Common Core Algebra I.Unit 2.Lesson 7.Solving Equations with Unknown Constants

Hello and welcome to another common core algebra one lesson by E math instruction. My name is Kirk weiler, and today we're going to do unit two lesson 7. Letters everywhere. They're one of my favorite titled lessons. Solving equations with unspecified constants. That's kind of a mouthful. I prefer to think of it as letters everywhere. Before we begin solving equations with letters everywhere, though, please note that you can find a copy of the worksheet used in this video, along with the homework set by clicking on the video's description, as well, don't forget about those QR codes that we have at the top of the worksheets. Those will allow you to take a smartphone or a tablet, scan the code and come right to this video. All right, let's get into it. There are many, many instances in science and engineering and other fields, where you're trying to solve an equation. But there are a lot of unknown constants. Now, there's a difference between a variable and an unknown constant. A variable is something you're solving for, a variable is something that you do not know, something that can vary. Constants are numbers that don't change, but they can be represented by letters. I know that that's kind of confusing. So let's jump into it. Exercise one says solve each of the following problems for the value of X in B, write your answer in terms of the unspecified constants a, B, and C okay. So let's jump into it. Now, I know you can solve number one letter a so pause the video just for a little bit and solve a. All right, that should have been pretty easy. It's a two step equation. I'm going to rewrite the problem right here, though. Start out today in red. All right? We're going to use the additive property of equality to add negative three to both sides. That's going to give us 5 X equals 30. Then we're going to use the multiplicative property to divide both sides by 5 or multiply by one 5th, and we're going to get X equals 6. No problem, right? Now let's do letter B now in B what we want to do is first and foremost, notice the similarity between these two problems, right? The big difference is that a takes the place of 5, B takes the place of three and C takes the place of 33. So what we want to do when we solve letter B is we want to draw parallels to what we did in a, we want to do exactly the same thing, but we also want to remember our limitations. So let's do exactly what we did. I'm going to rewrite the equation down because you know that's how I start at the last problem. The first thing I did was subtract three from both sides. Here, I'm going to subtract B now, here's a big, big difference. In the last problem, I was able to say that 33 minus three was 30. Here the absolute best I can do is say that C minus B is C minus B there's nothing more I can do with that. All right. Now, the next step was I divided both sides by 5. Here, I'm going to divide both sides by a okay, just like the 5s canceled the a's canceled, and my final answer is C minus B divided by a that's a great way to leave your answer, especially if you're working through a free response problem. If you're working on a multiple choice problem, there is a small chance that the division by a will be distributed, the distributive property applies not just to multiplication, but also to division. So it could also be that this is the final answer. I actually prefer this one. It's more compact, but you never know. All right? So that wasn't so bad. And really, the process that we went through on a and B is a good process whenever you're solving an equation with unknown constants in it. Make those constants into something that you know solve the problem that way, and then go back and work it with the unknown constants. Anyhow, why don't you pause the video? And write down what you need to, and then I'm going to scrub out the text. Okay, here I go. Let's move on. I think I might switch back to blue. It's my favorite color. At least in these presentations. I actually am kind of partial to purple, but it's not a choice here. All right, we're going to do exactly the same thing we did before, but now the equation is going to look a little bit more complicated. I don't want you to forget what we did a few lessons ago, though. When we solved equations that involved just one variable, just one variable by reversing the order of operations. So let's take a look at a remember what's been done to X as we've added 5, then we've divided by two, then we've subtracted 7. And we got three as a result. So what we're going to do is we're going to undo that. We can already predict the steps. The last thing we did was subtract 7, so the first thing that we're going to do is add 7. The previous thing that we had done was divided by two, so we'll multiply by two. And finally, since we added 5 the last thing that we'll do is subtract 5. So I'm going to rewrite the problem down just so that we have it sitting here. Easier to work with, let's add 7 to both sides. Okay. Three plus 7, we can do that. That's ten. Now what we need to do is undo that division by two, so we'll use the multiplicative property of equality to multiply both sides by two, which gives me X plus 5 equals 20. Finally, we'll employ the additive property of equality against subtract 5 from both sides. And we'll get X equals 15. All right, so the additive property of equality, the multiplicative property of equality of the additive property of equality. Again, looking at B look at how similar it is to a, right? Instead of a 5, we have an a instead of a two, we have a B instead of subtracting 7, we subtracted C instead of having a result of three, we have a result of D so let's solve for X now we're going to run into similar problems that we did last time. We just have to keep in mind what we're doing. So last time we added 7 as our first step this time, we're going to add C additive property of equality works, no matter what the numbers look like, the big differences, three plus 7 was ten, D plus C here. Is D plus C or C++ doesn't matter. Now, the next thing that we did was we multiplied by two on both sides. Here we're going to multiply by B on by both sides. Now this is a lot trickier because we have to remember we have to remember to have that be there. And quite frankly, it's most likely going to be written, not like that, but like this. With the B out front. Most often when we multiply a binomial by a monomial, the monomial gets set up front. And then the last thing that we do is subtract a from both