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Lesson Explainer: Multistep Inequalities Mathematics • 7th Grade

In this explainer, we will learn how to solve multistep inequalities.

Before we begin explaining how to solve multistep inequalities, it is a good idea to recall what we know about solving one-step and two-step inequalities.

We start by recalling that an inequality is a way of describing a set of values, which can be represented on a number line. We use the symbol < to mean β€œless than” and > to mean β€œgreater than.” For example, π‘₯>3 means all the values of π‘₯ that are greater than 3. If we have a number line with the integers marked, then we can represent all integers greater than 3 on a number line by marking them with a solid dot.

We use an arrow to show that this pattern continues. This is a graphical representation of the set {π‘₯∢π‘₯βˆˆβ„€,π‘₯>3}. We could also write this set as {4,5,6,…}.

We can solve inequalities by rewriting them into this form. For example, if we are told that π‘₯+1>4 and π‘₯βˆˆβ„€, then we want the set of values that are greater than 4 when added to 1. So, the values of π‘₯+1 must lie on the positive side of 4.

We see that the values of π‘₯+1 can be equal to any of 5,6,7,…. In other words, when we add 1 to π‘₯, it must be larger than 4.

We can note that adding 1 to π‘₯ and checking if it is larger than 4 is the same as checking if π‘₯ is larger than 4βˆ’1=3. Hence, our values of π‘₯ must satisfy the inequality π‘₯>3.

We can think of this as subtracting 1 from both sides of the inequality, π‘₯+1βˆ’1>4βˆ’1π‘₯>3, or as shifting the set on the number line one unit in the negative direction.

It is worth noting that this works for any constant; adding or subtracting from both sides of the inequality gives an equivalent inequality. The same would be true if we added π‘₯ or a multiple of π‘₯ to or subtracted it from either side, since this holds for all possible values. For example, if 2π‘₯>π‘₯+1, we can subtract π‘₯ from both sides to get the equivalent inequality: 2π‘₯βˆ’π‘₯>π‘₯+1βˆ’π‘₯(2βˆ’1)π‘₯>1π‘₯>1.

Let’s now recall what it means to multiply the value of π‘₯ by a constant. By considering a number line, we note that the position of π‘₯ is its displacement from 0. So, multiplying this value by a constant will multiply this displacement and the sign of this constant will determine the direction.

For example, if we multiply π‘₯ by 2, then we are doubling the displacement of π‘₯. To help see how this affects the solution set of an inequality, consider the inequalities π‘₯>1 and 2π‘₯>1 for integer values of π‘₯. We can sketch the solution set of π‘₯>1 as follows.

For the inequality 2π‘₯>1, we want to represent all integer values of π‘₯ that, after having their displacements doubled, are at a displacement greater than 1 from 0. We note that if we want the displacement, after doubling, to be greater than 1, then this is the same as saying the displacement must be greater than 12. We get the following.

This is equivalent to just dividing both sides of the inequality through by 2 to get 2π‘₯2>12π‘₯>12.

The result is similar if the coefficient of π‘₯ is negative; however, the process is more difficult. For example, if we multiply π‘₯ by βˆ’3, then we triple the displacement and switch the direction. To help see this, consider π‘₯<βˆ’1, for integer values of π‘₯.

If we instead had βˆ’3π‘₯<βˆ’1, we would be looking for the values of π‘₯ that, after having their displacements tripled and reversed, would still be less than βˆ’1. This is difficult to think about, so we approach this in steps.

First, reversing the displacement is the same as reversing the inequality. We can think of this as multiplying the inequality through by βˆ’1 or reversing the direction on a number line: βˆ’3π‘₯<βˆ’1βˆ’3π‘₯Γ—βˆ’1>βˆ’1Γ—βˆ’13π‘₯>1.

Second, if we want the displacement, after tripling, to be greater than 1, then this is the same as saying the displacement must be greater than 13. We can think of this as dividing the inequality through by 3: 3π‘₯>13π‘₯Γ·3>1Γ·3π‘₯>13.

