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## Topic C: Lessons 15-16 Compound inequalities

# Compound inequalities examples

CCSS.Math:

## Video transcript

Let's do some compound
inequality problems, and these are just inequality problems
that have more than one set of constraints. You're going to see what I'm
talking about in a second. So the first problem I have is
negative 5 is less than or equal to x minus 4, which is
also less than or equal to 13. So we have two sets of
constraints on the set of x's that satisfy these equations. x minus 4 has to be greater than
or equal to negative 5 and x minus 4 has to be less
than or equal to 13. So we could rewrite this
compound inequality as negative 5 has to be less than
or equal to x minus 4, and x minus 4 needs to be less
than or equal to 13. And then we could solve each of
these separately, and then we have to remember this "and"
there to think about the solution set because it has to
be things that satisfy this equation and this equation. So let's solve each of
them individually. So this one over here,
we can add 4 to both sides of the equation. The left-hand side, negative
5 plus 4, is negative 1. Negative 1 is less than
or equal to x, right? These 4's just cancel out here
and you're just left with an x on this right-hand side. So the left, this part right
here, simplifies to x needs to be greater than or equal to
negative 1 or negative 1 is less than or equal to x. So we can also write
it like this. X needs to be greater than
or equal to negative 1. These are equivalent. I just swapped the sides. Now let's do this other
condition here in green. Let's add 4 to both sides
of this equation. The left-hand side,
we just get an x. And then the right-hand
side, we get 13 plus 14, which is 17. So we get x is less than
or equal to 17. So our two conditions, x has to
be greater than or equal to negative 1 and less than
or equal to 17. So we could write this
again as a compound inequality if we want. We can say that the solution
set, that x has to be less than or equal to 17 and greater
than or equal to negative 1. It has to satisfy both
of these conditions. So what would that look
like on a number line? So let's put our number
line right there. Let's say that this is 17. Maybe that's 18. You keep going down. Maybe this is 0. I'm obviously skipping a bunch
of stuff in between. Then we would have a negative
1 right there, maybe a negative 2. So x is greater than or equal
to negative 1, so we would start at negative 1. We're going to circle it in
because we have a greater than or equal to. And then x is greater than that,
but it has to be less than or equal to 17. So it could be equal to
17 or less than 17. So this right here is a solution
set, everything that I've shaded in orange. And if we wanted to write it in
interval notation, it would be x is between negative 1 and
17, and it can also equal negative 1, so we put
a bracket, and it can also equal 17. So this is the interval notation
for this compound inequality right there. Let's do another one. Let me get a good
problem here. Let's say that we have
negative 12. I'm going to change the problem
a little bit from the one that I've found here. Negative 12 is less than 2 minus
5x, which is less than or equal to 7. I want to do a problem that has
just the less than and a less than or equal to. The problem in the book that
I'm looking at has an equal sign here, but I want to remove
that intentionally because I want to show you
when you have a hybrid situation, when you have
a little bit of both. So first we can separate this
into two normal inequalities. You have this inequality
right there. We know that negative 12 needs
to be less than 2 minus 5x. That has to be satisfied, and--
let me do it in another color-- this inequality also
needs to be satisfied. 2 minus 5x has to be less than
7 and greater than 12, less than or equal to 7 and greater
than negative 12, so and 2 minus 5x has to be less
than or equal to 7. So let's just solve this the
way we solve everything. Let's get this 2 onto the
left-hand side here. So let's subtract 2 from both
sides of this equation. So if you subtract 2 from both
sides of this equation, the left-hand side becomes negative
14, is less than-- these cancel out-- less
than negative 5x. Now let's divide both
sides by negative 5. And remember, when you multiply
or divide by a negative number, the inequality
swaps around. So if you divide both sides by
negative 5, you get a negative 14 over negative 5, and you have
an x on the right-hand side, if you divide that by
negative 5, and this swaps from a less than sign to
a greater than sign. The negatives cancel out, so you
