Solving Systems By Substitution Common Core Algebra I Homework Answers

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Hello and welcome to another common core algebra one lesson. I'm Kirk weiler, and today we're going to be doing unit number 5 lesson number two, solving systems by substitution. Before we begin, let me remind you that you can find the worksheet and a homework assignment that go with this video by clicking on its 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 upper right hand portion of any worksheet. These will allow you to use your smartphone or tablet to scan the code and come right to this video. All right, let's begin. In the last lesson, we saw how to solve a system of equations by graphing both equations and finding their intersection points, because at those intersection points of course, both equations we're going to be true, and thus satisfy the definition of a system, which is that both equations are true at the same time. Today, we're going to be looking at solving systems by doing something called substitution. And the first exercise walks us through that process, and why it works. So let's take a look. In the first problem, we have this system, two X plus Y equals 9, and Y equals X minus three. Let array asks us to show that four comma one is a solution to the system. So you've seen this many times. I'd like you to pause the video and work through part a. All right, let's do it. So what we have to do to always show that something is a solution to a system is show that the point makes both equations true. So we're going to take that four, put it in for X, put one in for Y we'll get two times four is 8, 8 plus one is 9, 9 is equal to 9. Check. Then we've got Y equals X minus three. This is going to be even easier. We'll put one in for Y, we'll put four in for X, four minus three is one. So yeah. True true gives me yes. All right. Now, let her be says substitute X minus three. X minus three is an expression. That's just an expression. Substitute X minus three in for Y in the first equation and show that the .4 comma one is still a solution to this new equation. In other words, I'm going to take this and I'm going to substitute it into this equation. So. And actually, let me put it in parentheses. Now what I want to do is show that this thing is still a solution to this new equation. Now that may seem weird primarily because Y isn't here anymore. Right But hey, whatever, that just means I don't put the one in anywhere. Anyway, two times four is 8. Four minus three is one. And 9 is equal to 9. So yeah, it is. Now, why is that a good thing Well, it's a good thing because it means any time we ever substitute one equation into another, it still means that those two equations have the same solutions. Okay This is what's known as the principle of substitution equals can be substituted for equals. And of course, the big advantage of substitution is as soon as I put that X minus three in for Y, there is no Y left, right There's no Y no Y left. And that allows me to solve for X, right I can use the associative property. Of real numbers to decide to add these two. That'll give me three X minus three is equal to 9. Using the addition property of equality, I can add three to both sides, giving me three X is equal to 12. And dividing both sides by three. Oh, I'm getting low down here. I get X equals four. Now, that's not the final answer. The final answer is four comma something. And of course we know the commas one to figure out the Y coordinate we can take four and we can substitute it then into either this equation or this equation whichever is easiest, I think this is the easier equation Y equals X minus three, so all I have to do is put the four in. And I'll find that Y is equal to one. And that means the .4 comma one solves this system. It's really quite nice, right And it works on a very simple principle. The principle of substitution of equals. Equals can be substituted for equals, very important, in fact, let me clear out this text, so pause the video if you need to. All right, here they go. Ready Wait for it. Equals maybe substituted for equal so important I had to light it on fire. And somehow then it appeared. I don't know. Anyway, as far as special effects go, it's kind of snappy. The point is, any time two things are equal to each other, they can replace one another. And we're going to use that to our advantage when we solve systems using substitution. So anyway, let's get some practice on this. Exercise two, solve the following systems of equations by substitution. All right, this is cool. So letter 8 is very, very easy. Notice the equality here. That means that two X plus 5 can be substituted for Y so I'm going to take the expression two X plus 5 and I'm going to substitute it down into this equation. For that Y sometimes, many times, in fact, teachers will call this setting equations equal. Now, I like that phrase and I use it a lot myself, but never, never look past the fact that what you're doing still is substitution. Now, once you have that, what's really great is you can just solve this equation the way you always do. I'll use the addition property of equality to add three X to both sides to X plus three X is 5 X all right, likewise, I'll use the addition property to add a negative 5 to both sides. Negative ten and negative 5 is negative 15, and then of course I use the multiplication property of equality to divide both sides by 5, giving me X equals negative three. I still need to get my Y value though. I can pick either two of these equations. It doesn't matter which one. But I'll go with the first one because it's just as easy as the second. And I'm going to put in negative three for X, right Now I just have to evaluate two times negative three is negative 6. Negative 6 plus 5 is negative one. So Y equals negative one. It's completely okay to leave your solution that way, or you could write it as coordinate pair, just be careful to make sure to get the order right. Negative three negative one. Now, letter B is actually a bit more challenging because the first equation isn't rearranged into Y equals four. But that's okay. The second one is, right So we can take this and we can substitute it in for that Y now I should be very careful, though. All right That Y is being multiplied by negative two. So I have to remember to put parentheses around the negative 5 X plus 13. Or I won't remember to multiply it all by negative two. And in fact, the first thing I want to do is use the distributive property and distribute that multiplication by negative two, so negative two times negative 5 X is positive ten X and then negative two times positive 13 is negative 26. I can now use the associative property, which tells me I can add in any order I want. So simple enough. The additive property of equality allows me to do something like that. 14 X is equal to what would that be 42. And now I can divide both sides by 14. And although it may not be obvious, X is equal to three. Now I've got to figure out which one it substituted into. In this case, I think it is much, much easier to substitute it into this. So I'll get Y is equal to negative 5 times three plus 13, so Y is negative 15 plus 13. Which is negative two. So again, a nice way of writing this would be three comma negative two. That's