Common Core Algebra I.Unit 5.Lesson 5.Modeling with Systems of Equations
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Learning the Common Core Algebra I.Unit 5.Lesson 5.Modeling with Systems of Equations by emathInstructions
Hello and welcome to another common core algebra one lesson by eMac construction. My name is Kirk Weiler, and today we're going to be doing unit 5 lesson 5 modeling with systems of equations. Before we begin, let me remind you that you can find a worksheet and a homework assignment that go with this video by clicking on the video's description or by visiting our website at WWW dot E math instruction dot com. Also, don't forget about the QR codes at the top of every worksheet.
These codes will allow you to scan with your smartphone or your tablet and bring you right to these videos. All right, let's begin. For the last few lessons, we've been talking about solving systems of equations by various methods, graphically, substitution, the method of elimination. But today, we're going to model real world scenarios with systems of linear equations and unsolved them. So let's jump into the first exercise. This is a classic. Jonathan has 9 bills in his wallet, okay? And it's really important to understand this 9 bills. In other words, if he took those things out, even if you didn't know what they were and you just counted them, there would be 9 of them. They are all. Either $5 bills or $10 bills. Okay? So he's got all 5s and tens in his wallet.
Let ray asks us to fill out the table to see how the dependence of the two variables are. And how then to determine how much money Jonathan has. All right? So for instance, if he's got zero 5s, then the number of tens he has must be 9, right? Because he's got 9 bills and the number of 5s and the number of tens has to add up to be that. Now, of course, the amount of money he has is 9 times ten or $90. Simple enough, let me do one more, and then you do the rest. If he has one $5 bill, he must have 8 $10 bills, which of course means that he has one times 5 plus 8 times ten, which will be $85. Okay? What I'd like you to do is pause the video now and really quickly fill out the rest of that table.
All right, let's go through it. Well, if he's got two 5s, then he must have 7 tens, and he must overall have two times 5. Plus 7 times ten, which will be ten plus 70 and he'll have $80. If he's got three 5s, then obviously he's got 6 tens. Which means he's got three times 5. Plus 6 times ten, which will be 15 plus 60, which will be $75. It's not a coincidence, obviously, that every time that we replace a $10 bill by a $5 bill, this goes down by 5. It's not particularly important to the problem, but you might notice the pattern. And again, it's not a coincidence. We're trading tens for 5, so we get 5 less dollars each time. Let's take a look at letter B if F right here, we're going to introduce a variable. If F represents the number of $5 bills. And T represents the number of $10 bills, then what does the following expression calculate? So what does that expression give you as an output? And then explain. I'd like you to pause the video for a minute and think about that, but keep in mind again that F represents the number of $5 bills and T represents the number of $10 bills.
Okay? Pause the video now and try to explain or figure out what that expression calculates. All right, let's go through it. Well, if we take the number of $5 bills and we multiply it by 5, let me change colors here. Actually, let me erase all this circle in here. All right. Let me go red. If we take the number of $5 bills and we multiply it by 5, this is how much money he has in 5s. Not how many 5s, but literally how much money think about this for a second, right if he had, if F was two, that would mean he'd have two $5 bills, multiplied by 5, you have $10. Okay? On the other hand, this, right? If we take the number of tens and we multiply it by ten, this is how much he has intends. So when we add the two together, what we're getting is how much total money in dollars. Jonathan has. And if you think about it, that's exactly what we did in our calculation over here. We kept taking the number of 5s that we had. Let me pick a random one.
The number of 5s we had multiplying it by 5, then we kept taking the number of tens that we had and multiplying it by ten, adding them together to get the total amount of money. Okay? Now let us see we're finally going to set up and solve a system of equations. Letter C says if Jonathan has a total of $55, set up a system of equations involving F and T that could be used to determine how many of each bill he has solved the system. Remember that he has only 9 total bills. Now a system is two or more equations, but we're only going to be dealing with two-equation systems in this course. So what are the two equations? Well, one equation is pretty easy. If we take the number of $5 bills he has and we add to it the number of $10 bills he has. Let me make that look a little more like an F then we have to get 9. 9. Not 55. Remember those of the number of bills.
On the other hand, if we multiply the number of 5s he has by 5. And the number of tens he has by ten and set it equal to 55, right? That makes sense because we're just using that expression from part B and setting it equal to how much money he has. So this is our system, right? We have now modeled the scenario with a system of equations. We haven't solved the model. We haven't seen what the model produces, but we've modeled this scenario with two equations.
