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1. Equations & Inequalities
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Algebra 1

1.06 Multi-step inequalities

Lesson
Practice

Multi-step inequalities

Some mathematical relations compare two non-equivalent expressions. These are known as inequalities.

We can solve inequalities by using various properties to isolate the variable, in a similar way to solving equations.

Asymmetric property of inequality\text{If } a>b, \text{then } b<a
Transitive property of inequality\text{If } a>b \text{ and } b>c, \text{then } a>c
Linear inequality

An inequality that contains a variable term with an exponent of 1, and no variable terms with exponents other than 1

Example:

3x - 5 \geq 4

Solving an inequality using the properties of inequalities results in a solution set.

Solution set

The set of all values that make the inequality or equation true

Example:

3 \leq x

We can represent solutions to inequalities algebraically, by using numbers, letters, and/or symbols, or graphically, by using a coordinate plane or number line.

Different notations can be used to represent algebraic solutions, such as inequalities, set notation, and interval notation.

Set notation uses the notation \left\{⬚ \middle\vert ⬚ \right\}, where the vertical line is read as "such that" and defines the variable used. Interval notation uses parentheses \left( \enspace \right) to indicate values that are never reached and not included in the solution set. Square brackets indicate\left[ \enspace \right] closed intervals that include the endpoints in a solution set.

-5-4-3-2-1012345

The inequality x \lt 3 is graphed on the number line. The unfilled circle means that 3 is not included in the solution.

Set notation writes the solution x \lt 3 as \left\{x \middle\vert x \lt 3 \right\}.

Since 3 is not included in the solution, the interval notation would be written as \left(-\infty , 3\right).

-5-4-3-2-1012345

The inequality x\geq-3 is graphed on the number line. The filled circle means -3 is included in the solution.

Set notation writes x \geq -3 as \left\{x \vert x \geq -3 \right\}.

Since 3 is included in the solution, the interval notation would be written as \left[-3, \infty \right).

Based on the context, some values might be calculated algebraically, but are not reasonable based on the restrictions of the scenario. For example, time and lengths generally cannot be negative, which can create restrictions on the possible values x and y can take on.

Viable solutions

A valid solution that makes sense within the context of the question or problem

Non-viable solution

An algebraically valid solution that does not make sense within the context of the question or problem

Exploration

Complete the following chart by performing the indicated operations:

Consider the inequalityPerform the operation on the inequalityWrite the new inequalityTrue or false?
1 \lt 4\text{Add } 2 \text{ to both sides}
6 \gt -2\text{Subtract } 2 \text{ from both sides}
3 \lt 10\text{Multiply by } 2 \text{ on both sides}
1 \gt -7\text{Multiply by } -2 \text{ on both sides}
4 \gt 2\text{Divide by } 2 \text{ on both sides}
-8 \lt 12\text{Divide by } -2 \text{ on both sides}

  1. Did any operations cause the given inequality to become a false inequality?

  2. Can you think of something to change about a false inequality without changing the operation performed?

When multiplying or dividing an inequality by a negative value the inequality symbol is reversed.

The properties of inequality are:

Addition property of inequality\text{If } a \gt b, \text{then } a+c \gt b+c
Subtraction property of inequality\text{If } a \gt b, \text{then } a-c \gt b-c
Multiplication property of inequality\text{If } a \gt b \text{ and } c \gt 0, \text{then } ac \gt bc \\ \text{or if } a \gt b \text{ and } c\lt 0, \text{then } ac \lt bc
Division property of inequality\text{If } a\gt b \text{ and } c\gt 0, \text{then } \dfrac{a}{c}\gt\dfrac{b}{c} \\ \text{or if } a \gt b \text{ and } c\lt 0, \text{then } \dfrac{a}{c}\lt \dfrac{b}{c}

Examples

Example 1

Consider the inequality \dfrac{-8-3x}{2} \leq 5

a

Solve the inequality

Worked Solution
Create a strategy

We want to isolate x on one side of the inequality and a number on the other.

Apply the idea
\displaystyle \frac{-8-3x}{2}\displaystyle \leq\displaystyle {5}Original inequality
\displaystyle -8-3x\displaystyle \leq\displaystyle 10Multiplication property of inequality
\displaystyle -3x\displaystyle \leq\displaystyle 18Addition property of inequality
\displaystyle x\displaystyle \geq\displaystyle -6Division property of inequality

The solution can be written as x \geq -6, \left\{x \middle\vert x \geq -6\right\}, or \left[-6,\infty\right).

Reflect and check

Solving an inequality is similar to solving an equation. However, we need to reverse the direction of the inequality when multiplying or dividing by a negative number.

b

Plot the inequality on a number line.

Worked Solution
Apply the idea

Plot the solution set of the inequality x \geq -6. Note that since we include -6 the point should be filled.

-10-9-8-7-6-5-4-3-2-1012345678910
Reflect and check

What if the solution was x \gt -6?

Endpoints included in the solution are filled points.

Endpoints not included in the solution are unfilled points.

c

Is x=3 a viable or nonviable solution to the inequality?

Worked Solution
Create a strategy

We can determine if x=3 is viable or non-viable by using the number line or algebraically substituting the solution into the inequality.

