Topics
4. Equations & Inequalities
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United States of AmericaVirginia
Grade 7

4.05 Solve one-step inequalities

Lesson
Practice

Solving one-step inequalities

An inequality shows the relationship between two unequal values.

Exploration

Consider the inequality: 9 \lt 15

  • Add 3 to both sides of the inequality.

  • Add -3 from both sides of the inequality.

  • Subtract 3 to both sides of the inequality.

  • Subtract -3 from both sides of the inequality.

  • Multiply both sides of the inequality by 3.

  • Multiply both sides of the inequality by -3.

  • Divide both sides of the inequality by 3.

  • Divide both sides of the inequality by -3.

  1. Was the inequaltiy true after each operation? If not, what could you do to make it true?

  2. Test this out with other numbers. Is it still true?

Solving inequalities is very similar to solving equations except when multiplying or dividing by a negative.

Instead of using the properties of equality like we do for equations, when solving inequalities we use the properties of inequality :

\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{Multiplication property of inequality} \\\text{ (by a negative number)} \text{If } a \lt b \text{ and } c \lt 0, \text{then } a \cdot c \gt b \cdot c \\ \text{If } a \gt b \text{ and } c \lt 0, \text{then } a\cdot c \lt 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{Division property of inequality} \\\text{ (by a negative number)}\text{if } a \lt b \text{ and } c\lt 0, \text{then } \dfrac{a}{c}\gt \dfrac{b}{c} \\ \text{if } a\gt b \text{ and } c\lt 0, \text{then } \dfrac{a}{c}\lt \dfrac{b}{c}

Let's compare the equation -2x=8 and the inequality -2x\gt 8.

We will start by solving the equation -2x=8

\displaystyle \frac{-2x}{-2}\displaystyle =\displaystyle \frac{8}{-2}Division property of equality (divide by -2)
\displaystyle x\displaystyle =\displaystyle -4Evaluate the division

This equation has exactly one solution, x=-4. We can graph the solution of -2x=8 on the number line like this:

-8-7-6-5-4-3-2-10

This number line shows that there is only one solution to the equation, x=-4.

Next, let's solve the inequality -2x \gt 8

\displaystyle \frac{-2x}{-2}\displaystyle <\displaystyle \frac{8}{-2}Division property of inequality (reverse the inequality symbol)
\displaystyle x\displaystyle <\displaystyle -4Evaluate the division

We can graph the solution of -2x \gt 8 on the number line like this:

-8-7-6-5-4-3-2-10

This number line shows that there are many solutions to the inequality, including any number that is less than, but not equal to, -4.

We can verify solutions algebraically, by substituting the value of x back into the equation or inequality.

Let's verify the solution to the equation -2x=8. We found that the solution was x=-2.

\displaystyle -2\left(-4\right)\displaystyle =\displaystyle 8Substitution
\displaystyle 8\displaystyle =\displaystyle 8Evaluate the multiplication

After substituting the value and evaluating the equation, we are left with 8=8. This is true, so the answer is correct.

To determine if a value is a solution to the inequality, we can substitute values to determine if they are true. Inequalities can have multiple solutions, so let's see if x=-2 satisfies the inequality -2x \gt 8.

\displaystyle -2\left(-2\right)\displaystyle >\displaystyle 8Substitution property
\displaystyle 4\displaystyle >\displaystyle 8Evaluate the multiplication

We can read this as "4 is greater than 8" which is not true. x=-2 is not a solution to the inequality -2x\gt 8.

Examples

Example 1

Consider the inequality 1+x\lt 2.

a

Solve the inequality for x.

Worked Solution
Create a strategy

We solve an inequality by isolating x on one side of the inequality.

Apply the idea
\displaystyle 1 +x\displaystyle <\displaystyle 2Start with the given inequality
\displaystyle 1-1+x\displaystyle <\displaystyle 2-1Subtraction property of inequality
\displaystyle x\displaystyle <\displaystyle 1Evaluate the subtraction
b

Now plot the solutions to the inequality 1+x \lt 2 on the number line below.

