Linear Inequalities - Video Tutorials & Practice Problems

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1

concept

Interval Notation

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5m

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Hey, everyone, we already know how to express sets in set notation. So something that looks like this, but there is actually a much more compact way to represent solution sets in what's called interval notation. Now, you might be thinking, why do we need a new way to write sets if we already have one that works just fine and not only is it going to be quicker and easier, but you're actually going to be asked explicitly to express your solutions to inequalities in interval notation. So I'm going to show you how to take this set and turn it into something much more compact using parentheses in square brackets based on whether you have a less than sign or a less than or equal to sign. So let's go ahead and take a look here. So there are different types of intervals you can have based on whether you're dealing with less than or less than are equal to signs. So let's first look at our closed interval. So when I have a set like X or zero is less than, or equal to X is less than or equal to five because I have these less than or equal to inequality symbols. I am going to write my end points in between square brackets. So these less than equal to signs tell me that I'm just going to take my end points and stick them in between square brackets. Now, you're also going to be asked to graph these intervals. And what that really means is just write your interval on a number line. So whenever I have square brackets on my interval, I'm going to express on my number line, my end points with closed circles. So I'm still going to label my end points here. And then I'm going to draw a line connecting my end points to show that my set or my interval includes everything from zero all the way to five including zero and five. So whenever we're dealing with a, less than, or equal to sign a square bracket and a closed circle that tells me that I'm dealing with a set with end points included because it's not just less than five, it's less than, or equal to five. So let's take a look at our open interval here. So an open interval happens whenever I have something like zero is less than X is less than five. Now, you'll notice that these are not less than, or equal to signs, they are just less than signs. And when I have that, I'm just going to take my end points and stick them in between parentheses. So when I have an open interval. Those less than signs become parentheses in interval notation. Now, whenever I'm asked to graph this, I'm going to take my parentheses and make them into open circles still have my end points and still connect my end. But the open circles say to me that my end points are not included, they are excluded from my set. So anything I can have any number here from 0.0001 all the way up to 4.999. But I can't have zero or five because they are not a part of my set. So that's an open interval. We can also have a combination of closed and open intervals. So let's take a look at that here. Here we have zero is less than or equal to X is less than five. So you'll notice that the first symbol I have is that less than or equal to sign, which tells me that I need to enclose that first endpoint in a square bracket because it is included in my set. Now, for my second inequality symbol, I just have a less than sign. So that tells me that my second endpoint is just going to be enclosed in parentheses. So that's my interval in interval notation. When it is half open, half closed, let's go ahead and graph that as well. So whenever I have that square bracket and my end point is included, I want to make sure and have a closed circle there and still label my endpoint. And then with a parenthesis, I know that it's going to be an open circle and then I need to fill in because it's all the way from zero, including zero up to five, but not including it. So that's how you express a half closed, half open circle. Let's take a look at one last example here. So here my set is X is greater than or equal to three. Now, just looking at that, having that greater than or equal to sign tells me that I'm going to be dealing with a square bracket somewhere. But let's actually graph this first and see what's happening. So when I have a greater than or equal to, I know I'm going to be dealing with a closed circle. So I'm gonna go ahead and plot my three with a closed circle. Now, this is telling me that X can be anything greater than, or equal to three. So that means that anything to the right of three is going to be fine all the way up until forever, up until infinity. So whenever we don't have an explicit end point, so here, this just says that X can be anything greater than or equal to three with no limit, I'm actually going to use infinity symbols to express this. So either a positive infinity or a negative infinity based on what direction it's going in. And whenever we have infinity in intervals, we're going to treat this as an open bound because you can't get up to infinity. It goes on to forever, right? So it's not a hard end point. So to write this in interval notation, my three gets enclosed in a square bracket because it is included in my set. But then I have infinity which just gets a parenthesis because that is not a hard end point. It's not less than, or equal to infinity because you can't be equal to infinity, right? So that's all there is to express sets in interval notation. Let's keep going.

2

Problem

Problem

Express the given set in interval notation and graph.

