Lesson Objectives

- Demonstrate an understanding of interval notation
- Demonstrate the ability to graph an interval on the number line
- Learn how to solve a linear inequality using addition and subtraction » addition property of inequality
- Learn how to solve a linear inequality using multiplication and division » multiplication property of inequality
- Learn how to solve any linear inequality in one variable using more than one property of inequality
- Learn how to solve three-part linear inequalities in one variable

## How to Solve a Linear Inequality in One Variable

In our Algebra 1 course, we learned how to solve a linear inequality in one variable, write the solution in interval notation, and graph
the interval. So far in our Algebra 2 course, we have discussed
interval notation and how to graph an interval on the number line.
Let's begin our lesson by reviewing how to solve a linear inequality using addition or subtraction.

Example 1: Solve each inequality

x - 7 > 4

When we solve an equation or an inequality with one variable, our goal is to isolate the variable on one side of the equation. In this case, we have x - 7 on the left side of the inequality. We can isolate x by adding 7 to both sides of the inequality.

x - 7 + 7 > 4 + 7

x > 11

Our solution for the inequality is x is greater than 11. This means x can be any number larger than 11. We normally notate the solution for a linear inequality using interval notation and by graphing the interval on the number line. To write our solution in interval notation, we follow the format of:

smallest number, largest number

In this case, our smallest number is anything larger than 11. Since 11 is not included, we use a parenthesis next to 11:

(11,

For the largest number, there isn't one. Since x can be any value larger than 11, we will use infinity in the place of the largest number. A parenthesis is always used next to infinity:

(11, ∞)

To graph the interval on the number line, we find 11 and place a parenthesis (or an open circle). This tells us that 11 is not part of the solution. We then shade everything to the right of 11 on the number line. We also shade the right arrow to indicate the solution continues forever in the right direction. Example 2: Solve each inequality

x + 5 ≤ -1

To solve this inequality, we subtract 5 from each side. This will allow us to isolate x on the left side of the inequality:

x + 5 - 5 ≤ -1 - 5

x ≤ -6

We can write this solution in interval notation as:

(-∞, -6]

The bracket is used next to (-6) since (-6) is included in the solution.

We can graph our solution as:

Example 3: Solve each inequality $$3x > -9$$ To isolate x, we will divide each side of the equation by +3: $$\frac{3}{3}x > \frac{-9}{3}$$ $$\require{cancel}\frac{\cancel{3}}{\cancel{3}}x > \frac{-3\cancel{9}}{\cancel{3}}$$ $$x > -3$$ Interval Notation:

(-3, ∞)

Graphing the Interval: Let's now think about the other situation which will occur. In some cases, we will have to multiply or divide an inequality by a negative number. This will reverse the direction of the inequality. Suppose we started with something such as:

2 < 5

2 is less than 5 is a true statement. If we multiply both sides by a positive number such as 3:

2 • 3 < 5 • 3

6 < 15

We still have a true statement, 6 is less than 15. What happens if we multiply both sides by a negative number such as (-3):

2 • (-3) < 5 • (-3)

-6 < -15 (false)

The statement is no longer true. Remember a bigger negative represents a smaller value. Therefore (-15) is less than (-6) because it is a larger negative. This is the impact of multiplying a negative by a smaller positive number and a larger positive number. The larger positive number becomes a larger negative number and is now the smaller number. To deal with this situation, we introduce a second part to the multiplication property of inequality. If we multiply or divide both sides of an inequality by a negative number, we must flip the direction of the inequality symbol.

2 • (-3) < 5 • (-3)

-6 > -15 (true)

When we change the less than symbol to a greater than symbol, we maintain a true statement. -6 is greater than -15. Let's look at an example.

Example 4: Solve each inequality $$-\frac{2}{5}x ≥ 20$$ Multiply both sides of the inequality by -5/2: $$\frac{-5}{2} \cdot \frac{-2}{5}x ≥ \frac{-5}{2} \cdot 20$$ $$\cancel{\frac{-5}{2}} \cdot \cancel{\frac{-2}{5}}x ≥ \frac{-5}{\cancel{2}} \cdot 10\cancel{20}$$ Since we multiplied both sides of the inequality by a negative number, we must flip the direction of the inequality symbol. This means we change our greater than or equal to symbol to a less than or equal to symbol: $$x ≤ -50$$ Interval Notation:

(-∞, -50]

Graphing the Interval:

Example 5: Solve each inequality, write the solution in interval notation, graph the interval

5(6 - 5x) - 8x ≥ -30 - 3x

Step 1) Simplify each side:

5(6 - 5x) - 8x ≥ -30 - 3x

30 - 25x - 8x ≥ -30 - 3x

-33x + 30 ≥ -30 - 3x

Step 2) Isolate the variable term:

