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## Algebra 1

### Course: Algebra 1 > Unit 2

Lesson 6: Compound inequalities# Double inequalities

Sal solves the double inequality -16≤3x+5≤20, which is the same as the compound inequality -16≤3x+5 AND 3x+5≤20. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- How do you know a statement is and/or?(12 votes)
- The statement will have an "or" in it if it is an or inequality. For "and" inequalities it will either be in the way it was in the video(-16 ≤ 3x + 5 ≤ 20) or say: -16 ≤ 3x + 5 and 3x + 5 ≤ 20.(26 votes)

- So I know you can separate the inequalities, but is there a way to answer −13 > −5x + 2 > −28 by leaving it together? What I mean is that when you divide by negative 5 three times, do you like switch both signs? Or is that not a thing, and you're supposed to just separate them?(12 votes)
- You would be dividing by a negative number, so you should switch both signs.(3 votes)

- I have this problem on my summer packet and I'm confused.

7-3x≥-5 and -2≤5-7x

I solved and I got:

X≤4 and x≤1

However, on my answer key only x≤1 is listed and graphed on a number line. What makes you use only one answer and how do you know when to check for this?(6 votes)- I guess the "and" makes a difference. If x<=1 it is <= 4 as well, so the latter is not needed.

It's like saying "The car is black and it is not red". If you have stated that it is black, there is no need to say it is not red.(8 votes)

- so can you do word problems(4 votes)
- How do you know whether a double inequality will be AND or OR?(5 votes)
- Double inequalities are always AND. Sal explains this very early in the video (@0:25) when he splits the double inequality into -16≤3x+5 AND 3x+5≤20. He tells you that "both" inequalities must be true. The is the basic definition of an AND compound inequalities.(6 votes)

- Does every problem have to be figured out with a number line. Doesn't that kinda complicate things in a way??(5 votes)
- it depends on what type of problem it is and what its asking(2 votes)

- Hi! I just have one small question! How would we solve a double inequality with an x variable on two or all of the sides? Is that possible or can x only be on one side (or the one in the middle) when it comes to double inequalities? I would really appreciate your help, because I'm curious and I can't seem to find an answer when I google the question. If you have by any chance found any explanation on the internet, I would also appreciate if you link it! Thank you so much! Please keep safe in these troubling times!(5 votes)
- Split the inequality into two inequality statements.

As such there is no exact answer to your question as it will depend on the expression how you go about finding the answer.(4 votes)

- what about a number outside of the absolute value in the front like 5|4-3x| (greater than or equal to) 30(2 votes)
- Hi Sabrina,

If you have a number outside of the absolute value sticks, get it to the other side of the inequality first then proceed. So in your example:

5|4 - 3x| >= 30

divide each side by 5 to get

|4 - 3x| >= 6

then finish the problem as you normally would.

If the problem had been:

|4 - 3x| + 5 >= 30

subtract the 5 from each side to get

|4 - 3x| >= 25

As I said above, the goal is to get the absolute value by itself on one side of the inequality and then proceed.

Hope that helps :-)(9 votes)

- how do I solve -2<x<or equal to 5(4 votes)
- It has already been solved. The solution set is all real numbers larger than -2 up to and including 5.(4 votes)

- And also how would I solve: -x <-2x and 3x>2x ? the only thing I can think to do is divide but not sure how to process this(3 votes)
- If you have variables on both sides, the way to solve is to move by opposites, so -x <-2x, add 2x to both sides to get x<0. With 3x>2x, subtract 2x on both sides to get x > 0.(4 votes)

## Video transcript

We're asked to solve for x. And we have this
compound inequality here, negative 16 is less than or
equal to 3x plus 5, which is less than or equal to 20. And really, there's two
ways to approach it, which are really the same way. And I'll do both of them. And I'll actually do both
of them simultaneously. So one is to just solve
this compound inequality all at once. And I'll just rewrite it. Negative 16 is less than or
equal to 3x plus 5, which is less than or equal to 20. And the other way
is to think of it as two separate inequalities,
but both of them need to be true. So you could also view it as
negative 16 has to be less than or equal to 3x plus 5. And 3x plus 5 needs to be
less than or equal to 20. This statement and this
statement are equivalent. This one may seem a little
bit more familiar because we can independently solve
each of these inequalities and just remember the "and." This one might seem a little
less traditional because now we have three sides
to the statement. We have three parts of
this compound inequality. But what we can see
is that we're actually going to solve it
the exact same way. In any situation, we really just
want to isolate the x on one side of the inequality,
or in this case, one part of the
compound inequality. Well, the best way to
isolate this x right here is to first get rid of
this positive 5 that's sitting in the middle. So let's subtract
5 from every part of this compound inequality. So I'm going to subtract
5 there, subtract 5 there, and subtract
5 over there. And so we get negative 16
minus 5 is negative 21, is less than or equal to 3x
plus 5 minus 5 is 3x, which is less than or equal to
20 minus 5, which is 15. And we could essentially
do the same thing here. If we want to isolate the 3x, we
can subtract 5 from both sides. We get negative 21. Negative 21 is less
than or equal to 3x. And we get, subtracting
5 from both sides. And notice, we're
just subtracting 5 from every part of this
compound inequality. We get 3x is less
than or equal to 15. So this statement and this
statement, once again, are the exact same thing. Now, going back here, if
we want to isolate the x, we can divide by 3. And we have to do it to
every part of the inequality. And since 3 is positive, we
don't have to change the sign. So let's divide every part of
this compound inequality by 3. You divide every part by 3. This is equivalent to
dividing every part of each of these inequalities by 3. And then we get negative 21
divided by 3 is negative 7, is less than or equal
to x, which is less than or equal to 15
divided by 3 is 5. You do it here. You get negative 7 is
less than or equal to x, and x is less than or
equal to 15/3, which is 5. This statement and this
statement are completely equal. And we've solved for x. We've given you
the solution set. And if we want to graph
it on a number line, it would look like this. This is 0. This is 5. This is negative 7. Our solution set includes
everything between negative 7 and 5, including
negative 7 and 5. So we have to fill in
the circles on negative 7 and positive 5. And it is everything in between. That's our solution set. And so we can verify
that these work. You could try out
a number that's well inside of our
solution set, like 0. 3 times 0 is 0. So you're just left
with 5 is greater than or equal to negative
16, which is true. And 5 is less than
or equal to 20. Or negative 16 is less than or
equal to 5, which is less than or equal to 20. So that works, and
that makes sense. You could try 5. If you put 5 here,
you get 3 times 5 plus 5-- well, that's just 20. Negative 16 is less than or
equal to 20, which is less than or equal to 20. That works. Negative 7 should also work. 3 times negative 7 is negative
21, plus 5 is negative 16. So you get negative
16, which is less than or equal to negative 16, which
is less than or equal to 20. And you could try other values. You could go outside
of our solution set. Try something like 10. 10 should not work. And you see here, if you put 10
here, you get 3 times 10 plus 5 is 35. Negative 16 is less
than or equal to 35, but 35 is not less
than or equal to 20. And that's why 10 is not
part of our solution set.