Common Core Algebra I.Unit .Lesson 3.Properties of Systems and Their Solutions

And you know, I think I see the deeper message and a lot of things and I had to read this quite a few times because and this is mostly for the math teachers out there, but the students, you know, see if this makes sense to you too. This particular standard is really the basis of what we call the method of elimination in solving a system of equations. And we're going to develop that method today. It's kind of a neat way to solve a system of equations. On top of the ones that you already know, which is graphing an equation, or a graphing a system, sorry. And then solving a system by substitution. But the weird thing, of course, is that there isn't any standard in the common core that specifically says, students will be able to solve a system of equations using the method of elimination. And yet, what this standard gets at is all the mathematics that's behind that method. So everything that we do today is going to actually be based on the properties of equality, the properties of equality. Now, we've seen these in earlier lessons, but the two most important properties of equality. The two that will be the basis of what we do today are the first one love the animation. 

The addition property. And again, these properties are sort of like self evident when you think about them. But here it comes, right? If I've got a equals B, two quantity is equal to each other. Well, then I can add a third quantity to both sides. So a plus C is equal to B plus C if I don't do it with letters, but I do it with numbers. It's really obvious, you know? Because 7 equals 7, well, then 7 plus two equals 7 plus two. I think we can all agree that 9 is equal to 9. Anyway, you also have the multiplication property of equality, right? It essentially says the same thing, but with multiplication. So if you've got two numbers, a equals B well, then if you multiply both of those two numbers by the same thing, let's call it C, then they're still equal to each other. So if 7 is equal to 7, then two times 7 is equal to two times 7. 14 is equal to 14. So strangely enough, everything that we do today with systems and substitution and showing that solutions are still solutions, all of that kind of work is still just based on these two essential properties of equality. All right, so let's jump into the lesson and see how it goes. Exercise number one says consider the system shown to the right and its solution one 5. All right? 

So I'm telling you that this is a solution to this system. Fair enough. All right, but letter a asks us to show that it is a solution. So remember, what it means for a point or what it means for a replacement pair of variables to be a solution to a system is that it's got to make both of the equations and if there were more than two, all of the equations true. All right? In other words, we got to be telling the truth. Four X plus two Y has to be 14. And X minus Y has to be negative four. So let's check that. First, I'll check the four X plus two Y equals 14. So I'm just going to take that point and I'm going to see if the if the promise is true. In other words, when I take X and I multiply it by four, and I take Y and I multiply it by two, and then I add the two results. That's easy. And this is pretty simple. Do I get 14? And it looks like I do. So that's a true equation. 14 equals 14. Let's check the other one. X minus Y is equal to negative four. So this one just says, hey, look, if I take the X and I subtract off the Y, do I get negative four? I know it might be a little bit tricky to think about, but remember whenever we subtract a larger integer from a smaller integer, they don't even have to be integers any numbers. Then we result in a negative and negative four equals negative four. 

So since both of these equations are made true by the fact that this, these are both true once you work out the expressions values, then that is in fact a system to a solution, a solution to the system. All right, by the way, it's the only solution that we don't know, but I know. But that's irrelevant. So now what we're going to do is we're going to take these two equations for X plus two Y equals 14, and X minus four X minus Y equals negative four. And we're going to play around with them a little bit. Letter B says find the sum of the two equations. So I'm going to write down the two equations. Four X plus two Y equals 14. And X minus Y equals negative four. So literally, I'm going to add these two together. Now remember combining like terms. I can combine X's, I can combine Y's. I can't combine X, Y's, right? So what do I have? I have four X's plus one X is 5 X. Here I have to be a little bit more careful. I have positive two Y plus a negative Y leaves me with just a positive one Y 14 plus negative four is equal to ten. All right, so that's the sum of the two equations. Now the question asks is the .15, the solution to this new equation. Is it a solution? I want you to pause the video right now and you do what we did in a to see if it's a solution. 