-33x + 3x ≥ -30 - 30

-30x ≥ -60

Step 3) Isolate the variable: $$-30x ≥ -60$$ $$\frac{-30}{-30}x ≤ \frac{-60}{-30}$$ $$\frac{\cancel{-30}}{\cancel{-30}}x ≤ \frac{2\cancel{-60}}{\cancel{-30}}$$ We flip the direction of the inequality symbol since we divided by a negative number. $$x ≤ 2$$ Interval Notation:

(-∞, 2]

Graphing the Interval on the Number Line: Step 4) Check

Checking the solution for a linear inequality in one variable is a lot more time consuming than checking the solution for a linear equation in one variable. First we will check the boundary. The boundary separates the solution region from the non-solution region. To check the boundary, we replace our inequality symbol with an equality symbol for our original inequality and our solution. We then plug in and check:

5(6 - 5x) - 8x = -30 - 3x

x = 2

5(6 - 5(2)) - 8(2) = -30 - 3(2)

5(6 - 10) - 16 = -30 - 6

5(-4) - 16 = -36

-20 - 16 = -36

-36 = -36

Now that we have checked the boundary, we want to check a number on each side of the boundary. Since x is less than or equal to 2, this means that 2 or any number that is less than 2 will work as a solution. Let's substitute 0 in for x in our original inequality. Since 0 is less than 2, it should work as a solution:

5(6 - 5x) - 8x ≥ -30 - 3x

5(6 - 5(0)) - 8(0) ≥ -30 - 3(0)

5(6) ≥ -30

30 ≥ -30 (true)

Additionally, we can check a number on the other side of the boundary (2). Anything greater than 2 should not work as a solution. This means if we plug a number such as 3 in for x, we should get a false statement.

5(6 - 5x) - 8x ≥ -30 - 3x

5(6 - 5(3)) - 8(3) ≥ -30 - 3(3)

5(6 - 15) - 24 ≥ -30 - 9

5(-9) - 24 ≥ -39

-45 - 24 ≥ -39

-69 ≥ -39 (false)

We can see that plugging in a value from the non-solution region did not result in a true statement. This concludes our check. Since the process is so tedious, we will omit the checks for all future inequalities.

Example 6: Solve each inequality, write the solution in interval notation, graph the interval $$\frac{29}{6}x + 2x < -15 + \frac{13}{3}x$$ Step 1) Simplify each side

We can clear the fractions by multiplying both sides of the inequality by 6: $$6\left(\frac{29}{6}x + 2x \right) < 6\left(-15 + \frac{13}{3}x\right)$$ Left side: $$6 \cdot \frac{29}{6}x + 6 \cdot 2x$$ $$\cancel{6} \cdot \frac{29}{\cancel{6}}x + 12x$$ $$29x + 12x$$ $$41x$$ Right Side: $$6 \cdot -15 + 6 \cdot \frac{13}{3}x$$ $$-90 + 2\cancel{6} \cdot \frac{13}{\cancel{3}}x$$ $$-90 + 26x$$ Our inequality becomes: $$41x < 26x - 90$$ Step 2) Isolate the variable term: $$41x - 26x < -90$$ $$15x < -90$$ Step 3) Isolate the variable: $$15x < -90$$ $$\frac{\cancel{15}}{\cancel{15}}x < \frac{-6\cancel{90}}{\cancel{15}}$$ $$x < -6$$ Interval Notation:

(-∞, -6)

Graphing the Interval on the Number Line:

Example 7: Solve each inequality, write the solution in interval notation, graph the interval

-11 ≤ 3x - 5 ≤ -2

We can still use our normal procedure, just remember our goal is to isolate the variable in the middle, not on one side.

Step 1) Simplify each part.

Each part here (-11), (3x - 5), and (-2) is already simplified.

Step 2) Isolate the variable term in the middle.

Since 5 is being subtracted away from x, we need to add 5 to each part:

-11 + 5 ≤ 3x - 5 + 5 ≤ -2 + 5

-6 ≤ 3x ≤ 3

Step 3) Isolate the variable in the middle.

