### 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.