Common Core Algebra I.Unit 5.Lesson 7.Solving Systems of Inequalities

Hello and welcome to another common core algebra one lesson by E map instruction. My name is Kirk weiler, and today we're going to be doing unit number 5 lesson number 7, solving systems of inequalities. Before we begin, let me remind you that you can find a worksheet that goes with this video along with a homework assignment by clicking on the video's description or by visiting our website at WWW dot E math instruction dot com. As well, don't forget about the QR codes on the top of every worksheet. Use your smartphone or tablet to scan those codes and bring you right to these videos. Okay. So previously, we've learned about systems of equations and systems of inequalities. We've also learned about how to graph linear inequalities into variables. Today, what we're going to be doing is combining all of those to graph the solution sets of systems of inequalities. All right, but first, let's jump in and exercise one. Take a look at a system of inequalities, all right? And try to figure out whether particular points are solutions to those systems. All right? So we want to do is we want to figure out if the .3 comma 8 is a solution to that system. Now it doesn't matter whether it's a system of equality or a system of inequalities or a mixed up system. The plane fact is a point is a solution to a system if it makes all the equations or in this case inequalities true. So let's take a look and let her a if I take that first inequality, X plus Y is greater than ten. And I put three in for X and I put 8 in for Y, right? I'll get 11 is greater than ten, and that's true. If I take the next one, Y is greater than or equal to three X -5. And I put 8 in for Y and I put three in for X, I'll get 8 is greater than or equal to 9 -5. 8 is greater than or equal to four. That's also true. Because I have those two trues. The answer is yes. So at the end of the day, in letter C, we're going to graph the solution set to this system of inequalities, three comma 8, better be in there. Why don't you pause the video and see if the .5 comma 9 is a solution to this system of inequalities. All right, let's go through it. Hopefully this time you found that the answer was no. Now let's take a look. I have X plus Y is greater than ten. So I'll put 5 plus 9 in there. Greater than ten, 14 is greater than ten. Well, that's true. Let's try this one. Y is greater than or equal to three X -5. So I'll put 9 in there. And I'll put 5 in there. Three times 5 is 15 -5. 9 is greater than or equal to ten. And that's false. It doesn't particularly matter that this one's true. The plain fact is if any of them are false, the answer is no. So just as we pointed out in letter a, when we ultimately graph this thing, that point shouldn't be in the solution set. So let's remember how graph inequalities and actually I'm going to start with this one. Y is greater than or equal to three X -5, because it's the easier one. Remember what we're going to do is we're actually going to first graph this equation. Y equals three X -5. That should be pretty easy, right? We have a Y intercept of negative 5, and we have a slope of three over one. So one, two, three, four, 5 all the way down here. We're going to go up three into the right one, up three to the right one, et cetera. Now we have to make a decision whether or not we then connect these with a dashed line or a solid line. Do you remember how that works? Hopefully. If there's an equal sign here, which there is, then we graph it with a solid line. Not my best solid line. And we label it with its equation. Y equals three X -5. Okay? The reason I wanted to do that one first was because the Y was also already isolated. In this one, there's a little bit more of a challenge because the Y whoops is not alone. I don't know why I put the equals sign in there. It's not there. X plus Y is greater than ten. Now that's not a problem. I can subtract X from both sides. Now remember, a property of inequality is that you can add or subtract anything you want from each side and it's not going to change that inequality symbol, that inequality is only going to flip right if we divide by negative. Now what we want to do is we want to graph Y equals ten minus X, okay? In this case, that B is ten, and that M is negative one over one, right? How do I know that? Because I can use the commutative property of equality. Sorry. Just the commutative property of addition to flip flop that ten minus X and make an intent negative X plus ten. So here we go. One, two, three, four, 5, 6, 7, 8, 9, ten, down one to the right one. Et cetera. All right. Now, in this case, though, right? In this case, since there is no equal sign, we need to draw this in dashed. All right, but now we've got some shading, okay? In the first one, it was Y is greater than or equal to three X plus 5. So we're going to shade above. Above this line. And actually, I'm going to change my color. To red for the other one. The other one was also greater than Y is greater than ten minus X so I'm going to shade above this line. Now what exactly is my solution set? Well, remember, my solution set will be all the points that make both of these inequalities true. While that is going to be this little triangular region right up here. Okay. That is my solution set. And if you notice, right? 