Apply the idea
\displaystyle \frac{-8-3x}{2}\displaystyle \leq\displaystyle {5}Original inequality
\displaystyle \frac{-8-3(3)}{2}\displaystyle \leq\displaystyle {5}Substitute x=3
\displaystyle \frac{-8-9}{2}\displaystyle \leq\displaystyle {5}Evaluate the multiplication
\displaystyle \frac{-17}{2}\displaystyle \leq\displaystyle {5}Evaluate the subtraction
\displaystyle -8.5\displaystyle \leq\displaystyle {5}Evaluate the division

Since x=3 leads to a true statement, we can confirm that x=3 is a viable solution to the inequality.

Reflect and check

By using the number line, we can see that the point x=3 is in the solution set of the inequality, meaning it is a viable solution to the inequality.

Any points that are not in the solution set are considered non-viable and will lead to a false statement when substituted into the inequality algebraically.

Example 2

Calandra charges \$ 37.72 to style hair, as well as an additional \$ 6 per foil. Pauline would like the total cost for her styling to be no more than \$ 95.86.

a

Write an inequality that represents the number of foils Pauline could get.

Worked Solution
Create a strategy

Pauline has no more than \$ 95.86 to spend. "No more than" means "less than or equal to."

Apply the idea

We can write an inequality in words that represents the cost to style Pauline's hair:

\text{cost of styling}+ \text{cost per foil} \cdot \text{number of foils} \leq \text{total Pauline can spend}

Translating that into an algebraic expression we get: 37.72+6N\leq95.86 where N represents the number of foils.

b

How many foils could Pauline get and still afford the styling?

Worked Solution
Create a strategy

Solve the inequality and then write the solution set.

Apply the idea
\displaystyle 37.72 + 6N\displaystyle \leq\displaystyle 95.86Original inequality
\displaystyle 6N\displaystyle \leq\displaystyle 58.14Subtraction property of inequality
\displaystyle N\displaystyle \leq\displaystyle 9.69Division property of inequality

According to the solution Pauline could get 9.69 foils or fewer, however since she can't get a partial foil a more realistic solution is that she can get 9 foils or less.

c

Determine whether N=-2 is a viable solution to the inequality in the context of the question.

Worked Solution
Create a strategy

Keep in mind it is not realistic to get part of a foil or a negative number of foils.

Apply the idea

Pauline can get a maximum of 9 foils and a minimum of 0 foils, so while -2 is mathematically part of the solution set for the inequality N \leq 9.69 it is not a viable solution in this context.

Reflect and check

Unlike a value that is not in a solution set of an inequality, this is an example of a solution that was mathematically valid and part of the original solution set but when considering the context we have found that it is non-viable.

Example 3

Solve the inequality 4\left(x+5\right) \lt 3\left(2-x\right).

Worked Solution
Create a strategy

To solve this inequality, we need to simplify each side, isolate x on one side, and then use the inequality to find the solution. We can start by distributing the 4 and 3 on each side of the inequality.

Apply the idea
\displaystyle 4\left(x+5\right)\displaystyle <\displaystyle 3\left(2-x\right)Original inequality
\displaystyle 4x+20\displaystyle <\displaystyle 6-3xDistributive property
\displaystyle 4x+3x\displaystyle <\displaystyle 6-20Add 3x to both sides and subtract 20 from both sides
\displaystyle 7x\displaystyle <\displaystyle -14Simplify both sides
\displaystyle x\displaystyle <\displaystyle -2Divide both sides by 7

The solution to the inequality can be written as x \lt -2, \left\{x \middle\vert x \lt -2 \right\}, or \left(-\infty,-2 \right).

Reflect and check

To check our solution, we can substitute a value less than -2 into the original inequality and see if it holds true. Let's take x = -3 as an example.

\displaystyle 4\left(-3+5\right)\displaystyle <\displaystyle 3\left(2-\left(-3\right)\right)Substitute x=-3 into the original inequality
\displaystyle 4 \cdot 2\displaystyle <\displaystyle 3 \cdot 5Simplify the expressions
\displaystyle 8\displaystyle <\displaystyle 15Continue simplifying the expressions

The inequality 8 \lt 15 is true, so our solution x \lt -2 is correct.

Idea summary

Just like the properties of equality, the properties of inequality can justify how we solve inequalities.

The multiplication and division properties of inequality change the meaning of an inequality when multiplying or dividing by a negative number, meaning we have to reverse the inequality symbol when applying the property:

  • If a\gt b and c\gt 0, then a \cdot c \lt b \cdot c

  • If a \gt b and c \lt 0, then \dfrac{a}{c} \lt \dfrac{b}{c}

The solution set of an inequality is the set of values that makes the inequality true.

Because inequalities can have infinitely many solutions, inequalities used to represent real-world situations often include solutions that are unreasonable in context and therefore non-viable.

Outcomes

A.EI.1

The student will represent, solve, explain, and interpret the solution to multistep linear equations and inequalities in one variable and literal equations for a specified variable.

A.EI.1a

Write a linear equation or inequality in one variable to represent a contextual situation.

A.EI.1c

Solve multistep linear inequalities in one variable algebraically and graph the solution set on a number line, including those in contextual situations, by applying the properties of real numbers and/or properties of inequality.

A.EI.1f

Verify possible solution(s) to multistep linear equations and inequalities in one variable algebraically, graphically, and with technology to justify the reasonableness of the answer(s). Explain the solution method and interpret solutions for problems given in context.

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