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Contents

- 1 Solutions to Equations
- 2 Solutions to Equations
- 3 Solutions to Inequalities
- 4 Solutions to Inequalities
- 5 Solution Set
- 6 or equal to
- 7 or equal to
- 8 Brackets
- 9 Matching Notation
- 10 Matching Notation
- 11 Reversing the Inequality Sign
- 12 Practice 1
- 13 Practice 1
- 14 Practice 2
- 15 practice 2
- 16 Practice 3
- 17 Practice 3
- 18 Practice 4
- 19 Practice 4

In the last lesson, we looked at solving problems and writing expressions. This time, we're going to look at solving inequalities. We can use inequalities to compare values. We can figure out what amount is less than or greater than some other amount. Before we start solving inequalities, let's look at an equation and think aobut how many solutions there are. For this equation, figure out how many solutions there are, and think about how do you know. Choose one of these answers.

If you said one solution, nice work. We can use our properties of equality to solve this equation, and we can figure out exactly how many answers there are. We subtract 2 from both sides, then we divide both sides by 3. So, x must equal

For equations we just saw that there's one solution, but sometimes there's no solution or there's infinitely many solutions. Let's look at the number of solutions for inequalities. So for this inequality how many solutions do you think there are? None, one, two, three, or more than three. And to help you out, think about what you could plug in for x, I'm going to try 0. I can plug in 0 for x and I know 3 times 0 is 0. I get 8 is less than 2 but I know this isn't true. 8 is bigger than 2. So 0 wouldn't be a solution to this inequality. Can you think of some other values of x that would work? Remember you are not after the value of x you want to know how many solutions there are.

All right, let's build off our knowledge from the last problem. >> We know that here we had an equal sign here last time, so when this was equal x had to be two. >> So there is one solution, but we don't have an equal sign here, we have an in equality sign. >> So I want this right side to be larger than eight. If x was two, I get eight. >> So that means if x was any number bigger than two, then this side would be larger than eight. >> I'm going to try plugging in This is great. >> This is a solution that works. >> And in fact, any number bigger than 3 would give me a number larger than 8 when I plug in. >> So, there's gotta be more than one solution. >> There's not just 1 or 2, there's infinitely many. >> There's more than three. >> If you choose this one, nice work. >> You might be skeptical just looking at this one solution. >> Let me show you four. >> When I plug in 4, I get 8 is less than 12 plus 2, or 8 is less than 14. >> Plugging in any number bigger than 2, like 3 4 5 or even 500 would work. >> That's why there's infinitely many solutions here.

From the last problem, we found out there is an infinite number of solutions to this inequality. We know a way to write those answers. And, I don't want to list them out because that could be tedious. So, let's use a solution set. A solution set is all the answers to this inequality. To find them, we solve this inequality just like we solve an equation. We subtract 2 from both sides to get the variable, we divide by 3. And the answer is x is greater than 2. So, any number larger than 2 will make this inequality true. Let's use the solution set to write our answer. We can show the solution in two ways. We can use a number line or interval notation. We use an open parentheses at x equals 2 because x can't actually equal 2, its got to be bigger than 2. And then, we draw an arrow in the direction of the numbers that are all bigger than 2. 2 is considered to be the lower bound of our solution set. We use an open parentheses for 2 so that we don't include it. We can remember, because is we plug in 2 here, we'll get 8 greater than 8, which isn't true. So, we don't include 2 in the solution. We use an open parenthesis for positive infinity because we can't actually reach infinity. Any number bigger than 2 will work. You might be used to seeing an open circle at 2, which is correct. For this class, though, we'll only use the parenthesis when showing answers on the number line.

Let's change this same problem a little bit. Instead of just being greater than, I'm going to have greater than or equal to. What additional value can x now be? Is it 0, 1, 2, 3, or do no additional values work?

If I solve my inequality, just like from before, I get x is greater than or equal to 2. >> Notice x can now equal 2. >> If I plug in 2 into my original inequality, I'll get 8 greater than or equal to 8, >> Which is a true statement. If you said 2, nice work.

There is a little bit of different solution this inequality. We could add 2 in our answer now, so we need a different way of writing the answer in a set of parenthesis. We're going to use brackets. We use a bracket to indicate that the variable x could take on that value. The same is true on the number line. x can now equal to, so we put a bracket in set of a parenthesis. And again, notice I have an open parenthesis for infinity. The interval is considered to be open on this end, because we can always count higher to a number. We can't really reach positive infinity, so we use a parenthesis. In general, for less than or equal to signs and greater than or equal to signs, you will always use a bracket. For less than and greater than, you'll use parentheses. And of course, parentheses always go around positive infinity or negative infinity. Let's put these symbols to use.

For this skills check, I want you to match each inequality with a number line and its interval notation. Think about whether brackets or parentheses are important and think about if you're less than or greater than. For example, for this inequality, if you think its number line was 2 and its interval notation was d, you put 2d in the box.