Well, I hope you found out that it was. Let's take a look. All right, to check to see if it's a solution, I'm going to put one in for X and I'm going to put 5 in for Y, right? 5 times one is 5. And 5 plus 5 is ten. Yeah. So absolutely. It is definitely still a solution, right? Yes, justify your yes no answer. Okay, let's take a look at what letter C asks us to do. It says multiply both sides of the second equation by two. I'm just going to switch colors really quick. So if I go up to here and I look at the second equation, it's raining here and so it's thundering in January and Sony. So I take my second equation X minus Y equals negative four and it says to multiply both sides of the equation by two to get an equivalent equation. I'm going to go back to blue. So I'm going to take this equation and remember that's just a property of equality, right? Property of equality is if I know that two things are equal. And I'm told that X minus Y equals negative four, then I can multiply both sides by the same quantity. I'm going to also simplify this a little bit by distributing, right? I'm going to use the distributive property of multiplication. And I'll also just multiply this out. So I'll get negative 8. It now asks, is the .15 a solution to this new equation justify your yes no response. 

So please pause the video again now and check to see if one 5 is a solution to that equation. All right, let's take a look again. I hope that you found out that it was because what I'm going to do is I'm going to take my X, put it in one, take my Y and put it in 5, right? Do the multiplication before we do the subtraction, right? Excuse me. And now two minus ten is negative 8. All right, great. So this is kind of cool. Now let's just think about this for a moment before we scrub the text and go on. We had a solution to a system of equations. And then we took those equations and we added them together, and it was still a solution. That was part B, right? Then we took the second equation, and we multiplied it by a constant on both sides, and we found that it was still a solution. And none of this should really be a surprise, right? What we're basically saying is if we've got a pair of values, X, Y, or X, Y values, and then again, I'm going to go to a new color. So if I've got this pair of X, Y values, one 5, and I know that it makes these two true, then if I use properties of equality, like adding the two things together or multiplying both sides of the equation by something, it's still remains true. Kind of makes sense, right? So as long as we're true to begin with, anything that we do with properties of quality will maintain that truth. 

It's a little deep in a certain sense. But then again, I'm feeling philosophical because it's thundering in January. Anyway, I'm going to scrub out the text in a moment. Why don't you copy down what you need to? Are you ready? Because here goes that problem. Or not. Here we go. All right, moving on to the next page. We'll be activate my pen. And here we go. Now, letter D, we're going to continue to do this manipulation. It says, take the equation you found in letter C, and add it to the first equation. What happens? How does this allow us to now solve for the variable X? If so, what did you find? Well, there's a lot here. And of course, we don't have the other part of the page on the screen right now. But it says take the equation you found in C uh oh, I'm going to go back to go back to blue, I think. Not that it really matters. So the equation we found in C was two X minus two Y equals negative 8. That's what we found. And it says add it to the first equation. Well, that's four X plus two Y equals 14. Now I want you to remember something. This .15 made this equation true. It definitely made this equation true because it was a solution to the system. And now it's asking us to add this together. Now look what happens when we add these two equations. We get two X and four X and that gives us 6 X and we have negative two Y and positive two Y but that gives us if you will zero Y I'm going to write it down for right now even though that looks like the word oi. And then

 we have negative 8 plus 14, which is 6. So let me just rewrite it over here. And gives us 6 X equals 6, because obviously zero Y is zero Y well, that's kind of cool, right? What happens is that the Y's got eliminated. And the reason why, of course, is because we had opposites, right? We had negative two Y and we have positive two Y and they eliminated when we added them together. But what's really cool now is we know this must be a true equation. Didn't we find that back in letter B? When we added to two true equations together, we got another equation that was true. But the cool thing now about this equation is that only has the X in it, so we can easily solve for it by dividing both sides by 6, and obviously 6 -6 divided by 6 is one. And we find that X must be one in order to make this equation true. But that means X must be one to make this equation true. And this equation true, and this one, and this one, all the way back. So now we can solve for Y we can take any of the equivalent equations along the way. Let me actually take this one. Four X plus two Y equals 14. This equation was made true by this value. So I'm just going to take one, substitute it into this equation. Four times one is four. So just a little easier equation solving, I hope. 

