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Contents

- 1 Saving Money Equation
- 2 Saving Money Equation
- 3 Saving More Money
- 4 Saving More Money
- 5 Saving Money Inequality
- 6 Boundary Line
- 7 Boundary Line
- 8 Dashed or Solid
- 9 Dashed or Solid
- 10 Finding the Shaded Region
- 11 Finding the Shaded Region
- 12 Graphing Linear Inequalities
- 13 Graphing Linear Inequalities
- 14 Graphing Part 2
- 15 Graphing Part 2
- 16 Graphing Part 3
- 17 Graphing Part 3
- 18 Graphing Any Linear Inequality
- 19 Practice 1
- 20 Practice 1
- 21 Practice 2
- 22 Practice 2
- 23 Practice 3
- 24 Practice 3

In the previous two lessons, we learned how to graph and write the equations of a line. In this lesson, we'll learn how to graph a linear inequality. So, instead of having an equal sign, we'll have a less than, greater than, less than or equal to, or greater than or equal to symbol. We'll learn how to do this by saving money. Let's say you have $200 in a savings account and each month you save exactly $100. What's an equation that you could write to model this situation? You can use this table to help you out. You can write your equation in here, and remember it should involve the variables x and the variables y. X stands for the number of months, and y stands for the amount of money you saved.

Well, if no time has passed, then we start with $200. That's how much was initially in our savings account. We know each month we save exactly $100. So, after one month I would have $300, and then, if another month goes by, I save another $100, and we can easily fill out the rest of the table. But remember, in algebra, we want to know for x number of months how much money would we have. Well we start with 200 and then we multiply the number of months by 100. So we also need to add 100x. Now we've related the amount of money we saved with the number of months, x. We have this equation, y is equal to 200 plus 100x. Congrats if you got this right. I know this isn't the easiest algebraic reasoning, but if you work from a table and build it up, you can get here and we can switch these two terms around. I could have the 100x come first, followed by

We know that we could graph the equation like before. Here's the graph and we can see if the y intercept 200 is right here. Our slope is 100 because you save $100 for every month that goes by. Another way of saying that is our y value increase by 100 for every month that passes. But this equation tells a very narrow story. It assumes that you save exactly $100 each month. But what if you saved more than that? Is there a way to represent this in a, on a graph? Let's try and figure that out together. First, let's consider a few points. Which of these points would represent a time when we saved more than $100 in a month? Check all the boxes that apply.

After one month I would have $250. Well, this is below our line so that means we saved less than $100 in a month. We started with $200 and we only gained $50. We should have gained more than a $100. For 2 months and $500, we can see that, that is above our line. This means we must have saved more than $100 in one of these months. I know this because after 2 months, I should expect to have exactly $400 if I were to have saved exactly $100 a month. But we have more. So this one works. This one's a little bit tricky. After 1 month, I saved exactly didn't save more than 100 dollars. So this one's out. This point means that, after 3 months, I saved 400 dollars. That would put me here.

There are other points above this line that represent saving more than $100 per month. For example, after 1 month, I could have also saved $400 or after 1 month I could have also saved $450 or even $500. Similar reasoning could be used at 2 months, I could have saved maybe $600. Or 700, so there's tons of points farther above this region. Because all these y values are the amount of money saved falls above this line, we can change our equation into an inequality. We want our savings amount to be greater than the exact amount of $200 plus the $100 per month. So this is the inequality we can write. Notice that linear inequalities contain 2 variables, just like linear equations. This allows us to find a point, x and y, that satisfies this condition. For example I can use this point and plug in 2 for x and 500 for y and I should get a true statement. And when I do, I get 500 is greater than 400, so yes this is true. Also remember from before, we said that one 300 couldn't be a point for our inequality. We didn't save more than $100 per month. So I need some way to represent that on the line. I can't actually include the points on the line, so I'm going to use a dashed line to exclude them. We know any of the points above this line satisfy our inequality. This shaded reason up above would represent every instance when we saved more than $100 in a month. Let's say you allowed yourself to save exactly $100 or more per month. So now we could be greater than or equal to. Well, this point can now be a point on our line. If you allow yourself to save exactly $100 or more in a month then now we can include this point in our inequality. We can see that by plugging it in we get a true statement. 300 is greater than or equal to I'll use a solid one. That will represent the equal to part. When graphing linear inequalities, we use a solid line for greater than or equal to, and less than or equal to. If we see the greater than or the less than sign, then we use a dashed line.

Let's walk through graphing a different linear inequality. When we graph a linear inequality we want to start by graphing a boundary line. The boundary line would be made up of the points if this inequality were an equation. I want you to use your knowledge of slope and the Y intercept to figure which of these points would be on the boundary line. Check all of them that apply.

The y intercept would be positive 1, so it would be here. Then, the slope is negative 3 4ths, so I move down 3 units and right 4 units. So I'd wind up here. I could have also gone up three units and then left four units, which would put me at this point here. So I know my line should go through these three points.

So here's the next question. Should the line be dashed, or should be solid?

We have a less than inequality sign, so it should be dashed.

The last part of graphing an inequality is figuring our which region to shade. We could either shade this top region, region 1, or this lower region, region 2. One of these has to be correct. To figure it out, we can use a test point. We can choose any point in one of the regions to determine whether or not it makes the inequality true. I want to make sure I don't choose a point along this line. Because these points have been excluded. It's dashed. The origin is usually a great point to check. It's zeroes, and it's easy to plug in. So you check it, and determine. Should we shade region 1 or region 2? If 0,0 makes a true statement here, we want region 2. If 0, 0 doesn't make a true statement in this inequality, then we want region 1.