These results allow us to solve inequalities with linear expressions on either side by rearranging.

Properties: Solving Inequalities

  • Adding or subtracting a constant from either side of an inequality gives an equivalent inequality. For example, if π‘Ž>𝑏, then π‘Ž+𝑐>𝑏+𝑐.
  • Multiplying or dividing both sides of an inequality by a positive number gives an equivalent inequality. For example, if 𝑐>0 and π‘Ž>𝑏, then π‘Žπ‘>𝑏𝑐.
  • Multiplying or dividing both sides of an inequality by a negative number reverses the direction of the inequality. For example, if 𝑐<0 and π‘Ž>𝑏, then π‘Žπ‘<𝑏𝑐.

These properties are referred to as axioms of inequalities, and they hold true for rational inequalities as well.

We can use these properties to rewrite inequalities. For example, imagine we are asked to find the set of rational numbers that satisfy the inequality 2π‘₯βˆ’1<4π‘₯+3. We can do this by using the above rules to isolate the variable, π‘₯. First, rearrange the inequality so that the π‘₯ terms are on the same side of the inequality. We can do this by subtracting 2π‘₯ from both sides, noting that, for any rational number π‘₯, 2π‘₯ is a rational number. So, we are just subtracting a rational number from both sides of the inequality. This will give us an equivalent inequality: 2π‘₯βˆ’1βˆ’2π‘₯<4π‘₯+3βˆ’2π‘₯βˆ’1<(4βˆ’2)π‘₯+3βˆ’1<2π‘₯+3.

We can now subtract 3 from both sides of the inequality to get βˆ’1βˆ’3<2π‘₯+3βˆ’3βˆ’4<2π‘₯.

Now, we divide both sides of the inequality through by 2, where we note 2 is positive, so this will give us an equivalent inequality: βˆ’42<2π‘₯2βˆ’2<π‘₯.

We could leave the inequality like this; however, we usually write the variable on the left. We can do this by switching the direction of the inequality: βˆ’2<π‘₯ is equivalent to π‘₯>βˆ’2.

Finally, if we want to give our answer as a set, we note that π‘₯βˆˆβ„š and π‘₯>βˆ’2, and we write this set in set-builder notation as {π‘₯∢π‘₯βˆˆβ„š,π‘₯>βˆ’2}.

We can continue following this process to solve inequalities with linear expressions on either side. In our first example, we need to solve a multistep inequality over the set of rational numbers.

Example 1: Solving Multistep Linear Inequalities over the Set of Rational Numbers

Solve the inequality 5π‘šβˆ’9(π‘š+3)<14 in β„š.

Answer

We should start by simplifying the given inequality. We can distribute the 9 over the parentheses on the left-hand side of the inequality to get 5π‘šβˆ’9(π‘š+3)<145π‘šβˆ’9π‘šβˆ’27<14.

Then, we simplify, giving us βˆ’4π‘šβˆ’27<14.

There are now two methods for solving this inequality: we can isolate the variable m either on the left-hand side of the inequality or on the right-hand side. Isolating the variable on the left-hand side of the inequality is often more efficient for single inequalities since we usually read these inequalities with the variable on the left. However, this will sometimes mean we need to divide the inequality through by a negative number, so isolating the variable on the right-hand side can also be a preferrable method.

If we want to isolate the variable π‘š on the left-hand side of the inequality, we add 27 to both sides of the inequality, where we note adding a constant to both sides gives an equivalent inequality. This gives βˆ’4π‘šβˆ’27+27<14+27βˆ’4π‘š<41.

Now, we want to divide both sides of the inequality by βˆ’4. We note that this is negative, so we also need to switch the direction of the inequality. We get (βˆ’4π‘š)Γ·(βˆ’4)>41Γ·(βˆ’4)π‘š>βˆ’414.

If we instead isolate π‘š on the right-hand side of the inequality, we add 4π‘š to both sides of the inequality to get βˆ’4π‘šβˆ’27+4π‘š<14+4π‘šβˆ’27<4π‘š+14.