We'll subtract four from both sides. To Y is equal to ten. Let me write that up here. Two Y is equal to ten. Divide both sides by two. And Y equals 5. And look at that. And of course, the point is that we were able to get our solution. So that's really neat. And what this entire exercise drives at is a really kind of powerful notion. I'm going to scrub out the text in a second before we go through that idea. So why don't you copy down anything you need to? All right? And now it's gone. Okay. The key lesson idea here, and it really is key. Is that solutions to a system, right? So X, Y values, let's say, that make a system true. Right? We'll continue to make the system true. When we start to play around with it with properties of equality. So if I take one of the equations and I multiply both sides of it by a constant, that equation will still be true for the solution, right? If I add the two equations together, the solution will still make the resulting equation equal. And the really, really neat thing is we can kind of take that logic and say, well, fine. I'm going to then manipulate the equations in ways that are useful in order to eliminate a variable. All right? So we're going to see that idea again in the next exercise. Don't worry if you still don't quite get it. It's a little bit tricky. And we're even spending the next lesson on the method of elimination. So this isn't your only shot at understanding it. 

Okay, let's take a look at exercise number two. It says consider the system shown at the right. So let's just get comfortable with ourselves. Oh no, we flipped the page too quickly. Technology. All right. I've got the pen back. Anyway, so we've got the system shown at the right. All right, just let's just get comfortable with it. No big deal, just a couple of equations that relate X to Y and it says show that three comma negative one is a solution to the system. You should feel comfortable with how to do that right now. So pause the video, take as much time as you need to. By the way, as a suggestion, try to do as much of the math as you can using your head, right? The numbers I've kept specifically fairly small and fairly easy to manipulate. If you need to, use your calculator, especially if you just want to check your calculations, especially if something didn't turn out the way that you thought that it might. Anyhow, pause the video now and work through that check. All right, let's go through it. So in order for three comma negative one to be a solution to this system, as we've seen a few times, what it means is that Scott to make both of the equations true. So I'm going to go through this a little bit quicker than I did before, given that it's the second time in this lesson that we've done it. 

All right, now it's a little bit trickier. We've got four times three, which is going to be 12. I would highly suggest looking at this as negative three times negative one that way we can go ah, we've got positive three. And now we have 12 plus three is 15, 15 equals 15. That checks. Let's check three X plus two Y equals 7. All right, we're going to take our point. And substitute it in. Three times three is 9, two times negative one is negative two. I think I'll write it like that. 9 plus negative two is 7. Great. All right, so that point is a solution to this system, which means that we can start to play around with these equations. We can play around with them, and it'll continue to be a solution. Now, let her be read over that for a second. Pause the video and I want you to see if you can use what we've been talking about in the lesson to write an answer down for that problem, okay? Pause the video now. All right, so we're taking it up a little notch here, and this is what the common core is all about. Thinking about things in illustrating them. Letter B says the .3 comma negative one will be a solution to the system shown below. 

In other words, this new system, I'm claiming three comma negative one will be a solution to. How can we determine this without doing what we did in point a? Well, what we should be doing is actually looking at structure. And hopefully what you can see is that this equation 8 X -6 Y equals 30 can be produced simply from the first equation. By multiplying both sides by two. And what we've already learned is that if something is a solution to a system, it will continue to be a solution if you multiply both sides of the equations by the same constant, so that's what happened there. This was this equation multiplied by two. And this equation, we simply multiply. By three, right? So if I had that equation, and I multiply both sides of it by three, just using that property of equality, distribute the three. Right? So because I used properties of equality to get a new system, the solution to the old system will also be the solution to the new system. All right, but what's cool now is I also know that if I take these two equations and add them together, the resulting equation also has to be true. Okay. Based on the fact that three comma negative one is a solution. 

So what I'd like you to do now is pause the video, add those two equations together and see if you can finish the problem. All right, let's go through it. Well, do you see what happens? So we have a positive 8 X and a positive 9 X and when we add them together, we get a positive 17 X we have a negative 6 Y and a positive 6 Y when we add them together, we get zero Y again, I'm going to write that down even though it's not necessary. We have a positive 30 plus a positive 21 and we get a positive 51. So what happens is that we have 17 X equals 51 and what happened was we eliminated. We eliminated the Y so what happens when we add them together? We eliminate the Y and what's really cool now is that allows us to solve for X by dividing both sides. 51 divided by 17, it may not be obvious what that is, but that is three. Okay? Now, in order to figure out what the final Y value is, we can take that three, and we can substitute it into any equation along the way, because three comma negative one is a solution to these equations in these systems, it's a solution to all of them. So take whichever one you like. You know what I like? I like, I like this equation. Everything's positive. I don't have to deal with subtraction at all. So pick your battles, right? Here we'll have 9 plus two Y equals 7. Subtract 9 from both sides. Be careful. 7 -9 will be negative two. 

