The portal has been deactivated. Please contact your portal admin.

Lesson Explainer: Inequalities and Interval Notation Mathematics • 9th Grade

In this explainer, we will learn how to solve simple and compound linear inequalities and how to express their solutions using interval notation.

Before we begin discussing solving inequalities over the real numbers, we should recall the properties and methods of solving inequalities over the rational numbers.

First, we recall that we can add or subtract rational constants from both sides of an inequality to get an equivalent inequality.

We can also multiply or divide both sides of an inequality by a nonzero rational number to get an equivalent inequality, where we switch the direction of the inequality if the number is negative.

Second, we recall that we can sketch inequalities on a number line by using an open circle to represent that an endpoint is not included in the set and a closed circle to represent that an endpoint is included in the set.

For example, we can represent π‘₯<βˆ’3 on a number line by using an open circle at βˆ’3 and an arrow pointing left to represent all numbers less than βˆ’3.

This indicates that π‘₯ can take any value less than βˆ’3 and it satisfies the inequality.

π‘₯β‰€βˆ’3 has the same representation as π‘₯<3 on a number line with the only difference being that we use a closed circle at βˆ’3 to show that this value is included.

This indicates that π‘₯ can take any value less than or equal to βˆ’3 and it satisfies the inequality.

All of these results also hold for real numbers. For example, we can add or subtract any real number from both sides of an inequality to find an equivalent inequality. We can also multiply or divide by any nonzero real number to find an equivalent inequality where we switch the direction of the inequality if the real number is negative. Finally, we can also represent solutions over the real numbers on a number line in the same way.

We can write the properties for rewriting inequalities over the real numbers formally as follows.

Properties: Equivalent Inequalities over the Real Numbers

  • Adding or subtracting a real 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 real 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 real number reverses the direction of the inequality. For example, if 𝑐<0 and π‘Ž>𝑏, then π‘Žπ‘<𝑏𝑐.

These results allow us to solve any inequality over the real numbers where both sides of the inequality have linear expressions in a single variable. We do this by isolating the variable on one side of the inequality.

Before we see our first example, it is worth recalling that we can also represent inequalities using interval notation. We use an open parenthesis, ) or (, to represent that the endpoint is not included and a closed bracket, ] or [, to show that the endpoint is included.

For example, we can represent the set of real numbers greater than 2 but less than or equal to 3 as the compound inequality 2<π‘₯≀3 or as the interval ]2,3]. We can also sketch the set on a number line as shown.

We also note that we can use the infinity symbol, ∞, to represent an unbounded interval. For example, we know that the inequality π‘₯>4 has solutions of π‘₯ that can take infinitely large values. To represent this in interval notation, we use a left open interval at 4, since 4 is not included in the interval, and a right open interval at infinity: ]4,∞[. On a number line, this is given by the following.

In our first example, we will rewrite the solution set of an inequality in interval notation.

Example 1: Representing an Inequality in Interval Notation

Find the solution set of the inequality π‘₯≀20 in ℝ. Give your answer in interval notation.

Answer

We first note that the solution set of the inequality π‘₯≀20 includes all real numbers less than or equal to 20. We can sketch this solution set on a number line using a closed circle at 20 to show it is included in the set. This gives us the following.

To write this set in interval notation, we recall that since the left side of the interval is unbounded, we will need to use the symbol βˆ’βˆž. We also note that the right endpoint is closed at 20. We use an open left parenthesis at βˆ’βˆž and a closed right bracket at 20 to get the solution set ]βˆ’βˆž,20].

In our next example, we will solve an inequality over the real numbers by using the properties of equivalent inequalities and we will represent the solution set on a number line.

Example 2: Representing the Solution Set of an Inequality on a Number Line

Which of the following diagrams represents the inequality √2π‘₯>√8?

Answer

To find the solution set of this inequality, we will start by isolating the variable π‘₯ on the left-hand side of the inequality. To do this, we want to divide the inequality through by √2. We can do this by recalling that we can divide both sides of an inequality through by a positive real number to find an equivalent inequality. This gives us π‘₯>√8√2.

We now want to rewrite √8 by noting that 8=4Γ—2, and we recall that we can take the square root of each factor separately. Thus, √8√2=√4Γ—2√2=√4Γ—βˆš2√2=2.

Therefore, the inequality simplifies to π‘₯>2.

We can sketch this on a number line by drawing an open circle at 2 to show that this value is not included and then including all numbers greater than 2. This gives us the following diagram.