get 14/5 is greater than x, or x is less than 14/5,
which is-- what is this? This is 2 and 4/5. x is less than 2 and 4/5. I just wrote this improper
fraction as a mixed number. Now let's do the other
constraint over here in magenta. So let's subtract 2 from both
sides of this equation, just like we did before. And actually, you can do these
simultaneously, but it becomes kind of confusing. So to avoid careless mistakes, I
encourage you to separate it out like this. So if you subtract 2 from both
sides of the equation, the left-hand side becomes
negative 5x. The right-hand side, you have
less than or equal to. The right-hand side becomes
7 minus 2, becomes 5. Now, you divide both sides
by negative 5. On the left-hand side,
you get an x. On the right-hand side, 5
divided by negative 5 is negative 1. And since we divided by a
negative number, we swap the inequality. It goes from less than or
equal to, to greater than or equal to. So we have our two
constraints. x has to be less than 2 and 4/5,
and it has to be greater than or equal to negative 1. So we could write
it like this. x has to be greater than or
equal to negative 1, so that would be the lower bound on our
interval, and it has to be less than 2 and 4/5. And notice, not less
than or equal to. That's why I wanted to show you,
you have the parentheses there because it can't be
equal to 2 and 4/5. x has to be less
than 2 and 4/5. Or we could write this way. x has to be less than 2 and
4/5, that's just this inequality, swapping the
sides, and it has to be greater than or equal
to negative 1. So these two statements
are equivalent. And if I were to draw it
on a number line, it would look like this. So you have a negative 1, you
have 2 and 4/5 over here. Obviously, you'll have
stuff in between. Maybe, you know, 0
sitting there. We have to be greater than or
equal to negative 1, so we can be equal to negative 1. And we're going to be greater
than negative 1, but we also have to be less than
2 and 4/5. So we can't include
2 and 4/5 there. We can't be equal to 2 and 4/5,
so we can only be less than, so we put a empty circle
around 2 and 4/5 and then we fill in everything below that,
all the way down to negative 1, and we include negative 1
because we have this less than or equal sign. So the last two problems I did
are kind of "and" problems. You have to meet both of
these constraints. Now, let's do an "or" problem. So let's say I have these
inequalities. Let's say I'm given-- let's say
that 4x minus 1 needs to be greater than or equal to
7, or 9x over 2 needs to be less than 3. So now when we're saying "or,"
an x that would satisfy these are x's that satisfy either
of these equations. In the last few videos or in the
last few problems, we had to find x's that satisfied
both of these equations. Here, this is much
more lenient. We just have to satisfy
one of these two. So let's figure out the solution
sets for both of these and then we figure out
essentially their union, their combination, all of
the things that'll satisfy either of these. So on this one, on the one
on the left, we can add 1 to both sides. You add 1 to both sides. The left-hand side just becomes
4x is greater than or equal to 7 plus 1 is 8. Divide both sides by 4. You get x is greater
than or equal to 2. Or let's do this one. Let's see, if we multiply both
sides of this equation by 2/9, what do we get? If you multiply both sides by
2/9, it's a positive number, so we don't have to do anything
to the inequality. These cancel out, and you get
x is less than 3 times 2/9. 3/9 is the same thing as
1/3, so x needs to be less than 2/3. So or x is less than 2/3. So that's our solution set. x needs to be greater than or
equal to 2, or less than 2/3. So this is interesting. Let me plot the solution
set on the number line. So that is our number line. Maybe this is 0, this is 1, this
is 2, 3, maybe that is negative 1. So x can be greater than
or equal to 2. So we could start-- let me
do it in another color. We can start at 2 here and it
would be greater than or equal to 2, so include everything
greater than or equal to 2. That's that condition
right there. Or x could be less than 2/3. So 2/3 is going to be right
around here, right? That is 2/3. x could be less than 2/3. And this is interesting. Because if we pick one of these
numbers, it's going to satisfy this inequality. If we pick one of these numbers,
it's going to satisfy that inequality. If we had an "and" here, there
would have been no numbers that satisfy it because you
can't be both greater than 2 and less than 2/3. So the only way that there's
any solution set here is because it's "or." You can
satisfy one of the two inequalities. Anyway, hopefully you,
found that fun.