We will divide each part by 3, the coefficient of x: $$\frac{-6}{3} ≤ \frac{3}{3}x ≤ \frac{3}{3}$$ $$\frac{-2\cancel{6}}{\cancel{3}} ≤ \frac{1\cancel{3}}{\cancel{3}}x ≤ \frac{1\cancel{3}}{\cancel{3}}$$ $$-2 ≤ x ≤ 1$$ Interval Notation:

[-2, 1]

Graphing the Interval on the Number Line:

### Addition Property of Inequality

The addition property of inequality allows us to add or subtract the same value to/from both sides of an inequality without changing the solution. Let's look at an example.Example 1: Solve each inequality

x - 7 > 4

When we solve an equation or an inequality with one variable, our goal is to isolate the variable on one side of the equation. In this case, we have x - 7 on the left side of the inequality. We can isolate x by adding 7 to both sides of the inequality.

x - 7 + 7 > 4 + 7

x > 11

Our solution for the inequality is x is greater than 11. This means x can be any number larger than 11. We normally notate the solution for a linear inequality using interval notation and by graphing the interval on the number line. To write our solution in interval notation, we follow the format of:

smallest number, largest number

In this case, our smallest number is anything larger than 11. Since 11 is not included, we use a parenthesis next to 11:

(11,

For the largest number, there isn't one. Since x can be any value larger than 11, we will use infinity in the place of the largest number. A parenthesis is always used next to infinity:

(11, ∞)

To graph the interval on the number line, we find 11 and place a parenthesis (or an open circle). This tells us that 11 is not part of the solution. We then shade everything to the right of 11 on the number line. We also shade the right arrow to indicate the solution continues forever in the right direction. Example 2: Solve each inequality

x + 5 ≤ -1

To solve this inequality, we subtract 5 from each side. This will allow us to isolate x on the left side of the inequality:

x + 5 - 5 ≤ -1 - 5

x ≤ -6

We can write this solution in interval notation as:

(-∞, -6]

The bracket is used next to (-6) since (-6) is included in the solution.

We can graph our solution as:

### Multiplication Property of Inequality

The multiplication property of inequality allows us to multiply the same positive number by both sides of an inequality and not change the solution. Since division can be written as a related multiplication statement, this property extends to division as well. It is very important to notice the "positive number" part of the definition. When we multiply by a negative number, we will require an additional step. Let's look at an example.Example 3: Solve each inequality $$3x > -9$$ To isolate x, we will divide each side of the equation by +3: $$\frac{3}{3}x > \frac{-9}{3}$$ $$\require{cancel}\frac{\cancel{3}}{\cancel{3}}x > \frac{-3\cancel{9}}{\cancel{3}}$$ $$x > -3$$ Interval Notation:

(-3, ∞)

Graphing the Interval: Let's now think about the other situation which will occur. In some cases, we will have to multiply or divide an inequality by a negative number. This will reverse the direction of the inequality. Suppose we started with something such as:

2 < 5

2 is less than 5 is a true statement. If we multiply both sides by a positive number such as 3:

2 • 3 < 5 • 3

6 < 15

We still have a true statement, 6 is less than 15. What happens if we multiply both sides by a negative number such as (-3):

2 • (-3) < 5 • (-3)

-6 < -15 (false)

The statement is no longer true. Remember a bigger negative represents a smaller value. Therefore (-15) is less than (-6) because it is a larger negative. This is the impact of multiplying a negative by a smaller positive number and a larger positive number. The larger positive number becomes a larger negative number and is now the smaller number. To deal with this situation, we introduce a second part to the multiplication property of inequality. If we multiply or divide both sides of an inequality by a negative number, we must flip the direction of the inequality symbol.

2 • (-3) < 5 • (-3)

-6 > -15 (true)

When we change the less than symbol to a greater than symbol, we maintain a true statement. -6 is greater than -15. Let's look at an example.

Example 4: Solve each inequality $$-\frac{2}{5}x ≥ 20$$ Multiply both sides of the inequality by -5/2: $$\frac{-5}{2} \cdot \frac{-2}{5}x ≥ \frac{-5}{2} \cdot 20$$ $$\cancel{\frac{-5}{2}} \cdot \cancel{\frac{-2}{5}}x ≥ \frac{-5}{\cancel{2}} \cdot 10\cancel{20}$$ Since we multiplied both sides of the inequality by a negative number, we must flip the direction of the inequality symbol. This means we change our greater than or equal to symbol to a less than or equal to symbol: $$x ≤ -50$$ Interval Notation:

(-∞, -50]

Graphing the Interval:

### Solving Linear Inequalities in One Variable using More than One Property of Inequality

When we put the addition property of inequality together with the multiplication property of inequality, we are able to solve any linear inequality in one variable. We can show a four-step method, which is similar to our four-step method used when solving equations:### Four-Step Method for Solving a Linear Inequality in One Variable

- Simplify each side separately
- Use the addition property of inequality to isolate the variable term on one side of the inequality
- Use the multiplication property of inequality to isolate the variable
- Check
- Since checking the solution for a linear inequality in one variable is a very tedious process, we normally skip this step

Example 5: Solve each inequality, write the solution in interval notation, graph the interval

5(6 - 5x) - 8x ≥ -30 - 3x

Step 1) Simplify each side:

5(6 - 5x) - 8x ≥ -30 - 3x

30 - 25x - 8x ≥ -30 - 3x

-33x + 30 ≥ -30 - 3x

Step 2) Isolate the variable term:

