A solution set to a system is the collection of all the pairs that are solutions to the system. So it's very, very simple. Something is going to be a solution to a system if it makes all the equations true. And the entire solution set to a system will be all the pairs that make them both true. All right. Let's take a look at exercise one. Determine if the .2 comma 5 is a solution to each of the systems provided. Show the work that leads to each to your answer for each. All right, I think that you know enough about how to do this already to take a shot at it. So pause the video right now and see if you can figure out whether two comma 5 is the solution to these two different systems. All right, let's do it. Now remember, it's got to make both of them true. So let's start with Y equals four X minus three. I'm going to put 5 and for Y I'm going to put two in for X I'm going to get 5 is equal to 8 minus three, 5 is equal to 5. That's true. That's good. It's not enough, but it's good. All right, now I'm going to check this one. I'll put two in for X 5 and for Y I'll get four plus 5 is equal to 9. 9 is equal to 9. 

That's also true. So that makes them both true and the answer is yes. Let's try that or B, Y minus X is equal to three. Put the 5 in. I put the two in. And I get three is equal to three. That's obviously true. Uh oh, fractions. We'll be okay. It's a fraction of one half. All right, so we put 5 and for Y one half in for X, course one half times two is one. And 5 equals 7. Oh, that's false. So that means that overall this is false. And the answer is no. So one way or another, if I give you a system of equations, which is simply two or more equations, and I ask you whether or not a given XY pair is a solution to that system, you should be able to tell me. Right? Does it make all the equations true at the same time? All right. I'm going to clear this out. So pause the video now if you need to. All right, here it goes. Let's see what we have next. All right. So now let's consider the system of equations shown below. Y equals two X plus 5 and Y equals two minus X letter asks us to both graph both equations on the grid shown, use tables on your calculator to make the process faster, if necessary.

Label each line with its equation. Well, I'm going to graph these without bringing the graphing calculator into it. I'm just going to use slope and Y intercept. So for this first line, I know my slope is two divided by one and my Y intercept is 5, so one, two, three, four, 5. Now unfortunately here, I'm probably going to have to go instead of going to the right one and up to, I'm going to have to go down to and to the left one, but we're sort of used to that. So let me graph that actually, I think I am going to use. My prefab line and it says label. So Y equals two X plus 5. There we go. Let me go red on the other one. This one's tougher. In fact, I may want to rewrite it using the commutative property as negative X plus two. That will allow me to see that the slope is negative one divided by one. And the Y intercept is two. So I can now go down one to the right one, down one to the right one, et cetera. Also left one up one. And then I think we'll put our line in red. 

All right. And nice. I should probably have arrows on that other one as well, but not really the point of the problem. Y equals two minus X letter B asks at what point do the two lines intersect. Let's go green. Intersect right there. At negative one comma three. And part C asks this whoops. Letter C asks us to show that this point is the solution to the system. All right, no problem, right? To show that this point's a solution to the system. Just like we did in the first exercise, I'm going to put three in for Y and negative one in for X two times negative one is negative two, negative two plus 5 is positive three, so that's true, three equals three. Let me go with this one. Now I've got to be careful here. Let me put in three for Y, extremely careful, negative one for X remember subtracting a negative is the same as adding. So I'll get three equals three, and that's true. And because of that, those two make the overall true and yes, solution. All right. But this is cool because it leads us to one of the first and most fundamental ways one of the first and most fundamental ways of solving the system of equations. And that's by graphing both equations. 

Now, the logic of why graphing the two equations and finding their intersection points works is absolutely beautiful. You know, you may have learned this. You should have learned this in 8th grade math, but you may never have walked through the exact logic. So I'm going to clear this problem out, pause the video if you need to. All right, there it goes. And let's walk through the logic of solving a system by graphic. All right, first, let's make sure that we remember something about the graphs of equations, right? Any coordinate pair XY that makes the equation true lies on the graph. The entire graph is the collection of all the XY pairs that make the equation true. All of them. So remember that. All graph is all a graph is is a picture of every XY pair that makes an equation true, and any XY pair that doesn't end up lying on that graph. Now you put that together with what we know about systems of equations, and you can understand why graphing two equations and finding their intersection points solves them the system. So let's take a look at exercise three. Exercise three says, so now you can put the definition of a graph of an equation. 

Together with the definition of a system. Fill in the blanks with one of the words shown. So this is like a word bank problem. We've got four words true, intersection, solutions, and both. All right? So each one of those words goes into one of these blanks only once. All right, what I'd like you to do is pause and see if you can fill in each one of those blanks with one of those words. All right. Take a little bit of time. Once you're done, read through all four bullet points as if they were an entire paragraph because what they should do is create an argument, a reason, a recent argument for why graphing and finding intersections works. Okay, pause the video now. All right, let's go through it. Number one, to solve a system's system of equations graphs, graphically, you find the. Intersection of the two graphs. This works because any intersection point must lie on both graphs. Because intersection points lie on both graphs, they must make both equations. True. And because intersection points make both equations true, they are solutions to the system of equations. Isn't that cool? And that really is the argument for why graphing two equations and finding their intersection point or points solves a system, right? The intersection point lies in both graphs, right? Because it lies on both graphs, it makes the graphs or it makes both equations true. And that is the definition of the solution of a system. 