Now the question is, how do you solve it? And you've really got three choices. You can solve a graphically, which doesn't work too well in this problem, at least because we don't have graph paper. You could certainly solve it by substitution, but you'd have to probably solve one of the two equations for one of the variables in order to substitute it into the other. I'm going to actually go with the elimination method here, let me go red. I'm going to go elimination. In fact, most of the problems that we do today will be elimination. But substitution works as well. So what do we do at elimination? Well, I need to eliminate one of the two variables. So what I'm going to do is I'm going to take the first equation. And I'm going to use the multiplicative property of equality and multiply both sides by negative 5. And I'm thinking I'm going to write the result right underneath here. I'll have negative 5 F -5 T is equal to negative 45.
Now, remember, the reason I do that is because when I add in the elimination method, these two now will cancel. Ten T plus negative 5 T will be 5 T and 55 plus negative 45 will simply be ten. Simple now to solve dividing both sides by 5, we get T is equal to two. Now think about what that means. That means that we have two $10 bills because T stood for how many tens we had. We can now easily figure out how many 5s we have by either doing common sense or by substituting that to back into this original equation and getting F plus two is equal to 9. Again, I'm sure you can. Probably solve this without doing this, but we find that we have 7 5s. So we have two tens and 7, 5s. The beautiful thing is we can easily check to see if that's right. If we have two tens, then that gives us $20. If we have 7 5s, that gives us $35, add them together, and we get $55. So it's a good check. It's a good check. All right, there's a lot written down on the sheet. We still have a little bit of the problem to finish up.
But I'm going to have you pause the video now, copy down anything you need to, and then we're going to clear out the text, okay? All right, here we go. Continuing with this problem. Letter D says, let's say that we were told that Jonathan has 7 bills. So really all of this is irrelevant now. Now Jonathan has 7 bills and they're all 5s and 20s. And we're told that he has a total of a $120. It says set up and solve a system to help evaluate whether we could have been told true information. That seems a little bit weird. I mean, why would this be false? Well, let's do it. Let's let what did we say? Let's let F be equal to the number of 5. Dollar bills. Let's let Y be equal to the number of 20. Dollar bills. You could use, you could use T again, but I'm going with Y so the fact that he's got 7 bills allows us to model that piece of information by saying F plus Y is equal to is equal to 7.
Sorry, I think about this. So the number of 5 plus the number of 20s is equal to 7. That's this piece of information. Now let's get this piece of information down in an equation. If we take the number 5, then we multiply by 5. Take the number of 20s and we multiply by 20, add together that should give us the total amount of money, which is one 20. What I'd like you to do now is pause the video, use the method of elimination or substitution your call. To solve this system of equations. Pause the video now. All right. Now what I'd also like you to do before we solve the system is make a decision. Could we have been told true information? Think about that. If you haven't solved the system, obviously, and you're just waiting for me to do it.
That's fine. We'll go through it in a second. But if you have solved the system, determine whether we could have been told true information. All right, let's go through it. So method of elimination, I think I'm just going to stick with what I did before. I think I'll multiply this first equation both sides by negative 5, so a little multiplicative property of equality. So I'll get negative 5 X -5 Y equals negative 35. Cancel. I get 15 Y is equal to what is that. 85. Yep. Great. Divide both sides by 15. And we get Y is equal to 5.6 repeated. Now, we go on and we could figure out how many $5 bills we have, but we don't need to. All right? The final analysis is we could not. Have been told. True information. Because we can not have. A fractional. Number of 20s. All right, in other words, this is what we call a non viable. Solution. There's absolutely no way you can have 5.6 repeated, even if it weren't repeated. There's no way you can have 5.6 $20 bills. You can have 5 $20 bills. You can have 6, $20 bills. You can't have 5.6 of them.
All right? So it's kind of interesting in that way. Math can clarify for us. It can tell us whether or not information somebody gave to us is true or false. Okay? Even when it seems like it could be true. All right, we're going to do some more modeling. So pause the video now, write down anything you need to. All right, here we go. Let's do some more. All right, let's take a look at exercise number two. Samantha went to a concession stand and brought three bought three pretzels and four sodas. Let's take a look. Three pretzels and four sodas and paid a total of 1125. Raz I went to the same stand and bought 5 pretzels and two sodas and paid a total of 8 25. All right, so we're being given some information about a particular concession stand.
