Common Core Algebra I.Unit 2.Lesson 2.Seeing Structure to Solve Equations.by eMathInstruction

Hello and welcome to another E math instruction common core algebra one lesson. I'm Kirk weiler, and today we're going to be looking at unit number two lesson number two. Seeing structure in order to solve equations. We're adding something new to our worksheets right now. We're adding a QR code, which can be found under the date field. Teachers and students are going to really like the QR code. Because what it allows you to do is to take a little app that you can get on your cell phones typically for free. Scan that code and it'll take you right to this video, right? Now, as always, the worksheet and homework for this video can be found by clicking into the description underneath the video. So feel free to do that, download the worksheet, and if you need to see the video again, all you have to do is scan that QR code and your smartphone will take you right here. All right, let's take a look at what we're going to do today, though. This is the first lesson where we're really taking a hard look at solving equations, right? Finding the values of variables that we don't know. And before we do this, I want to go through a little video where we take a look at something that happens to the number three. Let's say we take three and we multiply it by two and we add one to the result. Let's just keep track of three. Due to order of operations, right, we're going to get 6 first, and then we're going to add on the one. And of course, the result very simple. It's going to be set, right? All right. This is easy enough. So let's say you had a number that you multiplied by two, and you added one, two, and the result was set. You multiplied by two, you added one, and you got 7. How can you turn the 7 back into a three? Well, ultimately right, you would undo what you had done. So you'd subtract off the one, that would get you back to your 6, then you would divide by two, and that would get you back to your final result of three. Now you might say, all right, what does this have to do with solving equations? Well, ultimately, my belief is that the final step almost always in solving any algebraic equation. Is to undo what has been done to the variable in the opposite order in which it's been done. So if I've got a number that I've multiplied by two and added one two and I know the result is 7. I can just go backwards by subtracting one and dividing by two, right? So you got two times the variable X plus one equals 7, right? You get back to X equals three by subtracting one from both sides. Getting hit two X equals 6, then dividing by two on both sides, and obviously getting you X equals three. So I want you to keep this in mind throughout the lesson today and throughout this entire course. At the end of the day, every time you solve an equation, your final step ultimately will be to undo what has been done to X so look for that structure. And let's start to get some practice. Exercise number one says, consider the equation 5 X plus 25 X plus three equals 23. Letter a asks us to list the operations uh oh, what happened there? I thought we had our pen activated. Nope, let's get it going. All right, there we go. It says list the operations that have been done to the variable X on the left hand side of the equation. So what's been done to X on that left hand side just in the expression? I don't want you to worry about the equation. Pause the video right now and make a list. There's two things that have been done to X what I want you to do is list them. In the order in which they happen, pause the video now. All right, well, due to order of operations, right? The first thing that happened to X was that we multiplied by 5, right? 5 times X and then we added three. All right? It's important to note. You can't think that you've added three first and then multiply the result by 5. That would be misinterpreting order of operations. And that's key because in B, when it says solve the equation by reversing what's been done to it, well we say we've got a number think about that that little video that you saw, right? With the three, we've got a number that was multiplied by 5, and we added three, and we got 23. What we're going to do is we're going to subtract three from both sides. All right? Of course three minus three is zero, so we just get 5 X and 23 minus three is 20. And now what we'll do is we'll divide both sides by 5. 5 divided by 5 is one, so that leaves us one X equals four. That's the solution to our equation. Of course, we can check that it's a solution to our equation by simply taking the four and substituting it back into our equation to see if it's true. Remember the last lesson that's really all we did. We just talked about what solution to an equation was. A value of the variable that makes the equation true, so 5 times four is 20. Plus three equals 23. Yeah, I think we can say that. So that's true, right? So, ultimately, what this boiled down to is identifying what had happened to X and reversing it. I'm going to scrub the screen in just a second, so take a look. Pause the video if you need to. Here we go, scrubbed away. Let's go on to the next bra. All right, an exercise number two, we've got more complicated equations, right? And what I want you to do is in each case, I want you to identify what happened to the variable on the left hand side of the equation. And then we're going to reverse what happened to the variable in the opposite order in which it happened. If you think you already know what's going on, pause the video now and try let it right. All right, let's go through letter a because some of you probably are like, wow, man, that looks horrible. But watch, if we look at the structure of the left hand side here, we just key in on this. Then what we notice is that the first thing that's happened is that we've subtracted three. We subtracted three from X now, the next thing that we happened, the next thing that we have. And the next thing that happened was we divided by two. Right. We divide by two. And the last thing that happened was that we added 7. Did I just say that that was three? We added. 