All points that make all equations true at the same time. All right, so I'm going to clear those words out. Write them down, pause the video. All right, let's move on. So let's do a practical application of this. You already know how to graph lines. So graphing a bunch of lines finding out where they cross isn't going to be time well spent. Let's take a look at one last applied problem. Janelle and sweater are taking a 50 question true false test in their history class. Okay? Janelle started after swetha. And had already finished I misread that. Janelle started after sueta had already finished 12 questions, all right? So sweat is working along. She's got 12 questions done and Janelle starts. Janelle answers questions that a rate of two per minute while sweat that answers them at a rate of four questions for every I'm sorry 5 questions for every four minutes. I can't seem to read this problem today. All right? So we've got Chanel at two per minute, swath is a little more complicated, 5 questions every four minutes. Janelle eventually catches up to swap it. How many minutes does it take her? And what question are they on when Janelle, when Janelle catches up? All right, letter ray asks us to create two linear models for Janelle and swathi's questions answered since Janelle started. Now remember Janelle starts second. All right. It may help to plot some points on graph paper, show the work that you use. All right, so actually, let's do this a little bit. All right, remember, this is the number of minutes since Janelle started. 

So let's just track sweater. Swetha had finished 12 questions this access goes by two. So I'm going to put sweater in blue. All right, so what is going to be in blue? And she had finished 12, right? And what did she do? Two per minute. So every minute that goes by, she answers another two. Now, that means that we only go up one space, though, for every minute, because these are these go up by twos. So this is what this graph. And actually, coming up with a linear model for sweater is going to be very, very easy. Right? In fact, we could say for sota that the number of questions she'd answered, I'll just call it Y is equal to two times X plus 12. Right? Because her slope is two, two questions per minute. And she starts with 12 questions already answered. So now, let's go with Janelle. Now, Janelle has a more complicated situation, but she does start at zero. Right? So, I'm doing this question wrong. Janelle answers two per minute. Sweat that answers 5 questions every four minutes. So I botched that. Let me go back and erase this really quick. I don't want to start the video all the way over. That's okay. And actually, sorry? This entire thing is wrong. Let me go back into blue. 

So sweater, my apologies, swath and answers them at 5 questions every four minutes. Now, she is starting at that 12. So every four minutes she's going to answer another 5. So we go over four minutes and then two, four, 5. One, two, three, four, one, three, 5. I know this is a little bit tricky. One, two, three, four, two, four, 5, one, two, three, four, one, three, 5, one, two, three, four, two, four, 5. It's tricky because that Y axis goes the Y axis goes unfortunately here by twos. Okay, that's a little bit better. Now let's get our equation. Now think about this. She's answering questions at 5 questions every four minutes. That is, whoops. That is your slope right there. Okay? 5 questions every four minutes, that's your slope. 5 fourths X plus 12. Right? Now Janelle, Janelle's got the easy one. Now it's got the easy one because Janelle is going to answer questions at two per minute, but she starts off at zero. Okay, because she starts off, well, with none of the questions answered. This is the one now that's going to look like it's got a slope of one but really doesn't. Right? Because what we're really doing is going over one and up to. And we just move it along. And we keep doing it. And eventually we'll connect these with a line as well. 

So. Here's Janelle's line. Right? And Janelle's equation. Is why equals what? Just two X, right? We could say two X plus zero, or we could just say two X I think I'm just going to get the two X, all right? So we've graphed the two equations. So we've taken care of that. And solve the problem. What's the problem? How many minutes does it take her to catch up and what question are they on when Janelle catches up? All right, well here's where they catch up. And what time is that? Wow, it takes her 16 minutes to catch up. And they are on. Watch out. They are on. Question 32. All right? It's kind of neat, isn't it? Sorry about botching it there at the beginning. You know, it's very, very easy to misread problems. You know, the reason that I caught my error because I knew that in order for Janelle to catch up, she had to be answering questions faster than swatter. But Janelle was answering him at two questions per minute, sweat is actually answering them at less than two questions a minute. 

Sweat is almost pretty much just doing almost one question a minute. I mean, 5 questions every four minutes. Just a little bit more than a question a minute. So Janelle will catch up. The way I was originally doing it, Janelle wasn't going to catch up. And so the question really didn't make a lot of sense. All right. As is though, we've got it right this time. So I'm going to clear out the text. Please copy down anything you need to. All right, here we go. Let's finish up. So what we did today is we reviewed what it just means to be a solution to a system, and that is simply an XY point or whatever, a pair of values that makes both equations true at the same time. We also saw how one not easy way but straightforward way of solving a system is to graph the two equations and find out their intersection points. We'll visit this more in later lessons. For now, I'd like to thank you for joining me for another common core algebra one lesson by E math instruction. My name is Kirk weiler, and until next time, keep thinking. I keep solving problems.