Video: Inequalities and Interval Notation

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

17:37

Video Transcript

In this video, we will learn how to solve simple and compound linear inequalities and how to express their solutions using interval notation. Remember that an inequality is a statement which compares the size of two or more quantities which may be values or expressions. For example, the statement “𝑥 is greater than two” tells us that the variable 𝑥 can take any value as long as it is strictly bigger than two. The statement “𝑦 is less than or equal to four” means 𝑦 can take any value less than four, but it can also take the value four itself. So this time four is included in the possible answers or solution set.

A statement such as “zero is less than four 𝑧 minus three which is less than or equal to five” means that the value of the expression four 𝑧 minus three is strictly greater than zero, and it’s also less than or equal to five. So it can take any value from zero, not including zero itself, up to and including five. We’ll review how to solve linear inequalities in the context of examples. But first, let’s consider how we can express the solutions to inequalities using interval notation.

Suppose we have the statement 𝑥 is greater than seven. This means 𝑥 can take any value at all as long as it’s larger than seven. So that’s any value in the interval from seven all the way up to positive ∞. We can express this using interval notation where seven and ∞ are the endpoints of the interval. The type of brackets or parentheses used here are curved brackets, which are used for an open interval, which means that the endpoints of the interval themselves are not included.

Suppose, instead, we have the statement “negative two is less than or equal to 𝑦 is less than or equal to four.” This means that 𝑦 can take any value between negative two and four and the endpoints of the interval are included. We can express this as an interval with endpoints of negative two and four, but we use square brackets or parentheses. This type of interval is described as closed, and it means the end points of the interval are included. So negative two and four are each possible values for 𝑦.

Of course, we could also have a mixture of inequality signs, for example, the statement “zero is less than 𝑧 which is less than or equal to 10.” This means that 𝑧 can take any value from, but not including, zero up to and including 10. We would write this as an interval with endpoints of zero and 10, which is open at its lower end and closed at its upper end. So the value zero is not included in the interval, but the value of 10 is. Let’s now consider some examples of solving inequalities and then writing the solution set using interval notation.

Find the solution set of the inequality 𝑥 is less than or equal to 20 in the real numbers. Give your answer in interval notation.

Now, I said this automatically when reading the question, but the first thing to remember is that the inequality notation that’s been used here is a “less than or equal to” sign. So this inequality tells us that the value of 𝑥 is less than or equal to 20. This is known as a simple inequality as there is only one inequality sign involved. And actually, this inequality has already been solved for us as we have 𝑥 on its own on one side of the inequality. Picturing a number line, we know then that 𝑥 can take any value at all from 20 all the way down to negative ∞.

We can write this as an interval with negative ∞ and 20 as its endpoints. But we must be careful about the type of brackets or parentheses that we use at each end of the interval. As 𝑥 is less than or equal to 20, 20 is included in the possible values for 𝑥. So it is included in the solution set, which means our interval needs to be closed at its upper end. However, as negative ∞ does not exist as an actual number, our interval needs to be open at its lower end. We therefore have the solution set of this inequality in interval notation. It’s the interval from negative ∞ to 20, which is open at the lower end and closed at the upper end.

In our next example, we’ll review the steps needed to solve a double-sided inequality or a compound inequality and then write its solution in interval notation.

Find all the values of 𝑥 that satisfy seven 𝑥 plus four is greater than 11 and less than or equal to 32. Give your answer in interval form.

This is what’s known as a double-sided or compound inequality. In general, a compound inequality contains at least two inequalities separated by either the word “or” or the word “and”. Here, we have an expression, seven 𝑥 plus four, and the inequality tells us that the value of this expression is greater than 11 and it’s less than or equal to 32. There’re two approaches we can take to solving a double-sided inequality.

The first approach is to treat the two parts of the inequality separately. So we have one inequality telling us that seven 𝑥 plus four is greater than 11 and another telling us that seven 𝑥 plus four is less than or equal to 32. We then solve each inequality. For the inequality on the left, the first step is just subtract four from each side, giving seven is less than seven 𝑥. We can then divide each side of this inequality by seven to give one is less than 𝑥 or 𝑥 is greater than one. So, we’ve solved our first inequality.