Letter a is kind of a cool question and said, could the pretzels have cost one 75 each and the sodas one 50 each? How can you evaluate based on the information given? I'd like you to pause the video. See if this could be true information, all right? Take a little bit of time. Remember, pay attention to both Samantha and Raza. Okay? All right, let's go through it. So you got to really watch because if you just look at Samantha, let's take a look. She bought three pretzels at a dollar 75, plus she bought four sodas at one 50. And if you add those two together, you get 1125, which is good, right? I mean, that's consistent. We know she spent 1125. On the other hand, though, if you take Raza, who had 5 pretzels and were thinking their one 75 each, and then two sodas at one 50 each, we find that we get too much, right? We actually find what do we have? 1175.
All right. But that's not equal to the 8 50 that we know he spent. That doesn't look like an 8. So the answer there is a big fat no. All right? So it's fine. Fair enough. So that's not the cost of a soda and the cost of a pretzel. Now, what this gets into is what's known as unit costs, unit costs are simply the cost for one item of something. So the unit cost for a gallon of milk might be $2 and 25 cents. Although I think that's kind of cheap milk. The unit cost for lollipop might be 50 cents. Okay? And that's what we're going to do in this problem. We're going to figure out how much single pretzel costs and how much a single soda costs. So in letter B it says letting X equal the unit cost of a pretzel, so how much one pretzel costs? Letting Y equal the unit cost of a soda, write a system of equations that models the information given. See if you can do that.
Pause the video for a moment and see if you're good enough at this point with algebraic expressions and equations to model the two scenarios that for Samantha and that for Raza, given those variables X and Y. All right. Let's go through it. Well, what do we know? We want to do exactly what we did over here. Samantha bought three pretzels, and if we multiply that by the price of a pretzel, which is X, we get how much she spent for pretzels. Add to it, right? Four times the cost of a single soda, right? That's how much she spent on sodas. And we should get how much she spent overall. So this equation models Samantha's piece of information. Raza has one that's almost identical, right? On the other hand, he brought 5 pretzels times the cost of a pretzel. Plus two sodas, times the cost of a soda, and he got or had to pay a 25. So that models this physical scenario.
Now, letter C asks us to solve the system of equations using the method of elimination. We've done that quite a number of times, so I'd like you to pause the video now and work through a solution. See if you can solve that for X and Y using the method of elimination. All right, let's go through it. Now, we could try to get rid of the X's in this situation. And if you did, good for you. The problem with getting rid of the X's is that we'd have to multiply both equations. On the other hand, to get rid of the Y's, if I can turn this particular term into negative four Y, then it would cancel this one. And I can do that pretty easily pretty easily. By using that multiplicative property of equality and multiplying both sides of this equation by negative two. So I'm going to just write that first equation back down again. I'm going to get three X plus four Y is equal to 1125. And then I'm going to get negative ten X minus four Y is equal to negative 16 50.
All right, so when I add together, watch your positives and negatives, these two are going to cancel. These two will give me negative 7 X when I add these together. I'll get negative 5 25. I just had to check my cheat sheet. Of course, dividing both the sides by negative 7 will give me an X value. Of .75. Now, you always want to step back at this point even before you solve for Y and just think, is that reasonable. In order to understand that, you have to say, well, what was X? X was the unit cost of a pretzel. In other words, a pretzel costs 75 cents. That makes sense, right? I mean, not so much that it makes sense that it's 75 cents. I've been to a lot of different ballparks, sometimes it's more expensive than that. Sometimes it's less expensive. But that makes sense if we had gotten a negative number, it wouldn't have made sense if we got an answer like $25 and 36 cents, that would have, that would have been not made any sense, either that or we would have gone to see a Yankees game. I don't know, something like that. Or even if we had gotten something like this, if we had gotten something like .83 9. That wouldn't have made any sense, right? You can't have a price of a pretzel like that. What would you give them exactly?
All right. Now we still need to figure out what Y is. So we can take that X value when we can substitute it into either one of the original equations I wouldn't substitute into the transformed one. But I think I'm going to plunk it right in here. We don't have a lot of room, so I'm going to try to squeeze it in here and we have three times .75 plus four times Y is equal to 1125. Three times .75 is two 25. Plus four Y is equal to 1125. I'm sorry about the space on the screen. But I'll subtract two 25 from both sides. I'll get four Y is equal to 9, divide both sides by four, and we'll get Y is equal to two 25. Again, something that I want you to pause and think about for a minute. Why is the cost of a soda? Again, it makes sense. It's not negative, it's not outlandishly large. It doesn't have extra decimals that it shouldn't. Two 25 for a soda seems reasonable. All right, let's finish this problem off with letter D if Leah went to the same concession stand and she bought two sodas, okay, two sodas, fair enough. How much? Sorry, how many? How many pretzels, which you need to buy so that she spent the same amount on both. Pause the video now. See if you can figure out what the answer to this problem is. Do it any way you want, but make sure to show your work. All right, let's do it.
