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Welcome to Algebra I: Solving Real-World Problems Involving Systems of Linear Equations

34 minutes

Hey, guys. Welcome to Algebra 1. Today's lesson's going to focus on solving real-world problems involving systems of linear equation. All the practice that you've had solving systems of equations and real-world problems will come in handy in this lesson. You ready? Let's go.

(Describer) She uses a stylus on a screen.

(Describer) She uses a stylus on a screen.

(Describer) She uses a stylus on a screen.

Okay. Before we begin solving systems of equations that deal with real-world problems, let's just jump back and review real-world problems in general. If you look at this problem, I won't solve it, but I'll review the process to solve it, just to jog your memory. "Mateo's monthly cable subscription is $80. "As part of his service, "he can order newly released movies to view at home "at a cost of $6 per movie. "Mateo was charged $104 for his first month of service. How many movies did he order?" If you remember how we've solved this kind of equation, we would read through it, and then go back and highlight anything that sounded like math-- anything that helps us focus on the information we need to set up this equation and solve for the unknown. For example, "His cable subscription's $80." Sounds like math, it's a number. "As part of his service, "he can order newly released movies to view at home at a cost of $6 per movie." Let's get that. "Mateo was charged $104 for his first month of service." Then "how many movies did he order?" The question we're trying to answer. We'd get this information, we'd set up an equation. I'll switch to my pen. So $80, plus $6 per movie, if I let X represent the number of movies.

(Describer) She writes 6x.

(Describer) She writes 6x.

(Describer) She writes 6x.

And then first bill was $104. We'd set up that equation and then take the necessary steps to solve for X. At the end, we'd figure out how many movies Mateo ordered. If you need a memory jog on that, go look at that video and see how we worked out those problems. Now, to jog your memory on systems of equations. A system, in Algebra 1, we'd stick with 2 equations. In Algebra 2, we add more equations to a system. But for Algebra 1, you'd have two equations. There were three strategies we could take to solve a system of equations. We could use the elimination method, where we add equations together and try to cancel out a variable. For help with that, go back and check out that video on solving systems of equations using elimination. Another method was substitution, where we would take-- for example, in this system, it's set up well for substitution because I see my second equation is already solved for Y. So I'd insert--I'd substitute that 2X plus 10 back into the first equation. That would help me get through and figure out my solution. To see that worked out, watch that solving systems using the substitution method video; that would really help. The third way to solve systems of equations is graphically. We'd solve for this first equation, for Y, in this case-- put them both in the calculator, graph them, and look for the point of intersection to find the solution to our system. What we're going to do here is combine these two things that we know. We'll combine our knowledge of solving real-world problems, and of solving systems of equations, and we'll get through this lesson about solving real-world problems involving systems of equations, okay? Let's get started. Well attack this and see how we do it. "Tickets at the local movie theater "are $10 for adults and $8 for children. "Yesterday the manager "reported $1,360 in admission sales "for yesterday's 11 a.m. show. "If 150 tickets were purchased "for yesterday's 11 a.m. show, "how many were adult tickets? How many were tickets purchased for children?" I know is seems like a lot going on there, but we'll use our steps to solve it. We've read through it; now we'll go back and highlight everything that seems like it's math, anything that helps us solve the problem. Pull out your highlighter and I'll switch to mine up here, and let's get going. "Tickets at the local movie theater are $10 for adults--" seems like something I'll need to know-- "and $8 for children." More things I'll need to know. "Yesterday the manager reported $1,360 in admission sales--" Okay. Ooh, here we go, my highlight didn't want to stay there. "--for yesterday's 11:00 am show. "If 150 tickets were purchased "for yesterday's 11:00 a.m. show, "how many were adult tickets? How many were tickets purchased for children?" Okay, so now that we've highlighted the information, let's attack this and see how we'll set up this system. When you're solving real-world problems involving systems of equations, there will be more than one unknown you're solving for. If we go back and look at the information we highlighted, we see we're dealing with $10 for adults, $8 for children, $1,360 in admission sales-- so we're dealing with money, in one case. Then we see, we have "150 tickets were purchased. How many adult tickets, how many children's tickets?" So we're also dealing with tickets. We want to figure out how many adult tickets were sold and how many children's tickets were sold. To find that information, we need to involve the amount of sales, the money for the sales, and the number of tickets sold. That will help with this problem. Let's start getting some things down. That'll help make more sense to you. Since I'm trying to figure out how many adult tickets were sold and how many children's tickets were sold, those are my unknowns I'm solving. I'm going to come underneath my problem a little bit and start writing notes here. I'm going to let X represent the adult tickets, and let the Y represent the children's tickets. For one equation I want to involve the money, I want to involve the $10 for adults, the $8 for children and then the $1,360 in admission sales. That's going to get me one equation for my system. So $10 for adults. And I'm letting X represent the number of adult tickets. So 10X. $8 for children, and I'm letting Y represent the number of children's tickets. So plus 8Y equals... Then I had $1,360 in admission sales. So equals 1,360. So that's one equation from my system. Let's keep going. "150 tickets were purchased. "How many were adult tickets? How many were children's tickets?" So I had a certain number of adult tickets, I don't know how many. I had a certain number of children's tickets, don't know how many. But I know in total I had 150 tickets purchased.

