1. Introduction
Solving inequalities is similar to solving equations. To find the solution to an inequality, use methods similar to those used in solving equations. In addition to using properties of equality, we will also use properties of inequalities to change inequalities into simpler equivalent inequalities.
2.1.3: Solving Linear Inequalities

2. Key Concepts
The properties of inequality are the rules that allow you to balance, manipulate, and solve inequalities. The properties are summarized in the following table.
2.1.3: Solving Linear Inequalities

3. Key Concepts, continued Properties of Inequality
Property
Example
If a > b and b > c, then a > c.
If 10 > 6 and 6 > 2, then 10 > 2.
If a > b, then b < a.
If 10 > 6, then 6 < 10.
If a > b, then –a < – b.
If 10 > 6, then –10 < –6.
If a > b, then a ± c > b ± c.
If 10 > 6, then 10 ± 2 > 6 ± 2.
If a > b and c > 0, then a • c > b • c.
If 10 > 6 and 2 > 0, then 8 • 2 > b • 2.
If a > b and c < 0, then a • c < b • c.
If 10 > 6 and –1 < 0, then 10 • –1 < 6 • –1.
If a > b and c > 0, then a ÷ c > b ÷ c.
If 10 > 6 and 2 > 0, then 8 ÷ 2 > 6 ÷ 2.
If a > b and c < 0, then a ÷ c < b ÷ c.
c. If 10 > 6 and –1 < 0, then 10 ÷ –1 < 6 ÷ –1.
2.1.3: Solving Linear Inequalities

4. Key Concepts, continued
When solving more complicated inequalities, first simplify the inequality by clearing any parentheses. Do this by either distributing by the leading number or dividing both sides of the inequality by the leading number. Then solve the inequality by following the steps learned earlier, as outlined on the following slide.
2.1.3: Solving Linear Inequalities

5. Key Concepts, continued
Solving Inequalities
If a variable appears on both sides of the inequality, choose which side of the inequality you would like the variable to appear on.
Add or subtract the other variable from both sides of the inequality using either the addition or subtraction property of equality.
Simplify both expressions.
Continue to solve the inequality as you did in earlier examples by working to isolate the variable.
Check that your answer is correct by substituting a value included in your solution statement into the original inequality to ensure a true statement.
2.1.3: Solving Linear Inequalities

6. Key Concepts, continued
It is important to remember that when solving inequalities, the direction of the inequality symbol (<, >, ≤, or ≥) changes when you divide or multiply by a negative number. Here’s an example.
If we had the simple statement that 4 < 8, we know that we can multiply both sides of the inequality by a number, such as 3, and the statement will still be true.
4 < 8 Original inequality
4 • 3 < 8 • 3 Multiply both expressions of the inequality by 3.
12 < This is a true statement.
2.1.3: Solving Linear Inequalities

7. Key Concepts, continued
We can also divide both sides of the inequality by a number, such as 2.
4 < Original inequality
4 ÷ 2 < 8 ÷ Divide both expressions of the equation by 2.
2 < This is a true statement.
Notice what happens when we multiply the inequality by –3.
4 < Original inequality
4 • –3 < 8 • –3 Multiply both expressions of the inequality by –3.
–12 < – This is NOT a true statement.
2.1.3: Solving Linear Inequalities

8. Key Concepts, continued
To make this a true statement, change the direction of the inequality symbol.
–12 > –24 This is a true statement.
The same is true when dividing by a negative number; you must change the direction of the inequality symbol.
4 < Original inequality
4 ÷ –2 < 8 ÷ –2 Divide both expressions of the equation by 2.
–2 < –4 This is NOT a true statement. Change the direction of the inequality symbol.
–2 > –4 This is a true statement.
2.1.3: Solving Linear Inequalities

9. Common Errors/Misconceptions
not understanding why the inequality symbol changes direction when multiplying or dividing by a negative number
replacing inequality symbols with equal signs
forgetting to switch the inequality symbol when multiplying or dividing by a negative number
switching the inequality symbol when subtracting a number
2.1.3: Solving Linear Inequalities

10. Guided Practice Example 1 Solve the inequality .
2.1.3: Solving Linear Inequalities

11. Guided Practice: Example 1, continued
Isolate the variable by eliminating the denominator.
In this inequality, the denominator means “divide by 7.” Eliminate it by performing the inverse operation, multiplication. Multiply both expressions of the inequality by 7.
2.1.3: Solving Linear Inequalities

12. Guided Practice: Example 1, continued Isolate the variable.
Perform the inverse operation of adding 4 to both expressions of the inequality.
Now solve.
2.1.3: Solving Linear Inequalities

13. Guided Practice: Example 1, continued
Divide both sides of the inequality by the coefficient, –3.
Notice that the direction of the inequality symbol changed because we divided by –3.
2.1.3: Solving Linear Inequalities

14. Guided Practice: Example 1, continued
The solution to the original inequality
is all numbers less than –13.
To check this, choose any number less than –13 to show a true statement. Let’s try –20. Be sure to substitute the value into the original inequality.
2.1.3: Solving Linear Inequalities

15. Guided Practice: Example 1, continued
Original inequality
Substitute –20 for x.
Multiply, then subtract.
Simplify the fraction.
8 > 5
2.1.3: Solving Linear Inequalities

16. ✔ Guided Practice: Example 1, continued
8 > 5 is a true statement; therefore, all numbers less than –13 will result in a true statement. ✔
2.1.3: Solving Linear Inequalities

17. Guided Practice: Example 1, continued
2.1.3: Solving Linear Inequalities

18. Guided Practice Example 2 Solve the inequality 5x + 4 ≥ 11 – 2x.
2.1.3: Solving Linear Inequalities

19. Guided Practice: Example 2, continued
Move the variable to one side of the inequality.
Notice the variable x is in both expressions of the inequality. Begin by choosing which side you want your variable to appear on. Just like with equations, this is a choice, but most people choose to have all variables on the left side of the inequality. Continue by adding 2x to both expressions of the inequality.
2.1.3: Solving Linear Inequalities

20. Guided Practice: Example 2, continued Isolate the variable.
Subtract 4 from both expressions.
2.1.3: Solving Linear Inequalities

21. Guided Practice: Example 2, continued
Finally, divide both sides of the inequality by the coefficient of 7.
2.1.3: Solving Linear Inequalities

22. Guided Practice: Example 2, continued
The solution to the original inequality, 5x + 4 ≥ 11 – 2x, is all numbers greater than or equal to 1.
2.1.3: Solving Linear Inequalities

23. ✔ Guided Practice: Example 2, continued
To check this solution, choose a number greater than or equal to 1, such as 2, and substitute it for all instances of x in the original inequality.
Our check proved true, so we can be sure
that our solution of x ≥ 1 is accurate.
5x + 4 ≥ 11 – 2x
Original inequality
5(2) + 4 ≥ 11 – 2(2)
Substitute 2 for each instance of x.
≥ 11 – 4
Simplify each expression.
14 ≥ 7
This is a true statement. ✔
2.1.3: Solving Linear Inequalities

24. Guided Practice: Example 2, continued
2.1.3: Solving Linear Inequalities