WEBVTT

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Let's do some compound
inequality problems, and these

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are just inequality problems
that have more than one set of

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constraints.

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You're going to see what I'm
talking about in a second.

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So the first problem I have is
negative 5 is less than or

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equal to x minus 4, which is
also less than or equal to 13.

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So we have two sets of
constraints on the set of x's

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that satisfy these equations.

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x minus 4 has to be greater than
or equal to negative 5

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and x minus 4 has to be less
than or equal to 13.

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So we could rewrite this
compound inequality as

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negative 5 has to be less than
or equal to x minus 4, and x

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minus 4 needs to be less
than or equal to 13.

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And then we could solve each of
these separately, and then

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we have to remember this "and"
there to think about the

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solution set because it has to
be things that satisfy this

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equation and this equation.

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So let's solve each of
them individually.

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So this one over here,
we can add 4 to both

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sides of the equation.

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The left-hand side, negative
5 plus 4, is negative 1.

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Negative 1 is less than
or equal to x, right?

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These 4's just cancel out here
and you're just left with an x

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on this right-hand side.

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So the left, this part right
here, simplifies to x needs to

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be greater than or equal to
negative 1 or negative 1 is

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less than or equal to x.

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So we can also write
it like this.

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X needs to be greater than
or equal to negative 1.

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These are equivalent.

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I just swapped the sides.

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Now let's do this other
condition here in green.

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Let's add 4 to both sides
of this equation.

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The left-hand side,
we just get an x.

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And then the right-hand
side, we get 13 plus

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14, which is 17.

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So we get x is less than
or equal to 17.

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So our two conditions, x has to
be greater than or equal to

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negative 1 and less than
or equal to 17.

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So we could write this
again as a compound

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inequality if we want.

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We can say that the solution
set, that x has to be less

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than or equal to 17 and greater
than or equal to

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negative 1.

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It has to satisfy both
of these conditions.

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So what would that look
like on a number line?

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So let's put our number
line right there.

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Let's say that this is 17.

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Maybe that's 18.

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You keep going down.

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Maybe this is 0.

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I'm obviously skipping a bunch
of stuff in between.

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Then we would have a negative
1 right there, maybe a

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negative 2.

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So x is greater than or equal
to negative 1, so we would

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start at negative 1.

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We're going to circle it in
because we have a greater than

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or equal to.

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And then x is greater than that,
but it has to be less

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than or equal to 17.

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So it could be equal to
17 or less than 17.

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So this right here is a solution
set, everything that

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I've shaded in orange.

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And if we wanted to write it in
interval notation, it would

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be x is between negative 1 and
17, and it can also equal

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negative 1, so we put
a bracket, and it

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can also equal 17.

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So this is the interval notation
for this compound

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inequality right there.

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Let's do another one.

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Let me get a good
problem here.

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Let's say that we have
negative 12.

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I'm going to change the problem
a little bit from the

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one that I've found here.

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Negative 12 is less than 2 minus
5x, which is less than

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or equal to 7.

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I want to do a problem that has
just the less than and a

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less than or equal to.

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The problem in the book that
I'm looking at has an equal

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sign here, but I want to remove
that intentionally

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because I want to show you
when you have a hybrid

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situation, when you have
a little bit of both.

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So first we can separate this
into two normal inequalities.

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You have this inequality
right there.

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We know that negative 12 needs
to be less than 2 minus 5x.

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That has to be satisfied, and--
let me do it in another

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color-- this inequality also
needs to be satisfied.

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2 minus 5x has to be less than
7 and greater than 12, less

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than or equal to 7 and greater
than negative 12, so and 2

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minus 5x has to be less
than or equal to 7.

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So let's just solve this the
way we solve everything.

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Let's get this 2 onto the
left-hand side here.

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So let's subtract 2 from both
sides of this equation.

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So if you subtract 2 from both
sides of this equation, the

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left-hand side becomes negative
14, is less than--

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these cancel out-- less
than negative 5x.

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Now let's divide both
sides by negative 5.

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And remember, when you multiply
or divide by a

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negative number, the inequality
swaps around.

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So if you divide both sides by
negative 5, you get a negative

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14 over negative 5, and you have
an x on the right-hand

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side, if you divide that by
negative 5, and this swaps

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from a less than sign to
a greater than sign.

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The negatives cancel out, so you
get 14/5 is greater than

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x, or x is less than 14/5,
which is-- what is this?

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This is 2 and 4/5.

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x is less than 2 and 4/5.

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I just wrote this improper
fraction as a mixed number.

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Now let's do the other
constraint

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over here in magenta.