7. So what we're going to do is we're going to solve the equation now by just reversing those in the opposite order, right? So a lot of students would look at this and they'd go, oh man, that equation just looks horrible to solve, but not us, right? We're just going to say, well, we added 7 as our last step, so it was our first step. We're going to subtract 7 from both sides. Of course, 7 -7 is zero. You add zero to anything you want as many times as you want. And 13 -7 is 6. So we took care of that. Now let's divide by two to undo division by two. We multiply both sides by two. Here we can just do what's called cross canceling infractions. And we'll just get X minus three is 12. So we got rid of that dividing by two by multiplying by two. And the last thing that we did was we subtracted three, so we're going to undo that. Bypassing three to both sides. Some teachers, of course, will show that adding three underneath, that's quite all right, you know, I go both ways. But we get X equals 15. Right? So you look at this equation, and it looks absolutely horrible, but when you break it down when you look at the structure on the left hand side, which is completely dictated by order of operations, completely dictated, then all you have to do is understand in your head. What happened to X? And then how do you reverse it? So pause the video and try letter beyond your own. All right, let's go through it. Well, what happened to X? We look up at this equation. We've got these parentheses here. That means that we have to do what's in them first. So we added one, right? Now once we added one, then we have a choice that we subtract two, then multiply by four, or do we multiply by four and then subtract two? Well, order of operations is pretty clear here. That multiplication multiply by two oh, sorry, multiplied by four comes next. Our last step was the subtraction by two. All right, so now what we're going to do is we're going to undo each one of these steps by just reversing what what's been done to it. The first thing that we do is we reverse subtraction by two. All right, so we get four times X plus one. Minus two, and then I'll undo that subtraction by two by adding by two. Here we can really look at this as negative two plus positive two, which again ends up just canceling and giving us zero. So we get four times X plus one is equal to negative four. Let's undo that multiplication by four, and we'll just divide both sides by four. Four divided by four is one. So we'll just have X plus one. Negative four divided by positive four will be negative one. Think about that a little bit if you struggle with negative numbers. And then of course, we'll undo there we go. I just did it in the below setting. We'll undo that addition by one by subtracting. Watch out here. Real easy to think that negative one minus one would be zero. But negative one minus one would be negative two. All right? And again, I encourage you to go back and check each one of these by substituting them into the original equation. And verifying that it's true. All right, I'm going to scrub the text on the screen in a moment, so get ready. Pause the video. All right, here we go. Let's go on to the next problem. Now, sometimes choice is not a great thing in terms of ways that you do a problem. A lot of times students want like one and only one method to carry out an algebra problem. And math isn't like that, math opens up a lot of different paths for us. So I'd like to solve one problem in two different ways. I'd like to do it the way that we've been doing it and then also do it a different way to just show you that there's more than one way to get through a problem typically. All right, so let's take a look at the equation negative two times X minus four plus 8 equals two. We're going to solve this equation two different ways. In letter a, I'd like to do what we've been doing. I'd like to reverse what's been going on in terms of order of operations. And it let her be we're going to use the distributive property. So what I'd like you to do is pause the video now and letter a I'd like you to identify what's been done to X and then reverse it to solve for X pause the video now. All right, let's go through that. Again, this is very similar to the last problem when we look at what's been done to X, we see that we've subtracted four. Now, I would imagine that many of you will eventually not need to write these steps out that I multiplied by negative two. And then finally, I added 8. So when I solved the equation, let me do that down here. I'm going to undo that process, right? First things first we'll undo adding the 8 by subtracting 8 from both sides. That's simple, right? Watch out over here. But on this side, of course, every time we do this we get kind of a predictable answer 8 -8 is zero. Watch out here two -8 is negative 6. Let's undo multiplying by negative two by dividing both sides by negative two. All right, again, be careful. A negative divided by a negative remember is always a positive and 6 divided by two is three. And of course, we can undo subtracting by four by adding four on both sides. Again, notice all we do is identify what's been done to X and the order in which it's been done, and then we reverse that process. It's really, really a nice way to look at it. But let's take a look at letter B all right. You did a lot of equation solving in 8th grade math, especially if you were in the common core program. And probably many teachers would have said, hey, look, if you're solving an equation like this, what you're going to do first and foremost is you're going to look to distribute, use the distributive property to get rid of that parentheses. So pause the video now if you think you know how to do it that way, right? And then we'll take it from there. All right, let's go through it. So if I use the distributive property, right? And that's just all I'm doing when I do that. I'm not solving the equation on just writing an equivalent expression I would get negative two times X, which is, well, negative two times X, then I would have negative two times, if you will, negative four, and that will give me positive 8. That's a little bit tricky. Then I just have this other plus 8, right? I can now use the associative property of addition to decide, right? That's all the associated property just says, hey look, if I got a plus B plus C, I can add these two. Together, right? Now of course I might say, well, I'll just subtract 16 from both sides. Be careful here. Two -16 would be a negative 14. And then maybe I'll divide by negative two on both sides. Remember a negative divided by negative is a positive. And look at that. I love it when I do math in two different ways and I get the same answer. You know, in part a, what we do is we identify what's been done to the variable, and we undo