{$x$ | 14 ≤ $x$ < 26}

A

(14,26)

B

[14,26]

C

[14, 26)

D

(14,26]

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Problem

Problem

Express the given interval in set builder notation and graph. (−∞, 0]

A

{$x\left|x\right.$ ≤ 0}

B

{$x\left|x\right.$ < 0}

C

{$x\left|x\right.$ > 0}

D

$\left\lbrace x|x\ge0\right\rbrace$

4

Problem

Problem

Express the given set in interval notation and graph.

$\left\lbrace\right.$$x$$|$$x\le7$$\}$

A

(−∞,7]

B

[−∞,7]

C

(−∞,7)

D

[7, ∞)

5

concept

Linear Inequalities

Video duration:

6m

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Hey, everyone. So we've already learned how to solve linear equations like two X minus six equals zero by finding some value for X that we can plug back into our equation to make a true statement. But now you're going to start to see problems that instead of two X minus six is equal to zero, have two X minus six is less than or equal to zero. So there I have an inequality symbol instead of an equal sign and this is called a linear inequality. So linear inequalities are literally just linear equations but with an inequality symbol instead of an equal sign. Now I know what you're thinking, why are you taking something that I already know how to do and changing it? But don't worry everything that we know about solving linear equations can be used to solve linear inequalities. We're just going to see a slightly different solution and I'm going to walk you through everything that you need to know about inequalities. So let's go ahead and get started. So like I said, with linear equations, we were looking for some value of X and we could just move some numbers over and end up with a solution. Now let's go ahead and start solving our linear inequality. The same exact way we did our linear equation. So if I have two X minus six is less than or equal to zero, I can start by moving my six to the other side. The same way I would with a linear equation. So if I just add six to both sides that will cancel, leaving me with two X is less than or equal to six. OK. Now that I'm here, I can isolate X by dividing by two. Again, the same way I would with a linear equation canceling that two out and leaving me with X is a less than or equal to three. So I actually still even ended up with a three there. Now instead of being equal to three, it is less than N are equal to three. So that's the difference that we're going to see here. So let's take a look at one more inequality and just change something slightly. So here I had two, X minus six is less than or equal to zero. And over here I have negative two X minus six is less than or equal to zero. So let's see what happens here while I start the same way I can go ahead and move my six over leaving me with negative two X is less than or equal to six. And now I need to go ahead and divide both sides by negative two to isolate it. So that will cancel, leaving me with X is a less than or equal to negative three. And this is my solution. But if this is my solution, then that means that I should be able to plug any value that is less than or equal to negative three back into my original inequality and get a correct statement. So let's see if that works here. So some number less than or equal to negative three, if I try X is equal to negative four, that should work. So plugging that back into my original inequality, I have negative two times negative four minus six is less than or equal to zero. So negative two times negative four is positive eight minus six is less than or equal to 08 minus six is just two. So two is less than or equal to zero, but two is definitely not less than or equal to zero. So something is wrong here and this actually is not my solution at all. So the reason why that's not my solution is because any time that we multiply or divide by a negative number, I actually need to go ahead and flip my inequality symbol. So let's see what happens when we do that instead. So we're going to start again with this negative two, X is less than or equal to six. And then when I divide both sides by this negative two, I'm still canceling it out and left with X. But I'm going to take my inequality symbol and I'm going to flip the direction it's facing. So instead of a less than or equal to sign, I'm going to use a greater than or equal to sign. And then I have six divided by negative two, which gives me negative three. So if this is my solution and this worked here, that means that any number greater than or equal to negative three, I should be able to plug it. So if I take X is equal to zero and plug that in that should hopefully work. Let's go ahead and give that a try. So negative two times zero minus six is less than or equal to zero, negative two times zero just goes away because anything times zero is zero and I'm left with negative six is less than or equal to zero. Now, this is definitely a true statement. Negative six is less than or equal to zero. So this is my solution and X is greater than or equal to negative three. So let's talk about how we want to express these solutions. Well, whenever we were using doing linear equations, we just had one single value, we had something like X is equal to three. So if I were to graph that, it would literally just be a single point on my graph because my solution is just a single value. But whenever we're dealing with linear inequalities, I don't have that X is less or I don't have the X is equal to zero. I have that X is less than, or equal to zero. And that's because I'm no longer dealing with a single value, but I'm instead actually dealing with a whole range of values. So in order to graph that I still have that same end 0.3 and I know since it's less than or equal to, that means I need to have a closed circle. So I'm going to draw a closed circle at my end 0.3. And if X is anything less than or equal to three, that means it could be anything to the left of that three, all the way to negative infinity. And remember we're also going to want to write these in interval notation. So in interval notation, remember that whenever we have an infinity, it always gets a parenthesis. But should I enclose that three in a square bracket or parenthesis? Well, again, since I have that less than or equal to sign, that means that three needs to get a square bracket because it is included in my interval. So let's look at our final example over here and rewrite our solution on a graph and an interval notation. So since I have X is greater than or equal to a negative three, I still have my end point at negative three with a closed circle. Now, since X can be anything greater than, or equal to that negative three. It's going to go all the way to the right up to infinity and forever. So if we write that in interval notation, my negative three is again going to get a square bracket here because I have a less than, or equal to sign and not just a less than, or a greater than sign. So then I have infinity, which is of course, going to get enclosed in parentheses and I'm done here. So that's all you need to know about solving linear inequalities. Let's get some more practice.