(Describer) X plus y equals 150.

(Describer) X plus y equals 150.

(Describer) X plus y equals 150.

So there we go. That is my system of equations. Scroll down a bit. The first equation's representing the amount of money in sales. And the second equation's representing the number of tickets that were sold. Those two equations are going to help me solve this system. Remember we had different methods for solving systems: elimination, substitution, graphing? Here, the method I'm choosing is substitution because my numbers are large. I don't want to start multiplying by even larger numbers to get even larger numbers in my problem. So I'll solve that second equation. I'll solve it for Y--- I could also solve it for X-- but I'll solve this one equation for Y because I can solve for Y in one step, and from there, I'll use the substitution method to solve this system. Okay? Let's get a little more workspace. Because if you remember, substitution took a little more work than elimination. But in this case, it's the easier route to take. Let's just move that to the center and let's get a little more room here for work. Okay. The first thing I'm going to do is solve that second equation for Y. I'll number my steps so you can keep them straight. Get the pen back. I'm going to call this "Step 1." I'm going to take that X plus Y equals 150, and I'm going to solve this equation for Y. I'm going to subtract X from both sides, cancel out over there. And I have Y equals negative X plus 150. Now, maybe you wrote 150 minus X, it doesn't really matter. You should get into the habit of writing your X term first because it'll help you as we go along in the course, when we solve for Y in an equation with two variables. So you want that X term written first. For now, if you reversed it, it doesn't matter, but try to get into habit of writing it first. Now that I know that second equation when I solve for Y, that it's negative X plus 150-- what it equals-- I'm going to take that value and substitute it back into my first equation to see what I get when I can solve it for X. Let me show you what I mean. Let's get some more room here, because remember I said these do take a bit. Okay. My second equation is 10X plus 8Y equals 1360. Let me just write that down. 10X plus 8Y... equals 1360. I know that the first equation, when I solve it for Y, that Y equals negative X plus 150, what I'll do is substitute that negative X plus 150 into that equation for Y. Watch me do it. I'll have 10X plus 8. Open up parentheses. Negative X plus 150, close 'em. Equals 1360, okay? Did you see what I did? I literally just replaced Y with that negative X plus 150. If you haven't already, you can use a calculator because I'm not going to trust myself to do mental math with this problem because these numbers are large. That's going to come up next. If we start cleaning this up, simplifying this equation, we've got 10X. All right, 8 times negative X, that's negative 8X. So minus 8X. Then I have 8 times 150, which for that, I am going to pull out my calculator for 8 times 150. And I already cleared the memory on this one. So 8 times 150 is 1200. So back to my problem. So that is plus 1200... equals 1360. Okay. Let's move things out of our way, keep solving for X. I have like terms to combine right here, this 10X and this minus 8X. If I combine those, I'll get 2X plus 1200 equals 1360. Going to keep going-- we want to isolate the 2X, get it by itself. Let's subtract 1200 from both sides. Going to need more room here, see it happening. All right. So my 1200s, or my 1200 minus 1200, cancels out, bring down my 2X. 1360 minus 1200. Kind of large, but I trust myself--that's 160. If not, use the calculator and see what it is. I have 2X equals 160. All right, so I'll divide both sides by 2. X is 80. Okay. We have found out the solution to one of our variables. X is 80. Remember, we're dealing with a word problem. We want to go back and determine what that means as far as our word problem. We're scrolling. If X is 80, then that means that we sold 80 adult tickets. Right now, we know the answer to one of our unknowns. Let me go ahead and write, "80 adult tickets." Now we want determine how many children's tickets were sold. You don't have to go back to your equations in your system; you can go back to the problem. You sold 150 tickets, and 80 of them were adults. Let's figure out how many of them were children. We can just do 180 minus 50-- sorry, 150 minus 80. I'm going to move down so I can write down the work. This is our third step. Remember what we said: that 80 represented 80 adult tickets. And we know that we sold 150 tickets in total. So 150 minus 80. That leaves 70. That means that we sold 70 children's tickets. You've solved your system. I believe it was a local movie theater we were dealing with here. Let's scroll-- the 11 a.m. show, mm-hmm. There were 80 adult tickets sold and there were 70 children's tickets sold. Let's try another one just to make sure you got it. It's a lot of steps, but if you break it down, you'll have it in no time. Check this one out. "Tickets to the local high school sporting events "are $7 for students and $12 for adults. "During last week's event, 107 tickets were sold "for a total of $949. "How many student tickets were sold? How many adult tickets were sold?" We've read through the problem; now let's go back and let's highlight all of the key information. If you find you're able to read the problem and highlight the