For the first inequality, I have x is less than 3. If you have trouble remembering the less than symbol, remember that the arrow points to the left, or to numbers less than zero. I know I need a parenthesis because I can't be equal to. So, it must be the number line 2 or number line 4. I want numbers lower than numbers that are from negative infinity all the way up to 3. So, that must be b. For x greater than or equal to 3, I want to use a bracket, that's number line 1. The interval notation would be d because I get d equal to 3 and then have any number bigger or up to positive infinity. Here, I have 3 is greater than or equal to x. I'm used to seeing the variable on the left side, so I'm going to flip this in equality around. I need to make sure that the arrow still points to the x when I switch this inequality around. I want 3 to be greater than or equal to x, and I want x to be less than or equal to 3. For the number line, it must be at number 3. I need a bracket at 3 and my values could be less than or equal to 3. The interval notation would be a. I could have numbers from negative infinity all the way up, and equal to 3. This must mean that the last one is 2c. I want numbers greater than 3. Here are numbers greater than 3 and here are numbers between 3 and positive infinity.

We know inequality's can be solved just like equations. But there is one exception. An inequality sign behaves different when you multiply or divide both sides by a negative value. Let's look at that. I am going to start by not doing division or multiplication, I am going to do subtraction. First, I subtract one from both sides, and I get 1 is greater than 0. Then, I can subtract 2 from both sides and I get negative 1 is greater than negative 2. This causes a lot of trouble for students. Let's look at this. Here's negative 1 and here's negative line, so it must be bigger in value compared to negative 2. But we're not after comparing numbers here. We want to know about the inequality sign. I've rewritten negative 1 is greater than negative 2 and instead of doing the subtraction twice, what if we just would've multiplied? If we multiplied both sides of our inequality by negative 1, we'll get negative 2 and negative 1. But I can't just carry my inequalities straight down. We know negative 1 is greater than negative 2, so the sign has to reverse in direction. The same is true when dividing by a negative 1. We know when we divide by a negative 1, we change the signs of the numbers. So, I'll still need to flip my inequality sign. These inequalities might look different, but they're actually equivalent statements. Here, I have negative 1 is greater than negative 2, and here, I have negative 2 is less than negative 1. If we flip this entire inequality, we can see that we wind up with the same thing. We get negative 2 is less than negative 1. The key thing to remember when solving inequality problems is to reverse the direction of the sign when you multiply or divide both sides by a negative.

Let's try some practice. What do you think the answer is to this inequality? Enter your answer in this box here. When you type in the answer, you can use less than and the equal to symbol for less than or equal to. You could type in this for greater than or equal to. Good luck.

First, we subtract 6 from both sides, so I'm left with negative 3 x, make sure t carry the negative sign down with the term and negative 30 on the right. Here's where I want to be careful. I'm dividing by a negative 3, the coefficient in front of x. When I divide by a negative, my in equality sign must flip over, or reverse direction. So, x could be any number greater than or equal to 10. Just like with equations, we can check to make sure if we are right. Let's plug in any number greater than or equal to 10 into our original inequality to see if the condition is true. I'm going to plug in 10. And when I do that, I can see I get negative 24 is less than or equal to negative 24. That definitely checks.

For this question, I want you to write the answer in interval notation, >> So using parentheses or brackets. >> You can put your answer in this box here.

When solving an inequality, I want to solve it just like an equation. I'm going to clear factions, clear groupings, and then combine variables on both sides. First, I'm going to distribute this negative 3. So, I have negative 3 times x and negative 3 times 4. I'm going to combine like terms to get negative 10. Next, I'm going to add 10 to both sides. Then, I can add x to both sides to isolate my variable on one side of the inequality sign. So, I have negative 2x is greater than or equal to 18. So, I'm left with positive x over here on the left, and negative 9 on the right. I'm left with x is less than or equal to negative 9. Remember, I divided by a negative value here, so I have to reverse the inequality sign. Here's the number line solution, x less than negative 9, and here's the interval notation.

Here's out third practice problem. >> Try this one out. >> Write your answer in interval notation. >> Good luck.

We can start this problem by distributing the 4. We want to clear our grouping. So I get 4x plus 8 on the left. I want to isolate the variable on one side of the inequality sign. I'm going to choose to subtract 4x. I made this decision so I could keep my variable positive. I know when I subtract 4x here, I'll wind up with positive 2x. So, positive 8 is less than or equal to 2x minus 8. Next, I add 8 to both sides to isolate my 2x. So I have 16 is less than or equal to 2x. Finally, I divide both sides by 2, the coefficient in front of x. So I have 8 is less than or equal to x, which is really the same thing as x greater than or equal to 8. So I use a bracketed 8, and then I want any numbers greater than 8, so up to positive infinity. This would be the interval notation.

For this last question, I want you to find the number line that represents the answer, or the answer might be all real numbers, or none of these number lines are the answer. Type the letter of the answer in the box here. For example, if you think this is the number line, you'd write d in the box.

To start solving this inequality, let's clear our groupings by distributing. I want to distribute this negative 2 and this positive 2. I'm showing my distribution in this step here, being really careful with my negative 3. It's negative 3 times negative 1. I'm going to combine like terms on each side of the inequality. I have 5 plus 3, which gives me 8, and 6 plus 1, which gives me 7. I bring down the negative 3m, and the 2m. I add 3m to both sides to keep my variable positive. A positive 5m and no more M's on the left. Next, I subtract 7 from both sides to isolate the 5m. Finally, I divide by 5 on both sides to get 1 equal to 1 5th M is greater than or equal to 1 5th, so that must be answer choice A.