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# 聯立不等式. (英): 聯立不等式

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- Let's do some compound inequality problems, and these
- are just inequality problems that have more than one set of
- constraints.
- You're going to see what I'm talking about in a second.
- So the first problem I have is negative 5 is less than or
- equal to x minus 4, which is also less than or equal to 13.
- So we have two sets of constraints on the set of x's
- that satisfy these equations.
- x minus 4 has to be greater than or equal to negative 5
- and x minus 4 has to be less than or equal to 13.
- So we could rewrite this compound inequality as
- negative 5 has to be less than or equal to x minus 4, and x
- minus 4 needs to be less than or equal to 13.
- And then we could solve each of these separately, and then
- we have to remember this "and" there to think about the
- solution set because it has to be things that satisfy this
- equation and this equation.
- So let's solve each of them individually.
- So this one over here, we can add 4 to both
- sides of the equation.
- The left-hand side, negative 5 plus 4, is negative 1.
- Negative 1 is less than or equal to x, right?
- These 4's just cancel out here and you're just left with an x
- on this right-hand side.
- So the left, this part right here, simplifies to x needs to
- be greater than or equal to negative 1 or negative 1 is
- less than or equal to x.
- So we can also write it like this.
- X needs to be greater than or equal to negative 1.
- These are equivalent.
- I just swapped the sides.
- Now let's do this other condition here in green.
- Let's add 4 to both sides of this equation.
- The left-hand side, we just get an x.
- And then the right-hand side, we get 13 plus
- 14, which is 17.
- So we get x is less than or equal to 17.
- So our two conditions, x has to be greater than or equal to
- negative 1 and less than or equal to 17.
- So we could write this again as a compound
- inequality if we want.
- We can say that the solution set, that x has to be less
- than or equal to 17 and greater than or equal to
- negative 1.
- It has to satisfy both of these conditions.
- So what would that look like on a number line?
- So let's put our number line right there.
- Let's say that this is 17.
- Maybe that's 18.
- You keep going down.
- Maybe this is 0.
- I'm obviously skipping a bunch of stuff in between.
- Then we would have a negative 1 right there, maybe a
- negative 2.
- So x is greater than or equal to negative 1, so we would
- start at negative 1.
- We're going to circle it in because we have a greater than
- or equal to.
- And then x is greater than that, but it has to be less
- than or equal to 17.
- So it could be equal to 17 or less than 17.
- So this right here is a solution set, everything that
- I've shaded in orange.
- And if we wanted to write it in interval notation, it would
- be x is between negative 1 and 17, and it can also equal
- negative 1, so we put a bracket, and it
- can also equal 17.
- So this is the interval notation for this compound
- inequality right there.
- Let's do another one.
- Let me get a good problem here.
- Let's say that we have negative 12.
- I'm going to change the problem a little bit from the
- one that I've found here.
- Negative 12 is less than 2 minus 5x, which is less than
- or equal to 7.
- I want to do a problem that has just the less than and a
- less than or equal to.
- The problem in the book that I'm looking at has an equal
- sign here, but I want to remove that intentionally
- because I want to show you when you have a hybrid
- situation, when you have a little bit of both.
- So first we can separate this into two normal inequalities.
- You have this inequality right there.
- We know that negative 12 needs to be less than 2 minus 5x.
- That has to be satisfied, and-- let me do it in another
- color-- this inequality also needs to be satisfied.
- 2 minus 5x has to be less than 7 and greater than 12, less
- than or equal to 7 and greater than negative 12, so and 2
- minus 5x has to be less than or equal to 7.
- So let's just solve this the way we solve everything.
- Let's get this 2 onto the left-hand side here.
- So let's subtract 2 from both sides of this equation.
- So if you subtract 2 from both sides of this equation, the
- left-hand side becomes negative 14, is less than--
- these cancel out-- less than negative 5x.
- Now let's divide both sides by negative 5.
- And remember, when you multiply or divide by a
- negative number, the inequality swaps around.
- So if you divide both sides by negative 5, you get a negative
- 14 over negative 5, and you have an x on the right-hand
- side, if you divide that by negative 5, and this swaps
- from a less than sign to a greater than sign.
- The negatives cancel out, so you get 14/5 is greater than
- x, or x is less than 14/5, which is-- what is this?
- This is 2 and 4/5.
- x is less than 2 and 4/5.
- I just wrote this improper fraction as a mixed number.