sides. Now, again, be careful, right? This can be very visually confusing. That subtraction of a occurs to this entire quantity, right? This entire thing right here is a number. Really, if you want, it's this thing. So we need to subtract a from that. What we don't want to do is have that a end up in parentheses. Now again, this is a perfectly good answer. X equals B times the quantity D plus C minus a but you could see variations where that B has been distributed, so B times D plus B times C minus a and that would also be a perfectly good way to leave it. All right, either way, the real point is that we're mimicking exactly what we did when we knew the constants with these unknown constants. Letters everywhere. All right. Pause the video if you need to. I'm going to scrub out the text. Here it goes. Gone. Exercise three little multiple choice problem. Now, this one's interesting because we're mixing numbers we know with numbers we don't know, right? So when two times X minus H plus K equals 8, the solve for X in terms of H and K, its solution is which of the following. Show the outbreak manipulations you use to get your answer. Here's what I'd like you to do. At this point, some of you know exactly what's going on. If that's the case, or even if you're a little bit unsure of what's going on. I'd like you to pause the video and play around with this. See what you can do, remember, keep your eye on the ball, you're solving for X, okay? Pause the video now. Excellent. Let's go through it. So I'm going to rewrite the equation down here. We've got two times X minus H plus K is equal to 8. Before we even dive in here, let's think about order of operations. What did we do? We subtracted H that was the first thing, right? Then we multiplied by two. That's this piece. Then we added K that was this piece. So let's undo it in the opposite order. We're going to subtract K from both sides. Okay, that's going to give me two times X minus H equals and the best I can do here is 8 minus K, real important that you don't do 8 plus K or do K -8. It's 8 minus K but regardless we took care of that. Now we want to undo multiplication by two. So we're going to divide by two on both sides. That's going to give me X minus H equals 8 minus K divided by two. All right? The last thing that we're going to do is we're going to add H to both sides. Right? Use the additive property of equality. And we get 8 minus K divided by two plus H and then we. Don't see this answer. Oh, I see. Right? We're going to use the distributive property of division or multiplication. And we're going to divide the 8 by two, we're going to divide the K by two, and we're certainly not going to divide the H by two because it's not divided by two. So a divided by two is four. Obviously, K divided by two is K divided by two. And now we have that. I still don't see it. Wait a second. This one looks pretty similar. So does this one. Now that one's not right. I got to have a cade divided by two. It's got to be this one. I see. That's just the commutative property of addition. So we kind of swing all this addition and subtraction around. We eventually get H plus four minus K divided by two. Then we bubble into on our scan drawn sheet. Right kiddos. Scantrons. Easy to grade, not that much fun to do while you're taking a test. Okay, why don't you go ahead and pause the video if you need to to write this down? Then we'll go on to some more problems. Okay, here we go. Clear it out. Well, what do we have next? Nice applied problem. One of the things that you'll have to do on and off and common core algebra is deal with some basic geometry facts that you know, or that you're supposed to know at least. One of the easiest is the perimeter. Never forget that the perimeter of any object, any object is simply the distance around that object. With rectangles, which is what we're dealing with here, we simply add the width to the width to the length to the length, right? Because we have two widths and two lights, I guess. Anyhow, we have four sides. We have to add their lengths together. Let's see, it says for a rectangle, the perimeter can be found if the two dimensions the length L and the width W are known. If a rectangle has a length of 12 inches and a width of 5 inches, what's the value of its perimeter include units? Why don't you go ahead and do this really fast, pause the video, take like one minute and go ahead and do it. All right. Now, it's a good idea to actually write out the equation. Let's see. If the length is 12 inches and the width is 5 inches, right to the perimeter is the distance around. So what will I have? I'll have 5 inches. That's this. Plus another 5 inches. Plus 12 inches. Plus 12 inches. So that'll give me ten plus 24. I'm using the associative property to add here. And that'll give me 34 inches. All right. Don't forget the inches include units. I'm going to actually get rid of this right now. So that we clean out a little bit. All right. So letter B says, write a formula for the perimeter in terms of L and W well, keep in mind, right, that we should now be able to look at this. And what do we have? We had a W and a W and an L and an L so W plus W plus L plus L technically that's a good formula. But we're going to say that the perimeter is equal to twice W plus twice L there it is, that's it, nothing more than that. Now letter C says rearrange this formula so that it solves for the length L and then we need to determine the value of L when the perimeter is 20 and the width is four. So I want to solve for L. All right, so let's write down this equation. And think about what we have to do to solve for L now, let's see what happened to L well, it was multiplied by two, and then we added two times W to it. So to solve for L, the first thing I'm going to do is use the additive property of equality to undo a adding the two W now, the best I can do on that left hand side of the equations is say P minus two W nothing else I can do there. And then I'm going to get rid of the multiplying by two by dividing by two. All right, and eventually, I'm going to get L is equal to P minus two W divided by two. You can definitely distribute the division by two there, use the distributive property to divide by two, but I'm not going to. I'm going to just leave it like that. That's fine. And in fact, I can now do the second part of the problem, which says to figure out the value of L if I know that P is 20 and W is four. So I can now just substitute those into the equation, I'll get 20 -8 divided by two, which gives me 12 divided by two, which gives me 6. No