To solve the second, we also subtract four from each side, giving seven 𝑥 is less than or equal to 28 and then divide both sides by seven to give 𝑥 is less than or equal to four. We’ve found then that the value of 𝑥 is greater than one and less than or equal to four. So we must make sure to put these two parts of the solution back together again at the end. We can write our solution as the double-sided inequality 𝑥 is greater than one and less than or equal to four. As an interval then, this would have endpoints of one and four.

And now we need to consider the type of brackets or parentheses for each end. At the lower end, the sign is a strict inequality; 𝑥 is strictly greater than one. So the value one is not included in the solution set. And so, our interval is open at the lower end. However, at the upper end, we have a weak inequality, 𝑥 is less than or equal to four. So the value of four is included in the solution set, and our interval is closed at its upper end. And so, we have our answer to the problem. The solution set for this inequality is the interval from one to four, open at its lower end and closed at its upper end.

Now, notice that the steps involved in solving these two separate inequalities were exactly the thing. In each case, we subtracted four first and then divided by seven. So, in fact, there was no need for us to treat the two parts of this inequality separately. The second and probably more efficient approach, then, is to keep all three parts of the inequality together. In this case, we solve in exactly the same way but we must make sure that we perform the same operation to all three parts of the inequality. We begin by subtracting four from each part. 11 minus four is seven. Seven 𝑥 plus four minus four is seven 𝑥. And 32 minus four is 28. So we now have the statement seven 𝑥 is greater than seven and less than or equal to 28.

We then divide each part of our inequality by seven, giving the statement 𝑥 is greater than one and less than or equal to four, which we notice is identical to the solution we had using our first method. We would then express this in interval notation in exactly the same way. This method is certainly quicker, but we must make sure that we treat all three parts of the inequality identically. So if we’re subtracting four, we must make sure we do it from every part. And if we’re dividing by some number, in this case seven, again we need to do it to every part.

Let’s now consider a slightly more complicated example involving fractions.

Find the solution set of the inequality five 𝑥 minus one over 10 is less than negative two 𝑥 plus five which is less than 𝑥 plus three over two in the real numbers. Give your answer in interval notation.

Now, this is a relatively complex question. We have a double-sided inequality, we have some fractions, and we also have our variable 𝑥 appearing in each part of this inequality. Let’s first consider how we could solve this problem by treating the two parts of the inequality separately. So we have the single-sided inequality five 𝑥 minus one over 10 is less than negative two 𝑥 plus five and another single-sided inequality negative two 𝑥 plus five is less than 𝑥 plus three over two.

Through our first inequality, we’d begin by multiplying each side by 10 to eliminate the fractions. And we now have five 𝑥 minus one is less than negative 20𝑥 plus 50. We can then add one to each side of this inequality. Now, we still have 𝑥 terms appearing on each side. So next, we want to group all of the 𝑥’s on the same side. And if we add 20𝑥 to each side, we’ll have a positive number of 𝑥’s on the left-hand side of our inequality. Adding 20𝑥 to each side of the inequality, then, gives 25𝑥 is less than 51. Finally, dividing both sides of our inequality by 25 gives 𝑥 is less than 51 over 25. And we’ll keep this answer in fractional form.

So we’ve solved the first part, and now let’s consider how we can solve the second part. First, we can multiply both sides by two to eliminate the fractions, giving negative four 𝑥 plus 10 is less than 𝑥 plus three. We can then subtract three from each side giving negative four 𝑥 plus seven is less than 𝑥 and then add four 𝑥 to each side of the inequality, giving seven is less than five 𝑥. Finally, we divide by five, giving seven over five is less than 𝑥 or 𝑥 is greater than seven over five.

So, we’ve solved the two parts of our inequality separately. But actually, we can combine these answers. We can express this as a double-sided inequality, 𝑥 is greater than seven over five and less than 51 over 25. The question, though, asked us to give our answer an interval notation. We can write this then as the interval with endpoints of seven over five and 51 over 25. And as each of the inequalities are strict inequalities, meaning neither of the endpoints are included in the interval, this means that our interval will be open at both ends. So we have our answer to the problem. The solution set of this inequality is the open interval from seven over five to 51 over 25.