it. Equals can be substituted for equals. And that's a simple enough scenario. All right So I'm going to clear out the screen, write down what you need to. All right, here we go. Let's keep going. This is a fun kind of problem next. Max and his father Kirk are comparing their ages. They know that the sum of their ages is 52 and Kirk is 7 years older than four times max's age. If max's age is represented by M and Kirk's age is represented by K, write a system of equations that describes the two relationships from the problem. All right, the two relationships. Well, here's one of the relationships. Some of their ages is 52. That's going to be easy, right Max's age. Plus Kirk's age must be 52. No problem. But here's the other relationship. Kirk is 7 years older than four times max's age. So Kirk is 7 years older than four times max's age, while four times max's age looks like this, and then 7 years older looks like that. So this is a good way to model this problem using equations, right Or modeling the problem using equations. Now we can solve the system to find both their ages, right By doing substitution, we can in fact take the four M plus 7 and substitute it in right there. So M plus four M plus 7 equals 52. Do a little associative property and add those M terms together. Additive property of equality gets rid of the 7 on that side. And multiplicative property allows us to get max as H and he's 9. 9 going on 14. All right, so masses age is 9, right And Kirk's age is four times 9 plus 7, right Because I can just take this and plug it in here. I can't believe I just drew that era that well. So Kirk is 43. All right, max is 9 Kirk's 43. We're able to solve this almost like riddle like problem by doing it this way. It's kind of cool, isn't it All right. I'm going to clear out this text. Here we go. Keep going, keep going. All right, two cell phone plans offer different text packages. Thank you probably seen something like this before, not just this problem, but just text packages. The two planets are outlined below. Plan a charges $5 per month, along with a charge of three cents per text. Plan B, there is no monthly charge, but the texts are a lot more expensive, ten cents per text. Is there a certain number of texts when the two plans cost the same amount Determine your answer by setting up a system of equations that model the two plans. So really, at the end of the day, this is a yes, no question, right You know, is there a certain amount of tax 20 text, 52 texts, whatever, where the two plans will be the same. So let's do it. Let's set up equations. Shall we let's model it Why don't we say C for cost All right. Well, plan a is going to cost $5 right up front plus .03 times the number of texts that we that we make, right So if I take three cents and I multiply it by the number of texts and add 5, that's plan a. Plan B, on the other hand, the cost will simply be 0.10 times the number of texts. Okay. So to figure out where the two things are the same, I will literally solve the system. I will take this and I'll plug it in here or the other way around. 5 plus .03 T equals 0.10 T use the additive property of equality and subtract a .03 T from both sides. We'll get 5 is .07 T no problem there. I don't think right. Divide both sides by 7. And the number of texts because I don't really want to do this in my head. Is 71.428. All right, so what's the answer I'd like you to pause the video and think about what the answer is. All right, let's do it. The answer is no. That's the answer. Right Is there a certain number of texts from the two plants cost the same And the answer is no. Why Because there are no fractional texts. Right In other words, this answer is in a certain sense non viable. Nonviable. A viable answer is one that works. One that is realistic, one that we can use. But that's a non viable answer. What it's saying is that at 71.428 texts, these two plans are going to cost the same. But that doesn't make any sense. You don't make 71.428 texts. What would that mean You'd made 71 text and then you'd gotten part of the way through one before you bailed out on it. You know, there are some people I definitely changed my mind whether I'm going to text them in the middle of the text, but it doesn't. I don't get charged for a portion of text, right So the quite frankly, the answer here is no. But the algebra allows us to figure that out. All right I'm going to clear this out. And then we're going to do one more problem. Here we go. All right, it's gone. Okay. A man and a woman start 380 feet away from each other and walk in a straight line towards each other. If she is walking at a rate of 6 feet per second, and he's walking at a rate of two feet per second, when will they meet All right. Well, there are lots and lots of different ways of doing this problem. Okay I'd like you to pause the video and see if you can come up with one method. I don't want you to worry about whether you do it using something that is a system of equations. That's the way I'm going to do it eventually. But I just want you to figure out when they're going to meet. Okay All right, let's do it. So I don't know how you did it. But if you end up getting the same answer, I got that I'm sure you did it in a good way. What I'm going to do is I'm going to actually set up a number line. I'm going to say that she's at zero. And he is at 380. All right And what I'm going to do is I'm going to keep track of their position, I'll call it P right. So the woman starts at zero, right That's her Y intercept. So to speak. And she's walking at 6 feet per second. I'll use T to stand for time. So that's really her position. The man, on the other hand, his position, he starts at 380, but since his position is decreasing right, he's getting closer to zero. I then subtract two T all right, those are their two positions. When they meet the two positions must be equal. So I'm going to use substitution and I'm going to substitute this into here. I'll get 380 minus two T is equal to 60. I can use the additive property of equality and attitude to both sides. 380 equals 8 C using the multiplicative property of equality I could divide both sides by 8. And I'll find that they meet after 47.5. Seconds. So if you've got 47.5, whether you did it this way or not, you did a nice job. Okay Believe it or not, we actually did a problem almost identical to this on the first lesson that we taught in this curriculum. I think that we had a man and a daughter walking towards each other and I think they were 300 feet away. And we did it in a way that was a little bit different than this. All right So let me clear this screen up and get rid of whistling man. Here we go. All right, let's finish up. Bye bye whistling there. Okay, so today we saw yet another way of solving a system of equations. The first way we saw was by graphing the two equations and finding out where they intersect. The second way is now by substitution. This is also another method that you were expected to learn in common core 8th grade math. But if you, if you forgot how to do it, or if you never learned how to do it, hopefully this lesson got you there. All right. For now, let me 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. And keep solving problems. 153554b96e

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