123-123-4567, 8, there's my three comma 8. On the other hand, my 5 comma 912-345-1234, 5, 6, 7, 8, 9 is not in the solution set, right? It's not in the double overlap. So the solution set to a system of inequalities is where the two parts of the shading overlap if they overlap at all. But that's it. That's the whole story. Okay? So pause the video now, really think about what we did because essentially we did the whole lesson here. We really did. It's just a combination of other things that we've done. I'm going to clear this out in a moment, though, so pause the video. All right, here we go. Okay, that's gone. Let's move on to the next exercise. So exercise number two. On the grid shown below graph the solution set to the system of inequality shown below. State a point lies in the solution set and one that does. All right. Well, let's start off with this guy. But maybe not in green. We could. Now, this one also already has Y isolated. So I'm going to change it into an equation Y equals negative three halves X plus two. Where clearly the Y intercept is two. And the slope is negative three divided by two, so that's going to be pretty easy to graph on. Do that, go down three to the right two, reverse it, get the left three, up to, all right, should be dashed, right? Dashed because there is no equality there, so the points along that line are not included. The boundary points are not part of the solution. That's why we dash it. And then I think I'll do the shading right away. We're supposed to shade under it because it's less than. So any point in there. And I have to call label the equation of the line. Y equals negative three halves X plus two. All right. Let me change colors. Let me go to red now for the next one. Now what do we do with this guy? Well, the line it's based on is X equals negative two. All right. But that is a vertical line. It's the line that has all X coordinates of negative two. So for instance, that point's got an X coordinate of negative two, and that point, and that point at that point, right? So all points that have X coordinates of negative two. All right, and it should be solid. Because the equalities included. So I'm going to draw a nice solid red line as well as I can, not all that great. Label it X equals negative two, and now it's greater than. Now, that's a little weird, right? Because it's not above or below. It's right left. And of course, greater than in this case is going to mean that the right, because all X coordinates to the right of negative two are greater than it. Okay. So that's it. I mean, that's the whole deal. Now, what's interesting is, again, the solution set itself is sort of this overlapping region. Which includes this boundary, but doesn't include that one. So the last part of the problem says, give one point that's in the solution set. Well, there's tons of points. Tons of them. Here's a really good example. Here's a point that's in the solution set, and that's got the coordinates negative one one. On the other hand, there's also lots of points that aren't in the solution set. For instance, this point is not in the solution set and it's at one two one two three four. But that's it. That's the whole deal. So graph the lines that the inequalities are based on, shade according to greater than or less than, and wherever the shading overlaps of the two inequalities, if at all, that is your solution set. Okay. Pause the video now and I'm going to clear out the text. Here it goes. All right. Exercise three. Nice little multiple choice problem. Which of the following points is the solution to the system of inequalities shown below, show the work that leads to your answer. Well, we have no graph paper here. If we had a piece of graph paper, we could do the full on problem graph at all and then simply plot the four points. But here we really have to rely on truth values, and it's kind of annoying. It's kind of a guess and check thing. But remember, if either one of the two are false, then the whole thing is false. Okay? So let's do one. Let's walk through one that is false. And then have you worked to find the one that's true. Okay? So let's try number one. Okay? The three common negative 6. Let's try Y is less than or equal to negative four X plus two. So I'll put negative 6 in there, less than or equal to negative four times three. Plus two, negative 6 is less than or equal to negative 12 plus two. Negative 6 is less than or equal to negative ten. Well, that's actually false. Negative 6 is greater than negative ten. Remember that if you struggle with comparing two negative numbers, think about them in terms of temperature. Which one is warmer, right? They're both cold, but negative 6 is warmer than negative ten. So that's a false statement, which means I don't need to do any more work on number one. It can't be a solution to the system. So pause the video now, work through your other three choices and see which one makes both of those inequalities true. All right, let's go through it. The correct choice here actually ends up being choice three. And let's work through it really quick. Negative two comma ten. All right, here we've got Y is less than or equal to negative four X plus two. If I put ten in, right? I get negative four times negative two, plus two, negative four times negative two, and negative times a negative is a positive. And I get ten is less than or equal to ten. Had that just been a less than the answer would have been false. But because there's that equal there, it's true. Likewise, let's check this one. Y is greater than X divided by two plus 7. Whoops. Let's put in the ten for Y greater than negative two divided by two plus 7. Negative divided by positive and negative. And negative one plus 7 is positive 6, and that's all so true. So since both inequalities are true at that point, it is part of the solution set. The other two choices either fail in both of the inequalities or fail in at least one. All right? So pause the video now, write down anything you need to. All right, here we go. Let's move on to the last problem. Exercise four, I like this one a lot. Consider the system of inequalities given below. Determine which, if any of these points is a solution to the system. So I've got these two points. I want to know, are they solutions to the system? And I know it's now three inequalities. That's okay. We'll live with them. Two of them are very, very simple. So check. Pause the video now and see if