Well, first off, let's see how much she spent on sodas. What she buys she bought two sodas. And she spent two 25 on them. Right? So she spent $4 and 50 cents on sodas. Right. Now, we want to spend the same amount on pretzels. So I'm going to let P be equal to the number of pretzels. Now what do I know? Well, I know that if I take the number of pretzels and I multiply by how much a pretzel costs, which is 75 cents, that's got to be equal to $4 and 50 cents. So I'm going to divide both sides by .75. And I'm going to find the number of pretzels. To be 6. All right. You could also just sort of work it from a guess and check scheme too. You could figure out that you spent $4 and 50 cents on sodas. And then just keep adding up the 75 cents until you get to the four 50. Luckily, it's still a viable solution, right? It kind of given that last problem had we worked through this and gotten the number of pretzels. It's like 5.1, and that wouldn't have made a lot of sense. Unless they're selling portions of the pretzel.
All right. I'm going to clear out the text. So pause the video. Take as much time as you need to to really think about this problem. Unit cost problems are very common when it comes to modeling using systems of equations. So this is quite an important one. All right, here it goes. Let's move on. Exercise number three a rectangle has a perimeter of 204 feet, very important that you know what perimeter means. And we'll come back to that in a bit. It's length is 6 feet longer than twice its width. So I want you to notice something right away. We're basing the length on the width. Okay? If L stands for the length of the rectangle and W stands for the width, right a system of equations that models the information given in this problem and solve it to find the length and the width of this rectangle. All right. Well, again, I'd like to turn you loose right now, pause the video, spend some time, try to solve this system. And try to model the problem and then solve the system. All right, let's go through it. So let's see if I can get a rectangle on it. Here's a rectangle.
Oh, how pretty is that? Okay, so I've got a rectangle. Let's say the length is 6 feet longer than twice the width. So the width is the shorter one and the length is the longer one. So we'll just label them like this. Let's use this first piece of information to write an equation. The perimeter is 204 feet. And of course, define parameter what we do is we add the two lengths to the two widths, and we get 204. And I think it would make sense then, obviously, to say that that's two L plus two W is equal to 204. Now, let's model the second piece of information with an equation. The length is 6 feet longer than twice the width. Well, here's twice the width and adding 6 to it makes it 6 feet longer. Okay. So here's our system. There it is. We've just modeled the information given with two equations, and now we want to solve them. We could certainly solve this by elimination, but have you noticed elimination works well when the two variables are on the same side of the equation, which is true with the first equation but not with the second. It's no problem we could do a little bit of rearranging. But this one is ideally is ideally suited for substitution. So what I'm going to do is I'm going to take my two W plus 6, and I'm going to substitute it into this equation right there. Now I'm going to be careful, though.
Because I'm multiplying the width by two, I have to make sure to get parentheses around that two W plus 6. But from then on out, I'm off to the races, right? I can now use the distributive property to rewrite this as an equivalent binomial for W plus 12 plus two W equals two O four. I can now use the commutative property of addition and swing that two W and that 12 around like that. I can now use the associative property and add those two together. Anytime I can remind you of the properties, why not? A little additive property allows me to add a negative 12 to both sides, gives me 6 W is equal to one 92. Divide both sides by 6. Just making sure, and W equals 32 feet. Great, the width is 32 feet. Seems reasonable, right? I mean, it's not negative. It's not bigger than the 204. So I think it sounds reasonable. We haven't figured out the length yet. I think I'll just try to sneak that in right here. So the length is twice the width plus 6. So since we know the width is now 32, I can just substitute that in. I'll find 64 plus 6 and the length. Is 70 feet. All right, check my answers. They still agree with what I had originally. So I think I'm right. All right, and again, the length is reasonable. One obviously easy thing to check here is the perimeter.
You can add two of the lengths to the width and see if it's two O four, and in fact is. So we know we're right. All right, pause the video now, write down anything you need to. All right, here we go. Let's finish up the lesson. So today, what we saw was real world scenarios that we modeled with two equations that were both linear. These equations then formed a system which we could solve either using the method of elimination or the method of substitution. If we had graph paper, or if we even chose to graph them on our calculator and use the intersect command, which we learned how to use a while ago, we could also solve them graphically. All right, you're going to get a lot of practice on this homework. It's a very, very important skill. For now though, I'd like to thank you for joining me for another common core algebra one method. Common core algebra one lesson by E math instruction. My name is Kirk weiler. And until next time, keep thinking. I keep soccer props.