units here because there weren't any given in letter C but the length must be 6 if the perimeter is 20 and the width is four. That's it. Kind of cool. And that's, by the way, that gets to the heart of why we do this. Very often in physics and engineering. There's some formula, and it's solved for something. Solve for something. But we want to solve it for a different quantity. So we rearrange it in terms of these unknown constants until we solve for what we want. All right, I'm going to clear out the text so pause the video if you need to. Here we go. Moving on. Okay. Now the last thing that we're going to tackle is actually a little bit tricky. Now you've got the background. So don't worry about it. Even though it's tricky, you've got the background you will understand this but try to stick with me. We're going to now solve some equations where X is on both sides of the equation. Up to this point, every variable we solved for, we were able to just reverse the order of operations to isolate it. This will be a little bit different. Okay? But we're going to kind of approach it the way we did in the last ones, which is we're going to take one where we know the constants, and then we're going to look at one where we don't. Okay? So we're going to solve 8 X plus one equals 5 X plus 22. The first thing that I'm going to do is I'm going to use the additive property of equality to add a negative one to both sides. That's going to give me 8 X is equal to 5 X plus 21. Then I'm going to use the additive property of equality to add a negative 5 X to both sides. Giving me three X is equal to 21. Then I'm going to use the multiplicative property of equality to divide both sides by three and I'll get X equals 7. Okay, so I want to do exactly the same thing in B but again, this is going to be a little bit trickier. Okay, I'm going to I'm going to try to mimic it. Step by step. First thing I did was I subtracted one from both sides, here I'm going to subtract B. Now, right before we combine these two terms, I can't really combine these two. The best I can do is put D minus B it's the best I can do. Now, in this step, we added a negative 5 X to both sides, here I'm going to add a negative CX or subtract CX from both sides. Now B careful. Okay, that leaves us with D minus B on this side, which is the same as being left with 21 on this side. But here, just for the time being, let me write this down. It's very dangerous. Students who look at algebra is simply being symbol manipulation, et cetera might look at AX minus CX and go, well, the X is subtract. So you just get a minus C nope. The X's didn't subtract when we had 8 X -5 X, they're not going to subtract this time. The question is, what can we really do? Before we said 8 X -5 X is three X AX minus CX, well, you know what I'm going to do. I'm going to factor in X out. Okay, and I'm going to get a minus C in here. If you really think about it, that's exactly what we had over here. We had 8 -5 and we got three times X here we have AX minus CX and we have a minus C times X I think the biggest difference is that on this side it was written more like this a minus C times X equals D minus B but yet we finish it off by doing exactly the same thing. We're going to divide by a minus C on both sides. And that's it. Isn't that kind of a cool result? D minus B divided by a minus C. Okay. So that's what you do if X shows up on both sides. And we're going to see that in the next couple of problems, so we'll get a little bit of practice on that, but that one's a little bit trickier. It is an important equation solving technique when you don't know the coefficients, which are the numbers that are multiplying X pause the video now if you need to, then I'm going to scrub out the text. Okay, here it goes. Let's get a little bit more practice on that. Okay. Exercise number 6, multiple choice problem. Which of the following solves the equation? I'm not going to even read it for you. It's too much of a tongue twister. All right? Notice, by the way, similar to the last problem, there's an Exxon both sides. Okay? What I'd like you to do is pause the video right now, take your best shot at figuring out which of the following four choices is correct. And then we'll walk through it. All right, let's do it. AX minus K is equal to three times X plus H all right. Well, I think if it were me, the first thing I would do is I would distribute that three. So I'll get three times X and I'll get three times H now keep your eye on the ball when I'm trying to do is solve for X generally speaking, I will take everything that doesn't have an X in it. And I will put it on one side very often the right side. So I'm going to use the additive property of equality to add K to both sides. Negative K positive K will cancel out, and I'll get three X plus three H plus K generally speaking I try to move everything with an X to the other side. So I'm going to subtract a three X from both sides. And then I'm going to get rid of this X because it looks funny. And write it again. Okay. Now again, here's where we have to be very careful. Very, very, very tempting to be to just think that you can get rid of the X's, but that would be kind of a problem. Because we're solving for X and anytime you're solving for X if X vanishes, you either have a weird situation or you've made a mistake. So let me just write this down. AX minus three X is equal to three H plus K all right, now we're going to employ that trick that we had last time, which isn't a trick. It's actually just the distributive property. And we're going to factor an X out of both of those. So we're running that distributive property in reverse, if you will. And now to get rid of multiplying by a minus three, we're going to divide by a minus three. So looks like choice one. All right. That's it. So I'm going to clear out the text, pause the video if you need to. All right, here we go. Wonderful. Let's wrap. So there's a lot of instances in engineering especially. But lots of other fields, economics, biology, medicine, chemistry, lots of situations. Where you have formulas with lots and lots of different variables, lots of different constants, and you want to solve for one particular variable or another. That's where the skills that came in today pay dividends. You'll work with this a lot more as you move through math, both this year and in subsequent years. But for now, I want to thank you for joining me for another common core algebra one lesson by eMac instruction. My name is Kirk weiler, and until next time, keep thinking. I keep solving problems.