用到乘與除的不等式 (英)

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用到乘與除的不等式 (英) : 用到乘與除的不等式

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In this video, I want to tackle some inequalities that
involve multiplying and dividing by positive and
negative numbers, and you'll see that it's a little bit
more tricky than just the adding and subtracting numbers
that we saw in the last video.
I also want to introduce you to some other types of
notations for describing the solution set of an inequality.
So let's do a couple of examples.
So let's say I had negative 0.5x is less
than or equal to 7.5.
Now, if this was an equality, your natural impulse is to
say, hey, let's divide both sides by the coefficient on
the x term, and that is a completely legitimate thing to
do: divide both sides by negative 0.5.
The important thing you need to realize, though, when you
do it with an inequality is that when you multiply or
divide both sides of the equation by a negative number,
you swap the inequality.
you swap the inequality
Think of it this way.
I'll do a simple example here.
If I were to tell you that 1 is less than 2, I think you
would agree with that.
1 is definitely less than 2.
Now, what happens if I multiply both sides of this by
negative 1?
Negative 1 versus negative 2?
Well, all of a sudden, negative 2 is more negative
than negative 1.
So here, negative 2 is actually less than negative 1.
Now, this isn't a proof, but I think it'll give you comfort
on why you're swapping the sign.
If something is larger, when you take the negative of both
of it, it'll be more negative, or vice versa.
So that's why, if we're going to multiply both sides of this
equation or divide both sides of the equation by a negative
number, we need to swap the sign.
So let's multiply both sides of this equation.
Dividing by 0.5 is the same thing as multiplying by 2.
Our whole goal here is to have a 1 coefficient there.
So let's multiply both sides of this
equation by negative 2.
So we have negative 2 times negative 0.5.
And you might say, hey, how did Sal get this 2 here?
My brain is just thinking what can I multiply negative
0.5 by to get 1?
And negative 0.5 is the same thing as negative 1/2.
The inverse of that is negative 2.
So I'm multiplying negative 2 times both
sides of this equation.
And I have the 7.5 on the other side.
I'm going to multiply that by negative 2 as well.
And remember, when you multiply or divide both sides
of an inequality by a negative, you swap the
inequality.
You had less than or equal?
Now it'll be greater than or equal.
So the left-hand side, negative 2 times
negative 0.5 is just 1.
You get x is greater than or equal to 7.5 times negative 2.
That's negative 15, which is our solution set.
All x's larger than negative 15 will satisfy this equation.
I challenge you to try it.
For example, 0 will work.
0 is greater than negative 15.
But try something like-- try negative 16.
Negative 16 will not work.
Negative 16 times negative 0.5 is 8, which is
not less than 7.5.
So the solution set is all of the x's-- let me draw a number
line here-- greater than negative 15.
So that is negative 15 there, maybe that's negative 16,
that's negative 14.
Greater than or equal to negative 15 is the solution.
Now, you might also see solution sets to inequalities
written in interval notation.
And interval notation, it just takes a little
getting used to.
We want to include negative 15, so our lower bound to our
interval is negative 15.
And putting in this bracket here means that we're going to
include negative 15.
The set includes the bottom boundary.
It includes negative 15.
And we're going to go all the way to infinity.
All the way to infinity
And we put a parentheses here.
Parentheses normally means that you're not including the
upper bound.
You also do it for infinity, because infinity really isn't
a normal number, so to speak.
You can't just say, oh, I'm at infinity.
You're never at infinity.
So that's why you put that parentheses.
But the parentheses tends to mean that you don't include
that boundary, but you also use it with infinity.
So this and this are the exact same thing.
Sometimes you might also see set notations, where the
solution of that, they might say x is a real number such
that-- that little line, that vertical line thing, just
means such that-- x is greater than or equal to negative 15.
These curly brackets mean the set of all real numbers, or
the set of all numbers, where x is a real number, such that
x is greater than or equal to negative 15.
All of this, this, and this are all equivalent.
Let's keep that in mind and do a couple of more examples.
So let's say we had 75x is greater than or equal to 125.
So here we can just divide both sides by 75.
And since 75 is a positive number, you don't have to
change the inequality.
So you get x is greater than or equal to 125/75.
And if you divide the numerator and denominator by
25, this is 5/3.
So x is greater than or equal to 5/3.
Or we could write the solution set being from including 5/3
to infinity.
And once again, if you were to graph it on a number
line, 5/3 is what?
That's 1 and 2/3.
So you have 0, 1, 2, and 1 and 2/3 will be
right around there.
We're going to include it.
That right there is 5/3.
And everything greater than or equal to that will be included
in our solution set.
Let's do another one.
Let's say we have x over negative 3 is greater than
negative 10/9.
So we want to just isolate the x on the left-hand side.
So let's multiply both sides by negative 3, right?
The coefficient, you could imagine, is negative 1/3, so
we want to multiply by the inverse, which should be
negative 3.
So if you multiply both sides by negative 3, you get
negative 3 times-- this you could rewrite it as negative
1/3x, and on this side, you have negative 10/9 times
negative 3.
And the inequality will switch, because we are
multiplying or dividing by a negative number.
So the inequality will switch.
It'll go from greater than to less than.
So the left-hand side of the equation just becomes an x.
That was the whole point.
That cancels out with that.
The negatives cancel out. x is less than.
And then you have a negative times a negative.
That will make it a positive.
Then if you divide the numerator and the denominator
by 3, you get a 1 and a 3, so x is less than 10/3.
So if we were to write this in interval notation, the
solution set will-- the upper bound will be 10/3 and it
won't include 10/3.
This isn't less than or equal to, so we're going to put a
parentheses here.
Notice, here it included 5/3.
We put a bracket.
Here, we're not including 10/3.
We put a parentheses.
It'll go from 10/3, all the way down to negative infinity.
Everything less than 10/3 is in our solution set.
And let's draw that.
Let's draw the solution set.
So 10/3, so we might have 0, 1, 2, 3, 4.
10/3 is 3 and 1/3, so it might sit-- let me do it in a
different color.
It might be over here.
We're not going to include that.
It's less than 10/3.
10/3 is not in the solution set.
That is 10/3 right there, and everything less than that, but
not including 10/3, is in our solution set.
Let's do one more.
Let's do one more.
Say we have x over negative 15 is less than 8.
So once again, let's multiply both sides of this equation by
negative 15.
So negative 15 times x over negative 15.
Then you have an 8 times a negative 15.
And when you multiply both sides of an inequality by a
negative number or divide both sides by a negative number,
you swap the inequality.
It's less than, you change it to greater than.
And now, this left-hand side just becomes an x, because
these guys cancel out.
x is greater than 8 times 15 is 80 plus 40 is 120, so
negative 120.
Is that right?
80 plus 40.
Yep, negative 120.
Or we could write the solution set as starting at negative
120-- but we're not including negative 120.
We don't have an equal sign here-- going all
the way up to infinity.
And if we were to graph it, let me draw
the number line here.
I'll do a real quick one.
Let's say that that is negative 120.
Maybe zero is sitting up here.
This would be negative 121.
This would be negative 119.
We are not going to include negative 120, because we don't
have an equal sign there, but it's going to be everything
greater than negative 120.
All of these things that I'm shading in green would satisfy
the inequality.
And you can even try it out.
Does zero work?
0/15?
Yeah, that's zero.
That's definitely less than 8.
I mean, that doesn't prove it to you, but you could try any
of these numbers and they should work.
Anyway, hopefully, you found that helpful.
I'll see you in the next video.