We then subtract 14 from both sides of the inequality, giving us βˆ’41<4π‘š.

We then divide the inequality through by 4, yielding βˆ’414<π‘š.

We usually write the variable on the left-hand side of the inequality, so we will switch the sides and direction of this inequality. We get π‘š>βˆ’414.

In both cases, we want to write the inequality in set-builder notation. We note that we are told that π‘š must be rational, so we have ο¬π‘šβˆΆπ‘šβˆˆβ„š,π‘š>βˆ’414.

In our next example, we will solve a multistep inequality over the set of integers.

Example 2: Writing the Solution Set of Linear Inequalities with the Unknown on Both Sides over the Integer Numbers Set

Given that π‘₯βˆˆβ„€, write the solution set of 2π‘₯βˆ’6≀π‘₯βˆ’1.

Answer

To solve this inequality, we first need to isolate the variable, π‘₯. We can do this by subtracting π‘₯ from both sides of the inequality, where we note this will give an equivalent inequality. This gives us 2π‘₯βˆ’6βˆ’π‘₯≀π‘₯βˆ’1βˆ’π‘₯π‘₯βˆ’6β‰€βˆ’1.

We can now add 6 to both sides of the inequality to get π‘₯βˆ’6+6β‰€βˆ’1+6π‘₯≀5.

There are many different ways we can represent this set. We can use a number line by sketching solid dots over all integers that are less than or equal to 5, along with an arrow to show the direction.

We can also write this in set-builder notation as {π‘₯∢π‘₯βˆˆβ„€,π‘₯≀5}. Alternatively, we can just represent the solution set as {5,4,3,…}.

In our next example, we will solve a multistep inequality over the rational numbers.

Example 3: Solving Multistep Linear Inequalities over the Set of Rational Numbers

Solve the inequality βˆ’4(𝑦+4)βˆ’12<βˆ’50βˆ’(47βˆ’π‘¦) in β„š.

Answer

To solve an inequality with linear expressions on either side, we want to isolate the variable on the left-hand side of the inequality. To do this, we should start by simplifying the expressions on either side of the inequality. Let’s distribute both sets of parentheses and simplify: βˆ’4π‘¦βˆ’16βˆ’12<βˆ’50βˆ’47+π‘¦βˆ’4π‘¦βˆ’28<βˆ’97+𝑦.

We can now subtract 𝑦 from both sides of the inequality to combine like terms and produce an equivalent inequality. This gives βˆ’4π‘¦βˆ’28βˆ’π‘¦<βˆ’97+π‘¦βˆ’π‘¦βˆ’5π‘¦βˆ’28<βˆ’97.

Next, we add 28 to both sides of the inequality to get βˆ’5π‘¦βˆ’28+28<βˆ’97+28βˆ’5𝑦<βˆ’69.

To isolate 𝑦, we need to divide the inequality through by βˆ’5. Since this is negative, we recall that we will need to reverse the direction of the inequality. We get βˆ’5𝑦(βˆ’5)>βˆ’69(βˆ’5)𝑦>695.

Finally, since we are solving for π‘¦βˆˆβ„š, we can write the solutions to the inequality in set-builder notation as ο¬π‘¦βˆΆπ‘¦βˆˆβ„š,𝑦>695.

In our next example, we will solve an inequality over the set of natural numbers.

Example 4: Representing the Solution to a Linear Inequality on a Number Line

Which of the following number lines represents the solution set of the inequality 7(π‘₯βˆ’2)βˆ’8<5π‘₯βˆ’10 in β„•?

Answer

In order to solve an inequality, we want to use the axioms of inequalities to isolate the variable on one side of the inequality. We can start by distributing the factor of 7 over the parentheses on the left-hand side of the inequality to get 7(π‘₯βˆ’2)βˆ’8<5π‘₯βˆ’107π‘₯βˆ’14βˆ’8<5π‘₯βˆ’107π‘₯βˆ’22<5π‘₯βˆ’10.