either one of those points is a solution to that system, remember all three have to be true. All right, let's go through it. Well, let's check on one at a time. This inequality. Why is greater than or equal to negative two? Well, is four greater than or equal to negative two? Yeah, that's true. One down. Our next one is X is less than four. Well, this negative one, less than four. You bet it is. That's true. Our last one is Y is less than or equal to two X, that will be four is less than or equal to two times negative one, or four is less than or equal to negative two. That's a false, and therefore no. Came so close to out of the three. All right, let's try this one. So Y is greater than or equal to negative two. Well, that would be one is greater than or equal to negative two, and that's true. I guess one is bigger than negative two. All right, then we have X is less than four, that's three is less than four. That's certainly true. And Y is less than or equal to two X, that would give me one is less than or equal to two times three, or one is less than or equal to 6, and that's certainly true. So yes. So when I eventually announced graph this thing, well, we know that the first point won't be in the solution and the second well. So let's do it. I'm going to do some of the writing up here as we graph it down here. But why is greater than or equal to two? We'll do that one in blue. That's going to be based on this horizontal line, right? Horizontal line. It's the line where all Y coordinates are negative two. So for instance, this point, this point, this point. Et cetera, right? I don't think I really have to draw them all out, but every single point along that line has a Y coordinate of negative two, and yet here I am doing them all. All right, solid or dashed looks like it should be solid because the equalities included and greater than means that we should shade up here. All right. Y is equal to negative two. And keep forgetting the label in my lines. All right, let's do the next one. X equals four, that's going to be vertical. And that's one, two, three, four, any point. Along this line, right, as an X coordinate of four. Solid or dashed? Well, it's just less than so that seems like it's dashed. Right? That's definitely dashed. Okay, let's see. It's less than, so this is going to get to be a little bit of a complicated picture. But that means we've got a shade to the left. There we go. And let me go to the last one in green. Here we'll go with Y is equal to two X. Now, this one's interesting. Because this is definitely a slanted line, right? Both the X and the wire there, I put that plus zero in just for us to be able to think about the Y intercept of being zero and a slope of being whoops, that's not a Y intercept of zero. Oh, this is going to be tough to erase. I got to go super duper small. There we go. That worked. There's my Y intercept of zero. And then I'm going to go up to N over one. Up to an over one. Sorry about how messy the picture is getting, but we do have three boundaries on there. All right. Solid line. Because. Let me go back to red LEGO mad out of the line. X equals four. Now go back to green and do some shading. That's going to be less than. So it's going to be under that line. Oh. So where in the world is my solution set? Let me outline it in black. It's wherever the shading crosses each other. All of them together. That's all in here. Over to here and down to here. Everything in here is the solution set. Everything in there. Now letter C asks us to find the area of the portion of the XY plane that represents the solution. Well, let me pull that little puppy out. It's a nice right triangle. We can figure out its base one, two, three, four, 5, figure out its height, one, two, three, four. One, two, three, four, 5, 6, 7, 8, 9, all right, so not the nicest. Well, wait a second, 9 or ten. Oh, no, we're all the way up to here. All the way up to here. That was my mistake. Sorry about that. Yeah, the solution set extends all the way up there, so it's not 9. It's ten. So the area will be base times height, divided by two, so obviously 50 divided by two gives us an area of 25. Now I really like thinking about the last question, letter D why does the dashed line of one of the borders and that's hence this border, my X equals four border. Why does that make no difference in terms of the area? All right? Think about that for a second. This is actually a little bit of a tricky question. Let's just drop in. But think about why the dashed line doesn't make a difference in terms of the area. All right, pause the video now. Okay. I'm going to retrieve my pen. Oh, and I'm back, hey, hi. Okay, so here's the reason. Lines have no area. So in other words, the dashed nature of that line doesn't affect how much space is within the figure. Doesn't affect it whatsoever. All right. In fact, because the lines have no area, all three of the borders could have been dashed and the area that was shaded would have still been 25 units. Okay, so there's a lot on this page, lots of colors. I'm going to clear out the text pretty soon. So pause the video now. All right, here we go. Let's finish up the lesson. So today, what we did is we solved systems of inequalities and really what we did was we put together a lot of skills that we built been building up. We already knew how to evaluate the truth in the falsehood of an inequality at a given point. We also knew how to shade the solution set to a single inequality in two variables. And we also knew that a system of inequalities or equations for that matter consisted of all common points. In other words here where the two shading or in that last example, three shading areas overlap. Okay. For now, I'd like to thank you for joining me for another common core algebra one lesson by E math instruction. Until next time, my name is Kurt weymouth. Keep thinking. Thank you solving problems.