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Hello, and welcome to Bay College's video lectures
for Math 105, Intermediate Algebra.
In this video, we're going to look at Section 1.7.
1.7 is actually broken into three separate videos.
And in the first one, we're going
to look at compound inequalities in the terms of intersections.
That's part one.
Part two, we'll look at unions.
And the third part, we'll look at the absolute value
in inequalities.
So the first thing we're going to do
is define the difference between an intersection and a union.
Well, if we have two sets of number.
In set A, we have the values 1, 2, 3, and 4,
and I'm just going to write 1, 2, 3, and 4.
So this is the set of the values for A, 1, 2, 3, and 4.
In the set of B, we have 2, 4, 6, 8.
Well we have 2, 4, 6, and 8.
When we're talking about an intersection,
we use this symbol, and this is the symbol for an intersection.
It's also called an and statement.
So when we're looking at an intersection,
it says what is within set A and set B?
So it's what they both have in common.
So the intersection is where they overlap.
In set A, we have the values 2 and 4 and in set B,
we have the values 2 and 4.
So this area is called the intersection,
and we denote the intersection with this symbol.
Now if we're going to define a union,
we call that an or statement.
The union is an or statement.
So if we want to find the union of A and B,
what's in one or the other?
Well that would include what's in here or in here.
It would be all the values.
So we'd have 1, 2, 3, 4, 6, 8.
All six of these values would be in one or the other.
So we're looking at the entire set of A or B.
Let's look at some examples where
we look at an intersection here.
Now hopefully we recall and we reviewed
what these symbols are, we have less than.
So we have ax plus b is less than c,
which is a linear inequality.
We could have also the symbols greater than, less than
or equal to, greater than or equal to,
and we're also going to introduce interval notation.
And hopefully you've reviewed those.
Now if we have the set x such that x
is greater than negative 6 and, which is a union, and x such
that x is less than 1, we want to look at these two
together and see where they overlap.
Well let's take it one piece at a time initially.
If we look at the first set, x such that x
is greater than negative 6.
On a number line, if this value is negative 6
and if I know in interval notation,
I use a parenthesis here because it does not
include the endpoint, any value to the right
is going to be greater than negative 6.
So it says this value x is greater than negative 6.
So this would be the graph of this piece of our intersection.
If we look at this here, it says x such that x is less than 1.
Well, if 1 is over here, x is some value less than 1
but it doesn't include it because it's just
a less than symbol.
So we use that parentheses.
Now if we want to find the intersection, and notice I used
and-- they're interchangeable-- what
are the values that satisfy this one and this one
at the same time?
Well let's graph them both together.
If this is the value negative 6 and this is the value 1,
if we put them on the same graph,
the area in between their parentheses
would be the solution.
So their intersection, this one would continue that way,
this one continues this way.
Where they overlap are the solutions that solve this one
and this one at the same time, so any value in between.
Maybe I pick a test point like 0.
0 is between negative 6 and 1, and I could put it in here
and say 0 is greater than negative 6.
True statement.
0 is less than 1, also a true statement.
So that one solution solved this one and this one,
and that's why we call it an and statement.
When we have and statements or intersections,
we can write them as double inequalities.
So let's take a look at this one and write it
as a double inequality.
Well if we have this value x is greater than negative 6,
let's just switch that around for a moment.
Negative 6 is less than x.
That's the same thing as saying x is greater than negative 6.
Negative 6 is less than x.
We also have x is less than 1.
So x is less than 1 and x is greater than negative 6.
When we have and statements, we can write them
as a double inequality.
Now there are some things we always
have to be aware of when dealing with
and statements and double inequalities
is writing them in the proper format.
And one way to check it's in the proper format
is we always write it from the least number to the greatest
number from left to right.
Just like our number line.
These values are smaller.
These values are greater.
So when we look at this, one way to check
that it's in a proper notation is
to cover up the middle value.
Does it still make a true statement?
Negative 6 is less than 1.
That's a true statement.
If I cover up the other side, negative 6 is less than 1.
It's still a true statement.
So I know that my double inequality
is written in the proper format.
So let's go over here and check a few examples where
we have some double inequalities written
and to see if they're in the proper format.
Here we have negative 3 is less than x and x is greater than 4.
Well if we look at it, our values
are written from smallest to greatest.
Negative 3 on this side, positive 4 on that side.
It is from least to greatest.
But if I cover this up, is negative 3 greater than 4?
No.
So it makes a non-true statement.
If any part of this is not true, then it's
not in the proper format.
So this is not a true statement.
It's not in the proper format.
If we look at this one, we have negative 2 is less than x
and x is less than negative 3.
If I cover up this section, I can
see while it isn't written from least to greatest, negative 2
is greater than negative 3, but this statement
says otherwise so it's not true.
If I covered this up, it would still not be true.
So it's not true on either end of this.
So that's not true.
It's not in the proper format.
Now if we look at this, we have negative 4
is greater than or equal to x is greater
than or equal to negative 6.
Well, I notice that this is not in the proper format
for the reason that negative 4 is a larger
value than negative 6, and they have
to be written from least to greatest.
But if I cover this up, it still makes a true statement.
Negative 4 is greater than or equal to negative 6.
That's true.
If I cover that up, it says the same thing.
It's still a true statement.
So being that it's true, it's not in the proper format.
We can rewrite it to be in the proper format.
The least number is first, and the greatest number
is to the right.
So from left to right it goes from least to greatest.
Now if we think about this, x is greater than
or equal to negative 6.