This is given as option B.

In our next example, we will solve a compound inequality over the real numbers by using the properties of equivalent inequalities and we will represent the solution set on a number line.

Example 3: Representing the Solution Set of a Multistep Inequality on a Number Line

Which of the following shows the solution set of the inequality βˆ’βˆš3(π‘₯+1)<2√3π‘₯?

Answer

To sketch the solution set of this inequality, we first want to isolate the variable π‘₯ on the left-hand side of the inequality. We can do this in two ways.

We can divide the inequality through by √3 to get βˆ’(π‘₯+1)<2π‘₯.

Then, we can distribute βˆ’1 over the parentheses to get βˆ’π‘₯βˆ’1<2π‘₯.

Next, we recall that we can add any real number to both sides of the inequality to get an equivalent inequality. We then add π‘₯ to both sides to get βˆ’1<3π‘₯.

Then, we divide through by 3 to get βˆ’13<π‘₯.

Finally, we want π‘₯ on the left-hand side of the inequality. So, we switch the sides of the inequality and its direction to get π‘₯>βˆ’13.

We can also start by distributing on the left-hand side to get βˆ’βˆš3π‘₯βˆ’βˆš3<2√3π‘₯.

Thus, we can add √3 to both sides of the inequality to get βˆ’βˆš3π‘₯<2√3π‘₯+√3.

Similarly, we can subtract 2√3π‘₯ from both sides to get βˆ’βˆš3π‘₯βˆ’2√3π‘₯<√3βˆ’3√3π‘₯<√3.

We can now isolate π‘₯ on the left-hand side by dividing both sides of the inequality through by βˆ’3√3. We note that since this is negative, we need to switch the direction of the inequality. This gives π‘₯>√3βˆ’3√3π‘₯>βˆ’13.

We can sketch this on a number line by drawing an open circle at βˆ’13 to show that this value is not included and then including all numbers greater than βˆ’13. This gives us the following diagram.

This is given as option E.

In our next example, we will solve a compound inequality over the set of real numbers, and we will give our answer as an interval.

Example 4: Solving a Compound Inequality and Giving the Solution as an Interval

Find all the values of π‘₯ that satisfy 11<7π‘₯+4≀32. Give your answer in interval form.

Answer

Since the variable π‘₯ only appears in the middle of the compound inequality, we can solve this inequality by isolating π‘₯ in the middle. First, we subtract 4 from all parts of the inequality to get the equivalent inequality 11βˆ’4<7π‘₯+4βˆ’4≀32βˆ’47<7π‘₯≀28.

Next, we divide all parts of the inequality through by 7 to get 77<7π‘₯7≀2871<π‘₯≀4.

Therefore, the values of π‘₯ that satisfy this compound inequality are all of the real values between 1 and 4, excluding 1 but including 4. We can sketch this set on a number line by connecting an open circle at 1 to show that this value is not included and a closed circle at 4 to show that this value is included with a line.

Finally, we need to write this in interval notation. We use an open left parenthesis at 1 to show this value is not included and a closed right bracket at 4 to show this value is included in the solution set and satisfies the inequality.

This gives us that the values of π‘₯ that satisfy the inequality are any of the members of the set ]1,4].

In our next example, we will once again solve a compound inequality over the set of real numbers and we will give our answer in interval notation.

Example 5: Solving a Compound Inequality Where the Variable Appears on All Sides of the Inequality

Find the solution set of the inequality 5π‘₯βˆ’110<βˆ’2π‘₯+5<π‘₯+32 in ℝ. Give your answer in interval notation.

Answer

We first recall that we can solve compound inequalities by treating them as two separate inequalities. We have 5π‘₯βˆ’110<βˆ’2π‘₯+5 and βˆ’2π‘₯+5<π‘₯+32. We can solve each of these inequalities by isolating π‘₯ on the left-hand side of the inequality.

Let’s start with 5π‘₯βˆ’110<βˆ’2π‘₯+5. We can multiply the inequality through by 10 to write the coefficients of π‘₯ as integers. This gives 10Γ—ο€Ό5π‘₯βˆ’110<10Γ—(βˆ’2π‘₯+5)5π‘₯βˆ’1<βˆ’20π‘₯+50.

We can then add 20π‘₯ to both sides of the inequality to get 25π‘₯βˆ’1<50.

Next, we add 1 to both sides of the inequality, giving us 25π‘₯<51.