-4-3-2-101234
Worked Solution
Create a strategy

Use the \lt symbol to identify what type of endpoint to use and which direction to shade.

Apply the idea

The inequality x\lt 1 means that x can have any value less than but not equal to 1.

To show that 1 is not part of the solution, we will plot the point at 1 with an open (unfilled) circle. To show all values that are less than 1, we draw a ray from 1 pointing to the left.

-4-3-2-1012

Example 2

Consider the inequality -4 \lt 2m.

a

Solve the inequality.

Worked Solution
Create a strategy

Solve the inequality by isolating m on one side of the inequality.

Apply the idea
\displaystyle -4\displaystyle <\displaystyle 2mStart with the given inequality
\displaystyle \frac{-4}{2}\displaystyle <\displaystyle \dfrac{2m}{2}Division property of inequality
\displaystyle -2\displaystyle <\displaystyle mEvaluate the division
b

Now plot the solutions to the inequality -4 \lt 2m on the number line below.

-4-3-2-101234
Worked Solution
Create a strategy

Rewrite the inequality first so that we would have the variable on the left side. This would make understanding the variable easier.

Take note that when rewriting or switching the values on both sides of inequalities, we also switch the inequality symbol.

Then identify what the inequality means before plotting on the number line.

Apply the idea

Rewrite the inequality: m \gt -2

The inequality m \gt -2 means that m can have any value greater than but not equal to -2.

To show that -2 is not part of the solution, we will plot the point at -2 with an unfilled circle. To show all values that are greater than -2, we draw a ray from -2 pointing to the right.

-4-3-2-101234

Example 3

Determine whether c=-2 satisfies the inequality c+3 \lt 4.

Worked Solution
Create a strategy

Substitute the value of c into the inequality. The value on the left hand side should be less than 4.

Apply the idea
\displaystyle c+3\displaystyle <\displaystyle 4Start with the given inequality
\displaystyle -2+3\displaystyle <\displaystyle 4Substitution property
\displaystyle 1\displaystyle <\displaystyle 4Evaluate the addition

1 is less than 4 so c=-2 satisfies the inequality.

Reflect and check

Substituting the possible solution back into the inequality to see if the inequality remains true is a helpful strategy for checking your work.

Example 4

Solve -\dfrac{x}{2} \geq 5 for x.

Worked Solution
Apply the idea
\displaystyle -\dfrac{x}{2}\displaystyle \geq\displaystyle 5Start with the given inequality
\displaystyle -\dfrac{x}{2} \cdot \left(-2\right)\displaystyle \leq\displaystyle 5 \cdot \left(-2\right)Multiplication property of inequality (reverse inequality symbol)
\displaystyle x\displaystyle \leq\displaystyle 5 \cdot \left(-2\right)Multiplicative inverse
\displaystyle x\displaystyle \leq\displaystyle -10Evaluate the multiplication
Reflect and check

When we multiply both sides by -2, we must reverse the inequality symbol. If we had not reversed the symbol, we would have gotten:

\displaystyle -\dfrac{x}{2} \cdot \left(-2\right)\displaystyle \geq\displaystyle 5 \cdot \left(-2\right)
\displaystyle x\displaystyle \geq\displaystyle -10

Let's use the substitution property to check if a number in the solution set will satisfy the inequality. Let's use x=6, because it would be in the solution set for x \geq -10.

\displaystyle -\dfrac{x}{2}\displaystyle \geq\displaystyle 5Start with the original inequality
\displaystyle -\dfrac{\left(6\right)}{2}\displaystyle \geq\displaystyle 5Substitution property
\displaystyle -\left(3\right)\displaystyle \geq\displaystyle 5Evaluate the division
\displaystyle -3\displaystyle \geq\displaystyle 5Evaluate the multiplication

Because -3 is not greater than 5 or equal to 5, this solution set is invalid. This shows the importance of reversing the inequality sign when multiplying or dividing by a negative number.