It'd be tempting to put positive two. We all made that mistake. And we got our Y equals negative one. Right. Now we knew that at the beginning, I get it. So the amazing thing is, one more time. If we have a solution to a system, it continues to be a solution no matter how we manipulate the equations as long as we're using the properties of equality, adding two true equations together, multiplying both sides of an equation by a by a constant. All right? So I'm going to scrub out the text, get ready. All right? Here it goes. All right. So let's go on and do the last problem. Now, the idea in the last two is that we knew what the solution was, and we just wanted to verify that as we manipulated the equations using the properties of equality. We continued to get the same solution. What that now gives us permission to do is take a system such as this, and manipulate it in order to try to eliminate one of the two variables. Now, whenever we eliminate, we eliminate because we add two variables the same variable that have opposite coefficients of 5 Y and a negative 5 Y a positive two X and a negative two X so the first thing I always do when I use the system of elimination is I look at my two equations. Two X plus four Y equals two and 6 X plus three Y equals negative three and I just ask, well, are there any opposites? But two X and 6 X aren't opposite, four Y and three Y aren't opposite. 

So that's what I'm going to bring in the property of multiplication, right? The multiplication property of equality, which gives me permission to multiply. Both sides of the equation by anything I want and it will maintain the integrity of the system. The solution set will still be the solution set. So take a look at take a look at equation number one. I'm going to just try to give you a push in the right direction. What could you multiply equation one by? On both sides. That would produce a number that could cancel positive 6 X, pause the video and think about that a bit. Did you realize you could multiply both sides of the equation by negative three? That's our big one. That's what we want. Okay? So what I'm going to do is I'm going to take that first equation. And I'm going to multiply both sides of it by negative three. And again, because I'm using a property of equality, and I'll do this sort of as I talk, not always easy. I know that this equation now will have the same solutions as this equation. But the beautiful thing now here, I'm going to write down equation two again. The beautiful thing is I now have a system of equations, right? That when I add them together, the X variable gets eliminated, right? Negative 6 X plus positive 6 X is zero X not really. Negative 12 Y plus positive three Y is negative 9 Y all right? Negative 6 plus negative three be very careful is negative 9. 

So now that I've eliminated X Y becomes very easy to solve for. So I'll divide both sides by negative 9 and I'll get Y equals one. So Y equals one must be part of the solution to the system. Now the question is how do I get X? Well, I can take Y equals one and I can substitute it into either one of the two equations. Might as well substitute it into the first one. Nothing problematic about that. I'll do four times one, which is four, just watch your order of operations. Subtract four from both sides. Watch your math two minus four is negative two. Divide both sides by two. Watch out here, negative divided by a positive is a negative. And I get negative one. So that's my solution. Negative one positive one. By the way, really, really easy to check, right? We've already checked something a solution to a system many, many times. We could just go up, take two X plus four Y equals two. Substitute in the negative one. And the positive one, right? This will give me negative two plus four. I'm going to do the check pretty quickly. All right, that works. Now I can also substitute it into 6 X plus three Y equals negative three. 6 times negative one. Plus three times one. Equals negative three. Gives me negative 6 plus three equals negative three. 

All right. Wow. Sorry about that. You've seen the checks quite a few times in this lesson already, but I thought, I thought we would do due diligence. We should always check to make sure the solution to any algebra problem is correct. If we can. Sometimes you can't, but this case we could. I'm going to scrub out the text in a minute and we're going to finish up the lesson, so copy down anything you need to right now. All right, here it goes. All right. So we're going to work more with the method of elimination in the next lesson. All right, but you're going to play around with it also on the homework, hopefully. Thank you again for joining me for another E math instruction common core algebra one lesson. Click on the video's description if you need to get the homework or if you need to get the lesson itself. This has been a presentation of E math instruction. I'm Kirk Weiler, until next time, keep thinking and keep solving problems.