it in the order in which it's been done, and that is always my preference. But a lot of times teachers say, hey, look, if there's in multiplication left undone, like negative two times X minus four, they can easily be done using the distributive property. Why not do it? Ultimately speaking, by the time we get down to this step, we're still doing what we're doing over here anyway, right? We're undoing the addition of 16, then we're undoing the multiplication by negative two. So we still use this idea of reversing what's been done to X, we just do it a little bit later on in the problem. All right, so I'm going to scrub out the text, pause the video now if you need to. All right, here we go. Nothing like the clear sheet. Let's move on. Okay, the last thing that we're going to work on today is oftentimes not a student's favorite, but we're going to work on trying to translate verbal phrases into equations and then solving those equations. Now, we had a lot of work in the first unit on translating verbal phrases and math. Hopefully this is where it starts to pay off. So let's take a look at letter a. Why don't you read that to yourself for a moment? And then I'm going to read it out loud. All right. Ten less than 5 times the number results in 35. Ten less than 5 times the number results in 35. What is the number? That's like a riddle. I love these. I really do. But you have to be so careful, right? Let's take a look at the structure of the sentence. Ten less than right, 5 times a number. If I just had like ten less than 35, that would be 25, right? So ten less than means on subtracting, right? I'm making a making a number smaller. What number am I making smaller? While I'm making 5 times the number smaller, right? Results in 35. So let's take 5 times a number, let's make it ten smaller, and then let's make that result. In the number 35. Right? Think about that for a second. I encourage you to oftentimes think about the multiplication first. Think about like, are we multiplying the number? Or are we multiplying some kind of sum or difference? Here, we're taking a number, taking 5 times that number, finding ten less than an getting 35. Now, of course, what's the number? Well, why don't you go ahead and pause the video and solve that pretty easy equation by identifying what's been done to the number, and then reversing it. All right, let's go through it. Well, if we look at the structure of the left hand side, what we've done is we've taken a number, multiplied it by 5, and then decreased it by ten. We're going to undo that. I'm not going to even list the steps anymore. I'm going to undo that by adding ten on both sides. Right? Put it up here. That's going to leave me with 45 on the right hand side and then I'll undo multiplying by 5. By dividing by 5. And I get any side. All right. Notice it says check your answer for reasonableness. In other words, if I said, hey, take 9. You know, find 5 times it and then decrease it by ten, right? You should be able to say, hello, you know, okay. I can take 9 times 5, right? 5 times the number and I can get 45. And then if I find ten less than that, I get 35, right? And that's what I was supposed to get. So that's what I mean by check your answer for reasonableness, right? You often can't do that in these problems. Let's take a look at B it's a little bit trickier. Read over the text, and then I'll read it out loud to you. This one's a little bit harder. When three times the sum of a number, oh boy, is that an important phrase? The sum of a number and 7. When three times the sum of a number and 7 is increased by ten, the result is four. Let's deal with this. The sum of a number and 7. Right? And what are we doing to the sum of the number and 7? We've got three times I really need those parentheses, right? Because if I don't have the parentheses there, I'm only going to be multiplying the number N, right? Then we increase it by ten, so here it gives me this. This gives me this. The result is four. Wow, look at that. Right? The trickiest part here are these parentheses. Students will often often leave them out and they get the wrong answer. It's not even a reasonable answer. Anyway, pause the video now, identify what's been done to N, undo it, and get an answer. All right, let's go through it. So what's been done to it? Well, we added 7 to it. We multiplied by three, and we added ten. So let's undo that. Let's subtract ten from both sides. Be careful. Right? We have a four minus ten. Don't tell me that 6. I got four apples. I'm going to take away ten apples. I have all right, I have confusion, negative numbers are weird. But I owe some 6 apples, right? Anyway, I have negative 6. Let's undo the fact that we multiplied by three. Let's divide by three on both sides. It's going to leave us with our nice N plus 7, watch out here, a negative divided by a positive is a negative, 6 divided by three is two. So we get negative two. Let me take that. Take the mass up here. And then we'll subtract 7 from both sides. Again, be careful. Negative two -7 is going to be negative 9. What I would encourage you to do on all of the arithmetic is try your best to do it in your head. Really, try your best, and then check each step along the way with a calculator. I know that especially things with negatives and positives can be tricky. I know fractions can be tricky. It's really important that you be numerate, that you have the ability to work with numbers in your head independent of the black box. We'll talk about that more as the year goes on. But we get N equals negative 9, and look at that. I got a positive 9 I have a negative 9. Anyway, I'm going to scrub out the text. So take a look if you need to. All right, here it goes. Okay. Well, I want to thank you for joining me for another E math instruction common core algebra one lesson. Remember, you can get the worksheet and the homework for this lesson by simply clicking on the video's description. As well, try out that QR code and tell us what you think. You know, we think it's going to be really, really helpful in the future for students to be able to scan that code and for teachers to be able to scan it. Of course I'm a little bit nervous what happens when you have photocopy it with a bad copy machine and something a little squares get all mixed up. I got to think it won't work then, but I hope that you've got good photocopiers. Anyhow. For now, thank you again for joining me until next time, I'm Kirk weiler. Keep thinking. And keep solving problems.