Explainer: One-step Inequalities: Multiplication or Division

In this explainer, we will learn how to solve one-step linear inequalities by multiplication or division.

An inequality is a mathematical sentence containing an inequality symbol (<,≀,>,orβ‰₯); it shows that the value of one expression is greater than another, while an equation states the equality of two expressions. When a variable is involved in an inequality, solving an inequality means finding the range of values that the variable can take to make the inequality true. This range of values is given as an inequality in the form π‘₯Β (<,≀,>,orβ‰₯) 𝑐, where 𝑐 is a number.

We are going to learn in this lesson how to solve inequalities of the form π‘Žπ‘₯(<,≀,>,orβ‰₯)𝑐, where π‘Ž and 𝑐 are numbers.

Let us start with 3π‘₯<15. We can use the balance scale model here, interpreting this inequality by saying that 3π‘₯ is lighter than 15. This means that the side with 3π‘₯ is higher on the scale because it is lighter.

It makes sense here to divide the content of each scale side in three, so that we have π‘₯ on one side and 15Γ·3=5 on the other. So, we have

that is,

π‘₯<5.

If we have a fraction of π‘₯ in the inequality, for instance, 15π‘₯<5, then we can simply multiply both sides by 5 to solve for π‘₯:

that is,

π‘₯<25.

Let us solve the first question.

Example 1: Solving an Inequality by Multiplying or Dividing

Solve the following inequality: βˆ’2β‰₯π‘₯0.8.

  1. π‘₯β‰₯βˆ’2.5
  2. π‘₯β‰₯βˆ’2.8
  3. π‘₯β‰₯βˆ’1.6
  4. π‘₯β‰€βˆ’1.6
  5. π‘₯β‰€βˆ’2.8

Answer

To solve this inequality, we multiply both sides by 0.8 in order to have π‘₯ on its own on one side. We find

that is,

βˆ’1.6β‰₯π‘₯.

We want to have π‘₯ on the left-hand side of the inequality, so we write the equivalent inequality to βˆ’1.6β‰₯π‘₯, which is simply saying that if π‘Ž is greater than or equal to 𝑏, then 𝑏 is less than or equal to π‘Ž: π‘₯β‰€βˆ’1.6.

Our answer is π‘₯β‰€βˆ’1.6.

We can check that our answer is correct by first checking that when π‘₯=βˆ’1.6, then βˆ’2=π‘₯0.8. Then, using a value of π‘₯ that is less than βˆ’1.6, say βˆ’8, we plug it into π‘₯0.8. We find that βˆ’80.8=βˆ’10, which is less than βˆ’2. Therefore, the inequality is true with this value, so our answer is correct.

Now, let us consider the inequality βˆ’4π‘₯<28. Dividing our β€œweights” of βˆ’4π‘₯ and 28 by a negative number, βˆ’4, does not make any sense. So, we have a situation here where we cannot apply our balance-scale method. Instead, we are going to look at the meaning of an inequality on a number line.

Let us go back to a simple inequality, for instance, π‘₯β‰₯2. Its solutions can be represented on a number line: it is all the values greater than or equal to 2.

Knowing this, what can be said about the opposite of π‘₯, that is, βˆ’π‘₯? For this, recall that the opposite of a number is the number located on the number line at the same distance to zero, but on the other side of zero. Therefore, it is found by reflecting it in the origin. Here, we do not want to reflect a point in the origin but a ray. For this, we can take the endpoint of the ray (2) and another of its points, say 6, and reflect them in the origin: we find βˆ’2 and βˆ’6. The reflected ray starts at βˆ’2 and goes toward minus infinity. We see that reflecting our ray in the origin can be thought of as flipping it over the origin.

The blue ray represents all the values of π‘₯ that satisfy the inequality π‘₯β‰₯2 and the pink ray represents all the values of the opposite of π‘₯, βˆ’π‘₯. Looking at the pink ray, we see that all values of βˆ’π‘₯ are less than or equal to βˆ’2, which is written as βˆ’π‘₯β‰€βˆ’2.

With this example, we see that when we multiply both sides of π‘₯β‰₯2 by βˆ’1, which is equivalent to finding where the values of the opposite of π‘₯ lie, the inequality symbol changes from β€œgreater than or equal to” to β€œless than or equal to.” Visualizing the process on the number line makes it clear: values of π‘₯ greater than 2 make values of βˆ’π‘₯ less than βˆ’2.

Let us now consider the inequality βˆ’π‘₯>βˆ’3. Its solution can be represented with the blue ray on the diagram.

Now, we want to find where the values of π‘₯, the opposite of βˆ’π‘₯, lie. Remember that finding the opposite is reflecting in the origin, which can be thought of as flipping over the origin. We find here a ray whose endpoint is 3 and that goes toward minus infinity: it is where the values of π‘₯ lie.

We find here again that multiplying both sides of βˆ’π‘₯>βˆ’3 by βˆ’1 leads to changing the inequality symbol (here from greater than to less than). We find that π‘₯<3.

Let us now look at βˆ’9π‘₯>18.

We can split our process in two stages; first, we multiply both sides of the inequality by βˆ’1, meaning that the inequality sign will change from β€œgreater than” to β€œless than.” We find

that is,

9π‘₯<βˆ’18.

Then, we divide both sides by 9:

that is,

π‘₯<βˆ’2.

Both stages can be done in one by considering that whenever we multiply or divide both sides of an inequality by a negative number, the orientation of the inequality symbol changes. We would then go from βˆ’9π‘₯>18 to

These results can be summarized in the multiplication and division rule for inequalities.