-33x + 3x ≥ -30 - 30

-30x ≥ -60

Step 3) Isolate the variable: $$-30x ≥ -60$$ $$\frac{-30}{-30}x ≤ \frac{-60}{-30}$$ $$\frac{\cancel{-30}}{\cancel{-30}}x ≤ \frac{2\cancel{-60}}{\cancel{-30}}$$ We flip the direction of the inequality symbol since we divided by a negative number. $$x ≤ 2$$ Interval Notation:

(-∞, 2]

Graphing the Interval on the Number Line: Step 4) Check

Checking the solution for a linear inequality in one variable is a lot more time consuming than checking the solution for a linear equation in one variable. First we will check the boundary. The boundary separates the solution region from the non-solution region. To check the boundary, we replace our inequality symbol with an equality symbol for our original inequality and our solution. We then plug in and check:

5(6 - 5x) - 8x = -30 - 3x

x = 2

5(6 - 5(2)) - 8(2) = -30 - 3(2)

5(6 - 10) - 16 = -30 - 6

5(-4) - 16 = -36

-20 - 16 = -36

-36 = -36

Now that we have checked the boundary, we want to check a number on each side of the boundary. Since x is less than or equal to 2, this means that 2 or any number that is less than 2 will work as a solution. Let's substitute 0 in for x in our original inequality. Since 0 is less than 2, it should work as a solution:

5(6 - 5x) - 8x ≥ -30 - 3x

5(6 - 5(0)) - 8(0) ≥ -30 - 3(0)

5(6) ≥ -30

30 ≥ -30 (true)

Additionally, we can check a number on the other side of the boundary (2). Anything greater than 2 should not work as a solution. This means if we plug a number such as 3 in for x, we should get a false statement.

5(6 - 5x) - 8x ≥ -30 - 3x

5(6 - 5(3)) - 8(3) ≥ -30 - 3(3)

5(6 - 15) - 24 ≥ -30 - 9

5(-9) - 24 ≥ -39

-45 - 24 ≥ -39

-69 ≥ -39 (false)

We can see that plugging in a value from the non-solution region did not result in a true statement. This concludes our check. Since the process is so tedious, we will omit the checks for all future inequalities.

Example 6: Solve each inequality, write the solution in interval notation, graph the interval $$\frac{29}{6}x + 2x < -15 + \frac{13}{3}x$$ Step 1) Simplify each side

We can clear the fractions by multiplying both sides of the inequality by 6: $$6\left(\frac{29}{6}x + 2x \right) < 6\left(-15 + \frac{13}{3}x\right)$$ Left side: $$6 \cdot \frac{29}{6}x + 6 \cdot 2x$$ $$\cancel{6} \cdot \frac{29}{\cancel{6}}x + 12x$$ $$29x + 12x$$ $$41x$$ Right Side: $$6 \cdot -15 + 6 \cdot \frac{13}{3}x$$ $$-90 + 2\cancel{6} \cdot \frac{13}{\cancel{3}}x$$ $$-90 + 26x$$ Our inequality becomes: $$41x < 26x - 90$$ Step 2) Isolate the variable term: $$41x - 26x < -90$$ $$15x < -90$$ Step 3) Isolate the variable: $$15x < -90$$ $$\frac{\cancel{15}}{\cancel{15}}x < \frac{-6\cancel{90}}{\cancel{15}}$$ $$x < -6$$ Interval Notation:

(-∞, -6)

Graphing the Interval on the Number Line:

### Solving a Three-Part Linear Inequality

In some cases, we will see what is known as a "three-part" inequality. To solve a three-part inequality we isolate the variable in the middle.Example 7: Solve each inequality, write the solution in interval notation, graph the interval

-11 ≤ 3x - 5 ≤ -2

We can still use our normal procedure, just remember our goal is to isolate the variable in the middle, not on one side.

Step 1) Simplify each part.

Each part here (-11), (3x - 5), and (-2) is already simplified.

Step 2) Isolate the variable term in the middle.

Since 5 is being subtracted away from x, we need to add 5 to each part:

-11 + 5 ≤ 3x - 5 + 5 ≤ -2 + 5

-6 ≤ 3x ≤ 3

Step 3) Isolate the variable in the middle.

We will divide each part by 3, the coefficient of x: $$\frac{-6}{3} ≤ \frac{3}{3}x ≤ \frac{3}{3}$$ $$\frac{-2\cancel{6}}{\cancel{3}} ≤ \frac{1\cancel{3}}{\cancel{3}}x ≤ \frac{1\cancel{3}}{\cancel{3}}$$ $$-2 ≤ x ≤ 1$$ Interval Notation:

[-2, 1]

Graphing the Interval on the Number Line:

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