\text{Addition property of inequality}	\text{If } a \lt b, \text{then } a+c \lt b+c; \\\ \text{If } a \gt b, \text{then } a+c \gt b+c
\text{Subtraction property of inequality}	\text{If } a\lt b, \text{then } a-c \lt b-c; \\\ \text{If } a \gt b, \text{then } a-c \gt b-c
\text{Multiplication property of inequality}	\text{If } a \lt b \text{ and } c \gt 0, \text{then } a \cdot c \lt b\cdot c; \\\ \text{if } a\gt b \text{ and } c \gt 0, \text{then } a\cdot c \gt b\cdot c
\text{Division property of inequality}	\text{If } a\lt b \text{ and } c \gt 0, \text{then } \dfrac{a}{c} \lt \dfrac{b}{c};\\\ \text{ if } a \gt b \text{ and } c \gt 0, \text{then } \dfrac{a}{c}\gt\dfrac{b}{c}
\text{Asymmetric property of inequality}	\text{If } a\gt b, \text{then } b \lt a
\text{Transitive property of inequality}	\text{If } a \gt b \text{ and } b \gt c, \text{then } a \gt c

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

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

A number line ranging from negative 5 to 5 in steps of 1. An unfilled circle is on 4. An arrow is from 4 to the left of 4.

x \lt 4 is graphed on the number line. The unfilled circle means that 4 is not included in the solution.

A number line ranging from negative 5 to 5 in steps of 1. An unfilled circle is on 4. An arrow is from negative 3 to the right of negative 3.

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

Remember: when multiplying or dividing an inequality by a negative value the inequality symbol is reversed.

Examples

Example 1

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

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 10	Multiplication property of inequality
\displaystyle -3x	\displaystyle \leq	\displaystyle 18	Addition property of inequality
\displaystyle x	\displaystyle \geq	\displaystyle -6	Division property of inequality

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.

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.

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.

Is x=3 in the solution set for the inequality?

Worked Solution

Create a strategy

We can determine if x=3 is in the solution set 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 in the solution set for 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 becaue is in the shaded area.

Any points that are not in the shaded area on the number line are not in the solution set.

Example 2

Arlene charges \$ 42.75 for a pet grooming session, plus an additional \$ 5 for each special treatment. Clarisse wants the total cost for her pet's grooming to be no more than \$ 102.80.

Write an inequality that represents the number of special treatments Clarisse could get for her pet.

Worked Solution

Create a strategy

Clarisse wants the total cost to be no more than \$ 102.80. "No more than" means she could spend exactly that amount or less than that amount but not anything greater. This is "less than or equal to."

Apply the idea

We can write an inequality in words that represents the cost for a pet grooming session:

\text{cost of grooming}+ \text{cost per treatment} \cdot \text{number of treatments} \leq \text{total Clarrisse can spend}

Translating that into an algebraic expression we get: 42.75+5s\leq 102.80 where s represents the number of special treatments.

How many special treatments could Clarisse get for her pet and still afford the pet grooming?

Worked Solution

Create a strategy

Solve the inequality and then write the solution set.

Apply the idea

\displaystyle 42.75 + 5s	\displaystyle \leq	\displaystyle 102.80	Original inequality
\displaystyle 5s	\displaystyle \leq	\displaystyle 60.05	Subtract 42.75 from both sides
\displaystyle s	\displaystyle \leq	\displaystyle 12.01	Divide both sides by 5

According to the solution Clarisse could get 12.01 special treatments for her pet or fewer, however since she can't get a partial treatment, a more realistic solution is that she can get 12 special treatments or less.

Is s=-2 a viable solution?

Worked Solution

Create a strategy

When considering negative solutions to a contextual problem we need to decide whether a negative value is possible in the context.

Apply the idea

Keep in mind it is not realistic to get part of a special treatment or a negative number of special treatments.

Clarisse can get a maximum of 12 special treatments for her pet, so -2 is mathematically part of the solution set for the inequality s \leq 12.01 but it is not a viable solution in this context because the smallest number of treatments she could get is 0.

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.

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.

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

Outcomes

8.PFA.5

The student will write and solve multistep linear inequalities in one variable, including problems in context that require the solution of a multistep linear inequality in one variable.

8.PFA.5a

Apply properties of real numbers and properties of inequality to solve multistep linear inequalities (up to four steps) in one variable with the variable on one or both sides of the inequality. Coefficients and numeric terms will be rational. Inequalities may contain expressions that need to be expanded (using the distributive property) or require combining like terms to solve.

8.PFA.5b

Represent solutions to inequalities algebraically and graphically using a number line.

8.PFA.5c

Write multistep linear inequalities in one variable to represent a verbal situation, including those in context.

8.PFA.5e

Solve problems in context that require the solution of a multistep linear inequality in one variable.

8.PFA.5f

Identify a numerical value(s) that is part of the solution set of a given inequality.

8.PFA.5g

Interpret algebraic solutions in context to linear inequalities in one variable.

2.06 Solve multistep inequalities

Ideas

Multistep inequalities

Examples

Example 1

Example 2

Outcomes

8.PFA.5

8.PFA.5a

8.PFA.5b

8.PFA.5c

8.PFA.5e

8.PFA.5f

8.PFA.5g

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