Now, although the steps involved in solving our two separate inequalities were different here, it is actually possible to solve this inequality by keeping it all together, although this won’t always be the case. As the denominators involved in the fractions are 10 and two and as two is a factor of 10, we can eliminate each of the denominators by multiplying each part of our inequality by 10. Which takes us to five 𝑥 minus one is less than negative 20𝑥 plus 50, which is less than five 𝑥 plus 15. In that final part, the factor of two in the numerator has canceled with the factor of two in the denominator. But we still have a factor of five to multiply 𝑥 plus three by.

Now, we can subtract 50 from each part of the inequality to give five 𝑥 minus 51 is less than negative 20𝑥, which is less than five 𝑥 minus 35. But at this stage, we notice that we still have 𝑥 terms involved in each part of our inequality. We need to group them all together in the center. And in this instance, because the number of 𝑥’s on the far left of the inequality, that’s five 𝑥, is the same as the number of 𝑥’s on the far right of our inequality, we can do this by subtracting five 𝑥 from each part. However, had the number of 𝑥’s not been the same on the two sides of the inequality, we wouldn’t have been able to do this. And we’d need to have used our first method.

We now have negative 51 is less the negative 25𝑥, which is less than negative 35. Finally, we can divide each part of our inequality by negative 25. But we must be really careful here because we should remember that when we multiply or divide an inequality by a negative number, we have to reverse the direction of the inequality sign. So each of our less than signs must become greater than signs. We therefore have negative 51 over negative 25 is greater than 𝑥, which is greater than negative 35 over negative 25. In each of our fractions, the negatives in the numerators and denominators will cancel out. And the fraction 35 over 25 can be simplified to seven over five.

Writing our inequality the other way round so that the smallest number is on the left, we have seven over five is less than 𝑥 is less than 51 over 25, which is the same as the solution we found using our previous method. So we can convert our answer to interval notation in exactly the same way.

Let’s now consider one final example.

If 𝑥 is greater than negative six but less than or equal to four, then what interval does three 𝑥 minus five lie in?

In this question, we’re told the value of 𝑥 is between negative six and four. 𝑥 can be equal to four, but it can’t be equal to negative six. We then need to find the interval in which the expression three 𝑥 minus five lies, which is essentially the reverse process of solving an inequality. Let’s consider what operations have been performed to 𝑥 to give the expression three 𝑥 minus five. Well, firstly, 𝑥 has been multiplied by three and then we’ve subtracted five. We can do the same to the endpoints of the interval in which 𝑥 lies.

Firstly, multiplying each part of the inequality by three, we find that three 𝑥 is greater than negative 18 but less than or equal to 12. And then subtracting five from each part, we find that three 𝑥 minus five is greater than negative 23 and less than or equal to seven. We can therefore express this as an interval from negative 23 to seven. At the lower end of the interval, we have a strict inequality sign. Negative 23 is not included in the interval. So we use an open bracket. At the upper end of the interval, we have a weak inequality sign. Seven is included in the interval, so the interval is closed at this end.

And so, we’ve completed the problem. Three 𝑥 minus five lies in the interval from negative 23 to seven, which is open at the lower end and closed at the upper end.

Let’s now summarize the key points we’ve seen in this video. We can express the solution set of an inequality using interval notation. For weak inequalities, that’s less than or equal to or greater than or equal to, we use a closed interval as the values themselves are included in the interval. For strict inequalities, that’s strictly less than or strictly greater than, we use an open interval because the endpoints of the interval are not included. Although it’s not very common, we can also use outward-facing square brackets or parentheses to denote an open interval.

It’s possible for an interval to be open at one end and closed at the other. In the case of compound or double-sided inequalities, we can treat these either together or separately. And finally, a general reminder about solving inequalities, whenever we multiply or divide by a negative value, we must remember to reverse the direction of the inequality. A greater than sign must become a less than sign, and a less than sign must become a greater than sign.

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