6

Problem

Problem

Solve the inequality. Express the solution set in interval notation and graph. $2x+12>19$

A

$\left(-\infty,\frac72\right)$

B

$\left(-\infty,\frac72\right]$

C

$\left[\frac72,\infty\right)$

D

$\left(\frac72,\infty\right)$

7

concept

Linear Inequalities with Fractions & Variables on Both Sides

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3m

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Just as we saw happen with linear equations, sometimes linear inequalities will also have fractions variables on both sides of the symbol or maybe even both at the same time. But we're actually going to address them the exact same way we did with linear equations by getting rid of those fractions using the least common denominator and then getting all of our variables to one side. So let's go ahead and take a look at an example. So here I have 1/4 times X plus two is greater than or equal to 1/12 minus one third X. Now, the very first thing we want to do is get rid of those fractions, which we can do by multiplying the entire inequality by the least common denominator. Now, I have three denominators here, I have 4, 12 and three. So my least common denom is going to be 12. So let's go ahead and multiply our entire inequality by 12. Remember that whenever we're multiplying by our least common denominator, we need to make sure to carry that into every single term in our equation or in this case, our inequality. So let's go ahead and expand this out. So this becomes 12/4 times X plus two is greater than or equal to a negative 12/12. And then carrying that 12 to our very last term minus 12/3 X. Let's go ahead and simplify. So 12/4 is just three. So this becomes three times X plus two is greater than or equal to a negative 12/12 is negative one and then negative 12/3 is going to give me a negative four times X. OK. Now, we need to look at a couple of different things here. The first thing that we need to do from this point is to go ahead and distribute this three. So I need to distribute this three into my X and my two. So this is going to become three, X plus six is greater than or equal to negative one minus four X. OK? Now, since we have variables on both sides, this is where we need to move all of our variables to one side, all of our constants to the other, the same way we did with linear equations. So I want to move this four X to my left side and move my negative six to my right side. And I can do that by adding four X to both sides. Well, of course, cancel on that right side and then subtracting six from both sides, canceling it on the left and moving it to the right OK. So three X plus four X is going to give me seven X, I still have my inequality symbol greater than or equal to. And then on that right side, I have negative one minus +66, which will give me negative seven. OK. Last step here, we're going to go ahead and isolate X by dividing both sides by seven. Now I am simply left with X is greater than or equal to negative seven, divided by seven is going to give me a negative one. And this is my solution. But remember we want to express our solution set in interval notation and graph it. So let's go ahead and graph first so that we can better visualize this solution. So my solution is X is greater than or equal to negative one. I know that since it's greater than or equal to, I need to choose a solid circle, a closed circle for my end point of negative one. And then since it's anything that's greater than or equal to negative one, I'm just going to use an arrow to indicate that that goes all the way to infinity. OK. Now that I can visualize it better. What is this an interval notation? Well, interval notation for this, since I know that that negative one end point is included, I'm going to use a square bracket to enclose that end point of negative one. And then that goes all the way to anything greater than negative one which is just infinity which I always enclose with a parenthesis. And this is my final answer and my final graph, let me know if you have any questions.

8

Problem

Problem

Solve the inequality. Express the solution set in interval notation and graph.