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So let's subtract 2 from both
sides of this equation, just

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like we did before.

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And actually, you can do these
simultaneously, but it becomes

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kind of confusing.

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So to avoid careless mistakes, I
encourage you to separate it

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out like this.

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So if you subtract 2 from both
sides of the equation, the

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left-hand side becomes
negative 5x.

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The right-hand side, you have
less than or equal to.

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The right-hand side becomes
7 minus 2, becomes 5.

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Now, you divide both sides
by negative 5.

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On the left-hand side,
you get an x.

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On the right-hand side, 5
divided by negative 5 is

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negative 1.

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And since we divided by a
negative number, we swap the

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inequality.

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It goes from less than or
equal to, to greater

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than or equal to.

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So we have our two
constraints.

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x has to be less than 2 and 4/5,
and it has to be greater

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than or equal to negative 1.

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So we could write
it like this.

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x has to be greater than or
equal to negative 1, so that

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would be the lower bound on our
interval, and it has to be

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less than 2 and 4/5.

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And notice, not less
than or equal to.

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That's why I wanted to show you,
you have the parentheses

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there because it can't be
equal to 2 and 4/5.

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x has to be less
than 2 and 4/5.

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Or we could write this way.

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x has to be less than 2 and
4/5, that's just this

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inequality, swapping the
sides, and it has to be

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greater than or equal
to negative 1.

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So these two statements
are equivalent.

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And if I were to draw it
on a number line, it

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would look like this.

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So you have a negative 1, you
have 2 and 4/5 over here.

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Obviously, you'll have
stuff in between.

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Maybe, you know, 0
sitting there.

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We have to be greater than or
equal to negative 1, so we can

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be equal to negative 1.

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And we're going to be greater
than negative 1, but we also

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have to be less than
2 and 4/5.

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So we can't include
2 and 4/5 there.

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We can't be equal to 2 and 4/5,
so we can only be less

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than, so we put a empty circle
around 2 and 4/5 and then we

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fill in everything below that,
all the way down to negative

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1, and we include negative 1
because we have this less than

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or equal sign.

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So the last two problems I did
are kind of "and" problems.

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You have to meet both of
these constraints.

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Now, let's do an "or" problem.

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So let's say I have these
inequalities.

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Let's say I'm given-- let's say
that 4x minus 1 needs to

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be greater than or equal to
7, or 9x over 2 needs to

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be less than 3.

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So now when we're saying "or,"
an x that would satisfy these

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are x's that satisfy either
of these equations.

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In the last few videos or in the
last few problems, we had

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to find x's that satisfied
both of these equations.

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Here, this is much
more lenient.

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We just have to satisfy
one of these two.

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So let's figure out the solution
sets for both of

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these and then we figure out
essentially their union, their

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combination, all of
the things that'll

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satisfy either of these.

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So on this one, on the one
on the left, we can

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add 1 to both sides.

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You add 1 to both sides.

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The left-hand side just becomes
4x is greater than or

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equal to 7 plus 1 is 8.

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Divide both sides by 4.

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You get x is greater
than or equal to 2.

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Or let's do this one.

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Let's see, if we multiply both
sides of this equation by 2/9,

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what do we get?

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If you multiply both sides by
2/9, it's a positive number,

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so we don't have to do anything
to the inequality.

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These cancel out, and you get
x is less than 3 times 2/9.

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3/9 is the same thing as
1/3, so x needs to

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be less than 2/3.

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So or x is less than 2/3.

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So that's our solution set.

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x needs to be greater than or
equal to 2, or less than 2/3.

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So this is interesting.

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Let me plot the solution
set on the number line.

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So that is our number line.

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Maybe this is 0, this is 1, this
is 2, 3, maybe that is

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negative 1.

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So x can be greater than
or equal to 2.

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So we could start-- let me
do it in another color.

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We can start at 2 here and it
would be greater than or equal

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to 2, so include everything
greater than or equal to 2.

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That's that condition
right there.

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Or x could be less than 2/3.

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So 2/3 is going to be right
around here, right?

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That is 2/3.

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x could be less than 2/3.

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And this is interesting.

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Because if we pick one of these
numbers, it's going to

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satisfy this inequality.

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If we pick one of these numbers,
it's going to satisfy

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that inequality.

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If we had an "and" here, there
would have been no numbers

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that satisfy it because you
can't be both greater than 2

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and less than 2/3.

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So the only way that there's
any solution set here is

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because it's "or." You can
satisfy one of the two

00:11:40.740 --> 00:11:41.920
inequalities.

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Anyway, hopefully you,
found that fun.