Now, to isolate π‘₯, we divide both sides of the inequality through by 25. This gives us π‘₯<5125.

We also need the second inequality to be satisfied by solutions to the original compound inequality. We can rewrite the second inequality βˆ’2π‘₯+5<π‘₯+32 by first multiplying through by 2 to get 2Γ—(βˆ’2π‘₯+5)<2Γ—ο€Όπ‘₯+32οˆβˆ’4π‘₯+10<π‘₯+3.

Next, we subtract π‘₯ from both sides of the inequality. This gives βˆ’5π‘₯+10<3.

Then, we subtract 10 from both sides of the inequality. We get βˆ’5π‘₯<βˆ’7.

We can now divide the inequality through by βˆ’5; we recall that since this value is negative, we need to switch the direction of the inequality to get π‘₯>75.

Therefore, for π‘₯ to be a solution to the original compound inequality, we must have that π‘₯<5125 and π‘₯>75. It is worth noting that 5125=2.04 and 75=1.4. So, this is equivalent to saying that π‘₯ must be greater than 1.4 and less than 2.04. We can write this as an interval by using open parentheses to show that the endpoints are not included.

Hence, the solution set of the inequality is 75,5125.

In our final example, we will solve a compound inequality involving radical coefficients.

Example 6: Solving a Compound Inequality Involving Radical Coefficients

Find the solution set of the inequality π‘₯√2<βˆ’βˆš2π‘₯+√2<3√2π‘₯ in ℝ. Give your answer in interval notation.

Answer

We begin by recalling that we can solve compound inequalities by treating them as two separate inequalities. So, for π‘₯ to be a solution of the inequality, it must be a solution of both π‘₯√2<βˆ’βˆš2π‘₯+√2 and βˆ’βˆš2π‘₯+√2<3√2π‘₯. We can solve each of these inequalities by isolating π‘₯ on either side of the inequality.

Let’s start with π‘₯√2<βˆ’βˆš2π‘₯+√2. We can simplify the inequality by multiplying through by √2. This gives us √2Γ—π‘₯√2<√2Γ—ο€»βˆ’βˆš2π‘₯+√2π‘₯<√2Γ—ο€»βˆ’βˆš2π‘₯+√2.

We then distribute √2 over the parentheses in the right-hand side of the inequality and simplify to get π‘₯<√2Γ—ο€»βˆ’βˆš2π‘₯+ο€»βˆš2Γ—βˆš2π‘₯<βˆ’2π‘₯+2.

We can then add 2π‘₯ to both sides of the inequality. This gives us π‘₯+2π‘₯<βˆ’2π‘₯+2+2π‘₯3π‘₯<2.

Now, we divide both sides of the inequality through by 3. This yields π‘₯<23.

We also need the second inequality to be satisfied by solutions to the original compound inequality. We can rewrite the second inequality βˆ’βˆš2π‘₯+√2<3√2π‘₯ by multiplying through by √2 to get √2Γ—ο€»βˆ’βˆš2π‘₯+√2<√2Γ—3√2π‘₯√2Γ—ο€»βˆ’βˆš2π‘₯+√2<6π‘₯.

We then distribute √2 over the parentheses in the left-hand side of the inequality and simplify to get ο€»βˆš2Γ—βˆ’βˆš2π‘₯+ο€»βˆš2Γ—βˆš2<6π‘₯βˆ’2π‘₯+2<6π‘₯.

Next, we add 2π‘₯ to both sides of the inequality. This gives βˆ’2π‘₯+2+2π‘₯<6π‘₯+2π‘₯2<8π‘₯.

We can now divide the inequality through by 8 to get 28<π‘₯14<π‘₯.

We can combine these two inequalities to get 14<π‘₯<23. We can write this inequality as an interval by using open parentheses to show that the endpoints are not included.

Hence, the solution set of the given inequality is 14,23.

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

Key Points

  • Adding or subtracting a real constant from either side of an inequality gives an equivalent inequality.
  • We can also add or subtract constant real multiples of the variable from either side of an inequality to get an equivalent inequality.
  • Multiplying or dividing both sides of an inequality by a positive real number gives an equivalent inequality.
  • Multiplying or dividing both sides of an inequality by a negative real number reverses the direction of the inequality.
  • We can represent inequalities over 𝑅 in set-builder notation, on number lines, or in interval notation.
  • We can solve compound inequalities by treating them as two separate inequalities. So, for π‘₯ to be a solution of the inequality, it must be a solution of both.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.