It turns out, I want to shade region 2. If I plug in the values of 0 for x and for y, I can see that I get 0 is less than 1. This is a true statement. So, I know this point satisfies this inequality. So, all of these points also satisfy this inequaltity. Try testing some other points to be sure. You'll find that they all do.

Here is another linear inequality. This time I want you to choose one of these lines that you would start with to graph this inequality. Would you use line a, line b, line c or line d? Make your choice.

We should use line d. Remember, we can start graphing linear inequalities like a line. We start with the y intercept, which is that positive 3. Then we travel down 2 and right 3 units. We can continue doing that to get other points in the line.

So we know our line should be here, but should it be dashed or should it be solid? Make your choice.

We know the line should be solid, because we have a greater than or equal to symbol. Another way to confirm or think about this is to try a test point on the line. If we plug in x and y, we should get a true statement. If x is 0, and if y is 3, then we can see we get 3 is greater than or equal to 3. Because we can be greater than or equal to, this is a true statement. So this point works. And any of these points along the line will satisfy the inequality. That's why we keep it solid.

Finally, which region should be shaded? The lower region, region 1, or the higher region, region 2?

We should shade region 2. I can use the test point 0,0 and plug in these values to check. If I plug 0 in for y, and 0 in for x, I get 0 is greater than or equal to 3. But wait, this isn't true. I know zero is actually less than 3. Because this point doesn't satisfy our inequality and we wind up with a statement that's not true, we don't want any of this region. We only want the other region. Any points on this side of the line or on the line would work. So, here's the final graph of our linear inequality.

Sometimes we might run into inequalities that are not in slope intercept form. We can handle these inequalities much like we did linear equations. We can try and solve this inequality for y. First I subtract x from both sides, to get negative 4y is less than or equal to negative 12 minus x. I'm going to rearrange the terms here. So, I'll have negative 4y less than or equal to negative x minus negative, as before. And this 12 is still negative, as before. Finally, we divide every term by negative 4. Here's where we need to be very careful. We're dividing by a negative, and this is an inequality symbol. Just like from before, we know we need to reverse the inequality sign. The other thing we want to watch out for are the negatives on the other side. A negative divided by a negative makes a positive, and the same is true here. So we're left with, y is greater than or equal to 1 4th x plus 3. And don't forget about that 1 as the coefficient in front of x. That's how we can get positive 1 4th. Now that we have this inequality, we can graph it just like before. We'll start by plotting the line. Then we'll determine whether or not that line should be dashed or solid. And finally, we can determine the region by using a test point. We plug in a point to see if it makes the inequality true. If it does, we want that region. If not, we'll take the other one.

For the first practice problem, which of these graphs represents the linear inequality 4x plus 3y is greater than or equal to 12? Put the letter of the graph in the box.

The first graph was correct. Lets see why. Let's solve this inequality for y so we can get the boundary line. First, we subtract 4 x from both sides, then we divide each term by 3 to get y is greater than or equal to negative 4 3rd x plus represents the boundary. The y-intercept is 4, and my slope is negative 4 3rd. Down 4, right 3. That will get me to my next point. This is the boundary line for the inequality. I draw it solid, because the symbol is greater than or equal to. So, points on this line do satisfy this inequality. Next, I choose a test point, like the origin, to determine which region to shade. This region or this region. Plugging in zero for x and zero for y, I get, 0 is greater than or equal to 4. We know this is not true. So, points in this region do not satisfy the inequality. We want points in this region, so we shade above. Amazing work if you got that one right.

For the second practice question, I want you to find the graph that represents this inequality. Is it a, b, or c?

This one was c. If you got it right, excellent work. Just like in the first practice problem, we want to get the boundary line, so let's solve this inequality for y. First we subtract 3x from both sides, then we divide both sides of the inequality by negative 5. Remember, both these terms need to be divided by negative 5. And here we need to be very careful. We divided by a negative so we need to reverse the inequality sign. Now we're ready to graph. The y intercept is negative 3, which is here. 3 units down from the origin. Our slope is positive 3 5ths, so we rise 3 units up and run 5 units to the right. Now, I have two points and I can draw my boundary line. I have a less than symbol so I need to draw a dash line, because I can't include points on the line. And finally, we use a test point to determine which region should be shaded, the top region or the lower region. If I plug in 0 for x and 0 for y I get 0 is less than negative 3. I know this statement isn't true, so I don't want points in this region. I want this region. This should be my solution region and I can test another point in this region to be sure I'm going to use the point zero negative four and remember you could choose any point in this region. It wouldn't matter. If I plug in 0 for x and negative 4 for y, I get negative 4 is less than negative 3. This is a true statement, so yeah, I did this problem correctly.

Here's the third and final practice for this lesson. Which graph do you think represents this linear inequality? Good luck.

Here, the correct answer was b. Great job if you got that one right. First, let's solve this inequality for y. I can subtract 10 from both sides to get 5 y on the right, and 2 x minus 10 on the left. Then, I divide each term by 5. So, 2 side of the inequality, so let's rearrange this inequality. So, I have y is less than or equal to 2 5th x minus 2. Notice this inequality symbol is pointing towards the y in both inequalities. That's a great way that you can check to make sure you wrote it correctly. First, we plot our y-intercept at negative 2. And then, we use our slope, 2 5th. We rise 2 units and run to the right 5 units. So there's the next point. I use a solid line here because that could be equal to the line. We have y is less than or equal to, so points on this should satisfy this inequality. Again, I'm going to use the origin as a test point, and I get 0 is less than or equal to negative 2. This statement's not true, so I don't want this region. I need to shade below. And here's our graph.