Multiplication and Division Rule for Inequalities

An inequality still holds true when both sides of the inequality are multiplied or divided by the same positive number.

If π‘Ž<𝑏 and 𝑐>0, then π‘Žβ‹…π‘<𝑏⋅𝑐 and π‘ŽΓ·π‘<𝑏÷𝑐.

When both sides of an inequality are multiplied or divided by the same negative number, then the inequality symbol needs to change its orientation for the inequality to still hold true.

If π‘Ž<𝑏 and 𝑐=βˆ’|𝑐| (which means that 𝑐 is negative), then

and

Let us now look at some examples.

Example 2: Solving an Inequality by Multiplying or Dividing

Which of the following inequalities is equivalent to βˆ’4π‘₯β‰€βˆ’1?

  1. π‘₯=14
  2. π‘₯β‰₯4
  3. π‘₯<14
  4. π‘₯β‰₯14
  5. π‘₯≀14

Answer

To solve this inequality, we need to divide both sides by βˆ’4 to get π‘₯ on its own on one side.

Since βˆ’4 is a negative number, the inequality symbol will change its orientation when we divide both sides by βˆ’4. We find

that is,

π‘₯β‰₯14.

Now, we can check our answer. First, we check that βˆ’4π‘₯=βˆ’1 for π‘₯=14. Second, we take a value greater than 14, for instance, 1, and check that the inequality βˆ’4π‘₯β‰€βˆ’1 is true when we plug in this value. We find that βˆ’4β‹…1β‰€βˆ’1 is true; hence, our answer is correct.

Example 3: Solving an Inequality by Multiplying or Dividing

If βˆ’42π‘₯<βˆ’41, then .

  1. π‘₯<4142
  2. π‘₯>βˆ’4142
  3. π‘₯<βˆ’4142
  4. π‘₯>4142

Answer

To solve this inequality, we need to divide both sides by βˆ’42 to get π‘₯ on its own on one side.

Since βˆ’42 is a negative number, the inequality symbol will change its orientation when we divide both sides by βˆ’42. We find

that is,

π‘₯>4142.

Now, we can check our answer. First, we check that βˆ’42π‘₯=βˆ’41 for π‘₯=4142. Second, we take a value greater than 4142, for instance, 1, and check that the inequality βˆ’42π‘₯<βˆ’41 is true when we plug in this value. We find that βˆ’42β‹…1<βˆ’41 is true; hence, our answer is correct.

In the next examples, we are going to learn how to solve two-step inequalities.

Example 4: Solving a Two-Step Inequality by Multiplying or Dividing

Solve the following inequality: 𝑛5+1<6.

  1. 𝑛>25
  2. 𝑛<35
  3. 𝑛>35
  4. 𝑛<25
  5. 𝑛>12

Answer

To solve this inequality, we need to split the process into two steps. First, we isolate the 𝑛-term, and for this we subtract 1 from each side. We get

which is

𝑛5<5.

Second, we multiply both sides by 5 in order to have 𝑛 on its own on one side. We find

that is,

𝑛<25.

Now, we can check our answer. First, we check that 𝑛5+1=6 for 𝑛=25. Second, we take a value less than 25, for instance, 5, and check that the inequality 𝑛5+1<6 is true when we plug in this value. We find that 55+1<6 is true; hence, our answer is correct.

Example 5: Solving a Two-Step Inequality by Multiplying or Dividing

Solve the inequality 5βˆ’12π‘₯β‰₯10 for π‘₯.

  1. π‘₯β‰₯βˆ’10
  2. π‘₯≀103
  3. π‘₯≀10
  4. π‘₯β‰€βˆ’10
  5. π‘₯β‰€βˆ’15

Answer

To solve this inequality, we need to split the process into two steps. First, we isolate the π‘₯-term, and for this we subtract 5 from each side. We get

which is

βˆ’12π‘₯β‰₯5.

Second, we multiply both sides by βˆ’2 in order to have π‘₯ on its own on one side. As βˆ’2 is a negative number, we need to change the orientation of the inequality symbol. We find

that is,

π‘₯β‰€βˆ’10.

Now, we can check our answer. First, we check that 5βˆ’12π‘₯=10 for π‘₯=βˆ’10. Second, we take a value less than βˆ’10, for instance, βˆ’20, and check that the inequality 5βˆ’12π‘₯β‰₯10 is true when we plug in this value. We find that 5βˆ’βˆ’202β‰₯10 is true; hence, our answer is correct.

Key Points

  1. An inequality still holds true when both sides of the inequality are multiplied or divided by the same positive number.
    If π‘Ž<𝑏 and 𝑐 is positive, then π‘Žβ‹…π‘<𝑏⋅𝑐 and π‘ŽΓ·π‘<𝑏÷𝑐.
  2. When both sides of an inequality are multiplied or divided by the same negative number, then the inequality symbol needs to change its orientation for the inequality to still hold true.
    If π‘Ž<𝑏 and 𝑐=βˆ’|𝑐| (which means that 𝑐 is negative), then
    and
  3. To solve an inequality of the form π‘Žπ‘₯+𝑏(<,≀,>,orβ‰₯)𝑐, we first use the addition and subtraction rule for inequalities to have the π‘₯-term on its own on one side:
    and then we use the multiplication and division rule to have only π‘₯ on one side, keeping in mind that if a is positive, the inequality symbol will stay the same, while if it is negative, it will change its orientation.

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