 17

So the initial inequality was 16 is less than 20. By subtracting 3 on the left and on the right side, we can observe that the inequality remains true nevertheless. That is, the resulting inequality 13 is less than 17 is also true.

#### Example of Solving Inequalities Using the Subtraction Property

Now you can provide an example of how we use this property when solving inequalities. Write a simple linear inequality on the whiteboard that we want to solve, such as:

20 + x ≥ 15

Again, point out to children that the aim of solving an inequality is to get the variable alone on the left side. To do this, we’ll need to get rid of 20 on the left. How can we get rid of 20? Well, we can use the subtraction property of inequality.

We know that this property states that if we subtract the same number on both sides, the inequality stays true. So we’ll subtract 20 on both sides, and 20 and – 20 will cancel each other out on the left side. That is:

20 – 20 + x ≥ 15 – 20

x ≥ – 5

So the solution is x is greater than or equal to – 5. You can also choose some numbers and substitute them for the variable in the inequality 20 + x ≥ 15 in order to show students how to check if the solution is correct.

You may also consider enriching your lesson with multimedia materials, such as videos. For example, this free video contains step-by-step instructions on how to solve inequalities with the help of the addition and subtraction properties of inequality, as well as how to graph them.

In addition, this free video contains simple guidelines on how to solve inequalities with variables on both sides, and then draw the solution on a number line and select a few values in order to check the solution.

## Activities to Practice Solving Inequalities Using Addition and Subtraction

### Rags to Riches: Solving Inequalities

This is an online quiz that will help students practice their skills at solving inequalities using addition and subtraction. Implement this activity at the end of your lesson. Make sure you have enough technical devices for each student in your classroom.

Divide students into pairs. Make sure they’re seated across each other, that is, in a way that they aren’t able to see each other’s screens. Explain that they will play a game where they need to solve inequalities and provide instructions for the game.

Explain that in the game, students will have to solve inequalities by using the addition property of inequalities, as well as the subtraction property of inequalities. For each correctly answered question, they score a certain sum of money.

The sum they score will depend on the difficulty of the inequality – the more difficult the inequality, the higher the sum. If they get stuck, they can use a hint. In the end, the player with the highest sum is declared winner.

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