Example 5

Compare the process for solving 4.5x=27 and 4.5x\gt 27 and their solution sets.

Worked Solution
Create a strategy

Begin by solving the equation and the inequality, then compare their solutions.

Apply the idea

First, we will solve the equation 4.5x=27

\displaystyle 4.5x\displaystyle =\displaystyle 27Start with the given equation
\displaystyle \dfrac{4.5x}{4.5}\displaystyle =\displaystyle \dfrac{27}{4.5}Division property of equality
\displaystyle x\displaystyle =\displaystyle 6Evaluate the division

The solution is x=6

Next, we will solve the inequality 4.5x\gt 27.

\displaystyle 4.5x\displaystyle >\displaystyle 27Start with the given inequality
\displaystyle \dfrac{4.5x}{4.5}\displaystyle >\displaystyle \dfrac{27}{4.5}Division property of inequality
\displaystyle x\displaystyle >\displaystyle 6Evaluate the division

The solution is x\gt 6

Both processes are very similar in how they use properties of equality and inequality. They also both have the number 6 related to their solutions.

The difference is in the actual solution sets. The equation has only one value, x=6, in its solution set. On a number line it looks like this: 4.5x=27

-1012345678910

The inequality has a range of values in its solution set. It is all values greater than 6 so, unlike the equation, x=6 is not a part of the solution set. 4.5x \gt 27

-1012345678910

Example 6

You have \$18 and you want to buy some slices of cake.

a

If a slice of cake costs \$5, write an inequality to find how many cake slices you can buy.

Worked Solution
Create a strategy

Let x represent the number of cake slices. The total number of slices you can buy is limited by the money you have, so we should set up the inequality to show that the amount you can spend is less than your budget.

Apply the idea

5x \lt 18

b

Solve the inequality you wrote in part (a).

Worked Solution
Create a strategy

Solve the inequality by isolating x to one side of the inequality.

Apply the idea
\displaystyle 5x\displaystyle <\displaystyle 18Inequality from part (a)
\displaystyle \frac{5x}{5}\displaystyle <\displaystyle \frac{18}{5}Division property of inequality
\displaystyle x\displaystyle <\displaystyle 3.6Evaluate the division

Since you can't buy a portion of a cake slice, the number of cake slices you can buy is 3 at most.

Idea summary

The properties of inequality are used when solving inequalities:

\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{Multiplication property of inequality} \\\text{ (by a negative number)} \text{If } a \lt b \text{ and } c \lt 0, \text{then } a \cdot c \gt b \cdot c \\ \text{If } a \gt b \text{ and } c \lt 0, \text{then } a\cdot c \lt 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{Division property of inequality} \\\text{ (by a negative number)}\text{if } a \lt b \text{ and } c\lt 0, \text{then } \dfrac{a}{c}\gt \dfrac{b}{c} \\ \text{if } a\gt b \text{ and } c\lt 0, \text{then } \dfrac{a}{c}\lt \dfrac{b}{c}

Remember, when multiplying or dividing both sides of an inequality by a negative number, you must reverse the inequality symbol so the inequality stays true.

Outcomes

7.PFA.4

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

7.PFA.4a

Apply properties of real numbers and the addition, subtraction, multiplication, and division properties of inequality to solve one- and two-step inequalities in one variable. Coefficients and numeric terms will be rational.

7.PFA.4b

Investigate and explain how the solution set of a linear inequality is affected by multiplying or dividing both sides of the inequality statement by a rational number less than zero.

7.PFA.4c

Represent solutions to one- or two-step linear inequalities in one variable algebraically and graphically using a number line.

7.PFA.4f

Solve problems in context that require the solution of a one- or two-step inequality.

7.PFA.4g

Identify a numerical value(s) that is part of the solution set of as given one- or two-step linear inequality in one variable.

7.PFA.4h

Describe the differences and similarities between solving linear inequalities in one variable and linear equations in one variable.

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