Main content

## 7th grade

### Unit 5: Lesson 7

One-step inequalities- Plotting inequalities
- Inequality from graph
- Plotting inequalities
- Testing solutions to inequalities
- Testing solutions to inequalities
- One-step inequalities examples
- One-step inequalities: -5c ≤ 15
- One-step inequalities
- One-step inequality word problem
- One-step inequalities review

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# One-step inequalities examples

Our discussion of linear inequalities begins with multiplying and dividing by negative numbers. Pay attention for the word "swap." Super important! Created by Sal Khan and CK-12 Foundation.

## Video transcript

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. 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. 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. 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.