We can now subtract 5π‘₯ from both sides of the inequality to obtain 7π‘₯βˆ’22βˆ’5π‘₯<5π‘₯βˆ’10βˆ’5π‘₯(7βˆ’5)π‘₯βˆ’22<βˆ’102π‘₯βˆ’22<βˆ’10.

We then add 22 to both sides of the inequality, giving us 2π‘₯βˆ’22+22<βˆ’10+222π‘₯<12.

Finally, we divide both sides of the inequality through by 2 to obtain π‘₯<6.

We need to be careful, since our solutions must also be natural numbers. We can write this in set-builder notation as {π‘₯∢π‘₯βˆˆβ„•,π‘₯<6} or as the set {0,1,2,3,4,5}. We can represent this on a number line by using dots above the numbers in this set.

We see that this matches choice B.

In our next example, we will solve an inequality over integers by identifying the solution set on a number line.

Example 5: Representing the Solution to a Linear Inequality on a Number Line

Given that π‘₯βˆˆβ„€, which of the following number lines represents the solution set of 3π‘₯βˆ’5β‰₯4π‘₯βˆ’3?

Answer

To sketch the solution set of an inequality on a number line, we first want to isolate the variable on the left-hand side of the inequality. To do this, let’s first subtract 3π‘₯ from both sides of the inequality: 3π‘₯βˆ’5β‰₯4π‘₯βˆ’33π‘₯βˆ’5βˆ’3π‘₯β‰₯4π‘₯βˆ’3βˆ’3π‘₯βˆ’5β‰₯(4βˆ’3)π‘₯βˆ’3βˆ’5β‰₯π‘₯βˆ’3.

We can then add 3 to both sides of the inequality to get the equivalent inequality: βˆ’5+3β‰₯π‘₯βˆ’3+3βˆ’2β‰₯π‘₯.

It is worth noting that we usually write the variable on the left-hand side of the inequality. We can do this by also switching the direction of the inequality to get π‘₯β‰€βˆ’2.

We want all integers that are less than or equal to βˆ’2. That is {βˆ’2,βˆ’3,βˆ’4,…}.

Since we want to include βˆ’2 in this set, we use a solid dot at βˆ’2 and we include all integers below βˆ’2. This gives the following.

We see that this is answer A.

In our final example, we will solve a multistep inequality over rational numbers, where we use set-builder notation to represent the solution set.

Example 6: Solving Multistep Linear Inequalities over the Set of Rational Numbers

Solve the inequality 6π‘₯βˆ’274β‰₯45 in β„š.

Answer

In order to solve an inequality over the set of rational numbers, we want to use the axioms of inequalities to isolate the variable on one side of the inequality. We can start by multiplying both sides of the inequality by 20 simplify the fractions. We have 20Γ—6π‘₯βˆ’274β‰₯20Γ—455Γ—(6π‘₯βˆ’27)β‰₯4Γ—430π‘₯βˆ’135β‰₯16.

We can then add 135 to both sides of the inequality to get the equivalent inequality: 30π‘₯βˆ’135+135β‰₯16+13530π‘₯β‰₯151.

Finally, we divide both sides of the inequality through by 30 to obtain π‘₯β‰₯15130.

We can write this in set-builder notation as π‘₯∢π‘₯βˆˆβ„š,π‘₯β‰₯15130.

Let’s finish by recapping some of the important points from this explainer.

Key Points

  • We can represent solutions to inequalities over the set of integers or natural numbers on a number line by using solid dots to represent the solutions and an arrow to show the direction.
  • Adding a constant to or subtracting it from either side of an inequality gives an equivalent inequality.
  • We can also add constant multiples of the variable to or subtract them from either side of an inequality to get an equivalent inequality.
  • Multiplying or dividing both sides of an inequality by a positive number gives an equivalent inequality.
  • Multiplying or dividing both sides of an inequality by a negative number reverses the direction of the inequality.

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