- Now let's do the other constraint
- over here in magenta.
- So let's subtract 2 from both sides of this equation, just
- like we did before.
- And actually, you can do these simultaneously, but it becomes
- kind of confusing.
- So to avoid careless mistakes, I encourage you to separate it
- out like this.
- So if you subtract 2 from both sides of the equation, the
- left-hand side becomes negative 5x.
- The right-hand side, you have less than or equal to.
- The right-hand side becomes 7 minus 2, becomes 5.
- Now, you divide both sides by negative 5.
- On the left-hand side, you get an x.
- On the right-hand side, 5 divided by negative 5 is
- negative 1.
- And since we divided by a negative number, we swap the
- inequality.
- It goes from less than or equal to, to greater
- than or equal to.
- So we have our two constraints.
- x has to be less than 2 and 4/5, and it has to be greater
- than or equal to negative 1.
- So we could write it like this.
- x has to be greater than or equal to negative 1, so that
- would be the lower bound on our interval, and it has to be
- less than 2 and 4/5.
- And notice, not less than or equal to.
- That's why I wanted to show you, you have the parentheses
- there because it can't be equal to 2 and 4/5.
- x has to be less than 2 and 4/5.
- Or we could write this way.
- x has to be less than 2 and 4/5, that's just this
- inequality, swapping the sides, and it has to be
- greater than or equal to negative 1.
- So these two statements are equivalent.
- And if I were to draw it on a number line, it
- would look like this.
- So you have a negative 1, you have 2 and 4/5 over here.
- Obviously, you'll have stuff in between.
- Maybe, you know, 0 sitting there.
- We have to be greater than or equal to negative 1, so we can
- be equal to negative 1.
- And we're going to be greater than negative 1, but we also
- have to be less than 2 and 4/5.
- So we can't include 2 and 4/5 there.
- We can't be equal to 2 and 4/5, so we can only be less
- than, so we put a empty circle around 2 and 4/5 and then we
- fill in everything below that, all the way down to negative
- 1, and we include negative 1 because we have this less than
- or equal sign.
- So the last two problems I did are kind of "and" problems.
- You have to meet both of these constraints.
- Now, let's do an "or" problem.
- So let's say I have these inequalities.
- Let's say I'm given-- let's say that 4x minus 1 needs to
- be greater than or equal to 7, or 9x over 2 needs to
- be less than 3.
- So now when we're saying "or," an x that would satisfy these
- are x's that satisfy either of these equations.
- In the last few videos or in the last few problems, we had
- to find x's that satisfied both of these equations.
- Here, this is much more lenient.
- We just have to satisfy one of these two.
- So let's figure out the solution sets for both of
- these and then we figure out essentially their union, their
- combination, all of the things that'll
- satisfy either of these.
- So on this one, on the one on the left, we can
- add 1 to both sides.
- You add 1 to both sides.
- The left-hand side just becomes 4x is greater than or
- equal to 7 plus 1 is 8.
- Divide both sides by 4.
- You get x is greater than or equal to 2.
- Or let's do this one.
- Let's see, if we multiply both sides of this equation by 2/9,
- what do we get?
- If you multiply both sides by 2/9, it's a positive number,
- so we don't have to do anything to the inequality.
- These cancel out, and you get x is less than 3 times 2/9.
- 3/9 is the same thing as 1/3, so x needs to
- be less than 2/3.
- So or x is less than 2/3.
- So that's our solution set.
- x needs to be greater than or equal to 2, or less than 2/3.
- So this is interesting.
- Let me plot the solution set on the number line.
- So that is our number line.
- Maybe this is 0, this is 1, this is 2, 3, maybe that is
- negative 1.
- So x can be greater than or equal to 2.
- So we could start-- let me do it in another color.
- We can start at 2 here and it would be greater than or equal
- to 2, so include everything greater than or equal to 2.
- That's that condition right there.
- Or x could be less than 2/3.
- So 2/3 is going to be right around here, right?
- That is 2/3.
- x could be less than 2/3.
- And this is interesting.
- Because if we pick one of these numbers, it's going to
- satisfy this inequality.
- If we pick one of these numbers, it's going to satisfy
- that inequality.
- If we had an "and" here, there would have been no numbers
- that satisfy it because you can't be both greater than 2
- and less than 2/3.
- So the only way that there's any solution set here is
- because it's "or." You can satisfy one of the two
- inequalities.
- Anyway, hopefully you, found that fun.

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