Well since I moved this number to the side,
I have to flip that sign.
If I can lift it off the board and turn it around,
that's what I would have.
Now the same thing here.
Since we moved the 4 to the other side,
we can switch that around.
And we have negative 6 is less than or equal to x is less than
or equal to negative 4.
Both of these say the same thing,
but this one was not in the proper format.
It's always written least to greatest.
So even though it was a true statement,
it wasn't in the proper format.
So we have to be aware of that.
And we see different scenarios where well if any part of it's
not true, it's not a valid inequality.
If it is true, maybe we can rewrite it
from least to greatest to make it
in the proper format for a double inequality.
All right.
Let's look at two examples here.
Here we have x plus 1 is less than 4 and negative 2x plus 1
is less than 3.
So we might see this and or we might see the intersection
symbol.
Again, they're interchangeable.
So we're going to solve this.
And we're going to write our answer in three different ways.
We're going to write it in set notation.
We're going to write it in interval notation,
and we're going to graph it as well.
Not necessarily in that order.
So let's solve this one.
To solve it, we treat this just like an equal sign,
except we hopefully we recall that if we multiply or divide
by negative we have to change the sign.
So let's go ahead and solve this.
I'm going to subtract 1 from both sides.
And by doing that, I get x is less than 3.
So we've solved that.
It is in algebraic notation.
To put it into set notation, we simply
use the braces and x such that x is less than 3,
and we close it with a brace.
For this one here, we're going to subtract 3 from both sides.
And we see we get negative 2x is less than 2.
But now I have to get x by itself and by doing
so I divide by a negative.
Well when I divide by negative, it's going to change the sign.
Right now it's negative.
If I divide by negative 2, I get a positive x.
There was a sign change.
If I divide this side by that coefficient of negative 2,
it's positive.
But if I divide it by a negative 2,
it changes its sign to be a negative 1.
Because I changed the signs by dividing by a negative,
I have to remember to change the sign.
So we have x is greater than negative 1.
So we have this and statement, and we're
going to write this in set notation.
So this would be in set notation.
X such that x is less than 3 and x such that x
is greater than negative 1.
We can also graph this just like we've
seen in the previous examples.
Here we have x is less than 3.
Well if 3 is here, x is some value less than 3.
And on that same number line, I'm going to graph this.
X is greater than negative 1.
Well negative 1 would be somewhere over here,
and x is greater than negative 1.
We can see the interval that we have is between
negative 1 and 3.
And because it's an and statement,
I can write this in interval notation, which
would be a parenthesis at negative 1,
because it doesn't include the endpoints, and a parenthesis
at 3, because again it doesn't include the endpoint.
They were less than and greater than symbols, not equal to.
Now we also said an and statement
can be written as a double inequality.
If I combine these two into one statement,
the least value is negative 1, the greatest value is 3.
And if we look at this statement,
if I can just flip it over, x is greater than negative 1.
So negative 1 is less than x, x is less than 3,
and that statement is true.
So here we have our solution as a double inequality.
We have it in interval notation.
We have it in graphic notation, and we also
have it in set notation.
You can illustrate your answer in any one of these four
methods.
Be aware of that, and maybe your instructor
may ask you to do it a specific way.
Know how to do it in each of those.
What if we have a double inequality,
but it's not simplified?
Well just like we had used the properties of equality
here to solve for x, we can do the same thing here,
except what we do to one side we have
to do to all three sides instead of just two.
So to get x by itself, I'm going to subtract 9 from all sides.
So 0 minus 9 is negative 9.
This value minus 9 is 0, so 0 minus negative 5x
is just negative 5x.
And 29 minus 9 is 20.
Now to get x by itself I, divide all these values by 5
so negative 9/5 divided by negative 5
is a positive 9/5, a negative divided by a negative.
And negative 5 divided by negative 5
is a positive 1 for our x value.
20 divided by negative 5 is negative 4.
Now the reason why I didn't put these symbols in
is because I divided by a negative.
All the signs change.
This was negative.
Now is positive.
This was negative.
Now it is positive.
This was positive.
Now it's negative.
The signs changed, so I'm going to remember to change my signs.
Now we look at it, initially we might
think oh that's a double inequality.
And if I cover this up 9/5 is greater than negative 4,
and that statement is true no matter which one I cover.
But it's not in its proper format,
so we have to rewrite it from least to greatest.
Right now it's from greatest to least.
So I'm going to write negative 4 here
and 9/5 here, and negative 4 is less than x
because x is greater than negative 4.
And x is less than 9/5 because 9/5 is greater than x.
So we just rearrange it.
Now that we have it in it's algebraic notation,
to write it in set notation it's as
simple as saying x such that this algebraic statement.
So that's set notation.
The next thing we're going to do is maybe we want to graph it.
So I have negative 4 and I have 9/5,
and the values, because these symbols denote
the endpoints aren't included, it is the values
in between them.
It's an and statement, their intersection
where they would overlap.
They overlap in between their values.
And then finally interval notation.
Well we can take these parentheses
and we can have negative 4 to 9/5.
So we have set notation, graphic notation,
and interval notation, all three representations.
All right.
We're going to go to the last board here,
and I have two examples.
Here's your opportunity to try these yourself.
Make sure you check your work.
Write your answer in set notation,
graph it, and write in interval notation.
And do that for both of these.
You have this intersection, and you
have this count, or double inequality.
Try these yourself.
Work them out.
Choose test points, and make sure
that you have the correct answer.
So this has been section 1.7, part 1.
Thank you for watching.