Linear Inequalities - Online Tutor, Practice Problems & Exam Prep

Understanding interval notation is essential for expressing solution sets compactly, especially for inequalities. Closed intervals use square brackets to indicate included endpoints, while open intervals use parentheses for excluded endpoints. When solving linear inequalities, remember to flip the inequality sign when dividing by a negative number. Graphing these solutions involves using closed circles for included values and open circles for excluded ones. Expressing solutions in interval notation enhances clarity, allowing for effective communication of ranges, such as \( x \geq 3 \) represented as [3, ∞).

Interval Notation

Hey, everyone. We already know how to express sets in set notation, so something that looks like this, but there's 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? 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 and 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.

There are different types of intervals you can have based on whether you're dealing with a less than or less than or equal to sign. So let's first look at our closed interval. When I have a set like x ≥ 0 and x ≤ 5, because I have these less than or equal to inequality symbols, I am going to write my endpoints in between square brackets. These less than or equal to signs tell me that I'm just going to take my endpoints 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 points with closed circles. I'm still going to label my endpoints here, and then I'm going to draw a line connecting my endpoints to show that my set or my interval includes everything from 0 all the way up to 5, including 0 and 5. 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 endpoints included because it's not just less than 5, it's less than or equal to 5.

So let's take a look at our open interval here. An open interval happens whenever I have something like 0 < x and x < 5. You'll notice that these are not less than or equal to signs. They are just less than signs. When I have that, I'm just going to take my endpoints 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 endpoints, and still connect my endpoints. But the open circles say to me that my endpoints 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 4.9999. But I can't have 0 or 5 because they are not a part of my set. 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 x ≥ 0 and x < 5. 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 endpoint is included, I want to make sure to 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 0, including 0, up to 5, 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 ≥ 3. 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. I'm going to go ahead and plot my 3 with a closed circle. Now this is going to be fine all the way up until forever, up until infinity. So whenever we don't have an explicit endpoint, so here this just says the x can be anything greater than or equal to 3 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 our intervals, we're going to treat this as an open bound because you can't get up to infinity. It goes on forever. Right? So it's not a hard endpoint. So to write this in interval notation, my 3 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 endpoint. 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.

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] 

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

• A. {$x\left|x\right.$ ≤ 0}
• B. {$x\mid x$ < 0}
• C. {$x\mid x$ > 0}
• D. $\left\lbrace x|x\ge0\right\rbrace$

Express the given set in interval notation and graph.

$\{$$x$ $\mid$$x\le 7$$\}$

• A. (−∞,7]
• B. [−∞,7]
• C. (−∞,7)
• D. [7, ∞)

Linear Inequalities

Hey, everyone. So we've already learned how to solve linear equations like 2x-6=0 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 2x-6=0 have 2x-6≤0. 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 2x-6≤0, I can start by moving my 6 to the other side the same way I would with a linear equation. So if I just add 6 to both sides, that will cancel, leaving me with 2x≤6. Okay. Now that I'm here, I can isolate x by dividing by 2, again the same way I would with a linear equation. Canceling that 2 out and leaving me with x≤3. So I actually still even ended up with a 3 there. Now instead of being equal to 3, it is less than or equal to 3. 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 2x-6≤0 and over here I have -2x-6≤0 . So let's see what happens here. Well, I start the same way. I can go ahead and move my 6 over, leaving me with -2x≤6. And now I need to go ahead and divide both sides by negative 2 to isolate it. So that will cancel, leaving me with x≤-3, 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 3 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 3, if I try x=-4, that should work. So plugging that back into my original inequality, I have -2⁢x-6≤0. So -2⁢-4=8, minus 6 is less than or equal to 0. 8-6=2, so 2 is less than or equal to 0. But 2 is definitely not less than or equal to 0, so something is wrong here and this actually is not my solution at all. So the reason why that's not my solution is that any time we multiply or divide by a negative number, I 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 -2x≤6. And then when I divide both sides by this negative 2, 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 6 divided by negative 2, which gives me negative 3. So if this is my solution and this worked here, that means that any number greater than or equal to negative 3, I should be able to plug it. So if I take x=0 and plug that in, that should hopefully work. Let's go ahead and give that a try. So -2⁢0-6≤0. -2⁢0=0 and I'm left with -6≤0. Now this is definitely a true statement. Negative 6 is less than or equal to 0. So this is my solution, and x≥-3. So let's talk about how we want to express these solutions.

Well, whenever we were using linear equations, we just had one single value. We had something like x=3. 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 equal to 3, I have that x≤3. And that's because I'm no longer dealing with a single value, but I am instead actually dealing with a whole range of values. So to graph that, I still have that same endpoint 3. And I know that since it's less than or equal to, that means I need to have a closed circle. I'm going to draw a closed circle at my endpoint 3. And if x is anything less than or equal to 3, that means it could be anything to the left of that 3 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 3 in a square bracket or parentheses? Well, again, since I have that less than or equal to sign, that means that 3 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 in an interval notation. So since I have x≥-3, I still have my endpoint at negative 3 with a closed circle. Now since x can be anything greater than or equal to that negative 3, 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 3 is again going to get a square bracket here because I have a greater than or equal to sign and not just a greater than or a less 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.

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

Linear Inequalities with Fractions & Variables on Both Sides

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 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 14 × x + 2 ≥ 112 - 13 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 3 denominators here. I have 4, 12, and 3. So my least common denominator is going to be 12. Let's 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 inequality. Let's expand this out. This becomes 12 ÷ 4 × x + 2 ≥ - 12 ÷ 12 + - 12 ÷ 3 × x. Let's simplify. 12 ÷ 4 = 3, so this becomes 3 × (x + 2) ≥ - 1 - 4 × x. Now let's distribute the 3. This is going to become 3x + 6 ≥ - 1 - 4x. Since we have variables on both sides, 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 4x to my left side and move my - 6 to my right side. And I can do this by adding 4x to both sides and subtracting 6 from both sides. 3x + 4x = 7x and - 1 - 6 = - 7. Isolate x by dividing both sides by 7. Now I am simply left with x ≥ - 1, and this is my solution. But remember, we want to express our solution set in interval notation and graph it. So let's graph it first so that we can better visualize this solution. My solution is x ≥ - 1. I know that since it's greater than or equal to, I need to use a solid circle, a closed circle, for my endpoint of negative one. And since it's anything that's greater than or equal to negative 1, I'm just going to use an arrow to indicate that it goes all the way to infinity. Now that I can visualize it better, what is this in interval notation? Well, interval notation for this, since I know that the negative one endpoint is included, I'm going to use a square bracket to enclose that endpoint 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.

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

$\frac13\left(x+1\right)\ge\frac15\left(3+2x\right)$

• A. $\left(-4,\infty\right)$
• B. $\left[4,\infty\right)$
• C. $\left[-4,\infty\right)$
• D. $\left(-\infty,-4]\right.$