information all at one time, that's fine. Go ahead and do that. But you may want to take one time to read through it with no pressure, no extra work to do, then go back and attack the problem, pull it apart, highlighting what you need. If you find you can do that all at once as you do more problems and get more comfortable, that's okay. This time, since I've read through it once, I'll highlight the information I need to know. Just highlight anything that sounds like math. "Tickets to the local high school sporting events are $7 for students and $12 for adults--" I got that. Highlighter didn't want to stay. "During last week's event, 107 tickets were sold for a total of $949--" I got that. "How many student tickets were sold?" And "how many adult tickets were sold?" Okay. Didn't want to highlight. Okay, there we go. I'll pass through it one more time. $7 for students, $12 for adults. 107 tickets were sold for a total of $949. How many student tickets? How many adult tickets? I have a general idea of what I'm trying to solve and what information will help me get to that solution. Let me switch to my pen to get to work. Since I'm dealing with student tickets and adult tickets as far as what I'm trying to solve in this problem, I'll let X represent the number of student tickets. I'll make myself a note here. And I'll let Y represent the number of adult tickets. I've got that to the side. Then I see I'm dealing with the cost of the tickets, and the total. So I'll keep all the money things together in one equation; so $7 for students, so 7X, plus $12 for adults, so plus 12Y, equals $949. So there is one. The other situation in my problem is just the number of tickets that were sold. I don't know how many student tickets were sold, but I'll let X represent that. Y is the adult tickets, so X plus Y equals 107, because it said that I sold a total of 107 tickets. There lies my system. Now that I've got it, it's time to start solving it. I'm going to do a little rearranging here, just to get us some extra workspace. I'm going to move-- ooh--do want that. Going to move this down a little bit to the center, and I'm going to scroll down, and just make this page a little longer for us. Because I know it's going to take multiple steps, I'll label this as "Step 1." get the pen back-- so Step 1. I'm going to solve this system using substitution. I'll take that second equation and go ahead and solve it for Y, like on the first one. So X plus Y equals 107. I'm going to solve it for Y. Maybe you chose X, that's fine, but I'm going to choose Y. I'll subtract X from both sides. Cancel out over there, bring down my: Y equals negative X plus 107. Now that I have that second equation solved for Y-- I'm using the substitution method to substitute that expression, that negative X plus 107, into my first equation for Y. All right, watch me do that. I need a little more space here. Make the page longer again. That equation, that first one, is 7X plus 12Y equals 949. So 7X plus 12Y equals 949. So I'm going to substitute that negative X plus 107 into this equation for Y. So 7X plus 12... times negative X plus 107... equals 949. Again, it just took the place of Y. Let's get some more space here. Those numbers are large. We'll pull out a calculator. But to simplify this first, let's bring down that 7X. We've got 12 times negative X. So that's negative 12X. 12 times 107-- let's pull out the calculator for that. 12 times 107. Okay, so 12 times 107. That is 1,284. Let's write that down: 1284. So plus 1284 equals 949. Let's get some more space here to work. So we can combine those two like terms in the front, that 7X minus 12X-- that's negative 5X. Bring down that plus 1284. Equals 949, okay? So I'm trying to solve for negative 5X, so I'm going to subtract 1284 from both sides. So cancel out right here. Let's bring down that negative 5X and pull out the calculator again. 949 minus 1284. Okay, 949 minus 1284. Okay, that is negative 335. So let's go back to our paper. So that equals negative 335. And we need more space again. So now we can divide both sides by negative 5. So X equals-- back to the calculator-- negative 335 divided by negative 5. Okay, negative 335 divided by negative 5. And we've got 67. Back to our work. All righty, so X equals 67. So if we bring that back to our word problem to get the meaning of that answer... Let's go back and remember what X represented. Scrolling. So X represented the number of student tickets sold. Right now we know that there were 67 student tickets sold. And since we know that 107 tickets were sold in total, we can make a shortcut; we don't have to go back to our equation. Let's just subtract, let's do 107 minus 67 and see what's left over for the adult tickets. Let's scroll a bit and write that work down. Scroll, scroll, scroll, more room, okay. We know we have 67 student tickets, and there were 107 tickets sold in total. So 107 minus 67. That's 40. So there were 40 adult tickets sold. For our word problem, we got it. I think we were dealing with sporting events in this problem. Let's scroll back up. We were. We sold-- let's write that down, get this out of the way. There were 67 student tickets sold, and there were 40 adult tickets sold. That gives us our total of 107 tickets, and that gives us also our total of $949. We got through this system. I do believe it's time for you to try one. Press pause, take your time, go through this problem, and see what you get. When you're ready to compare your answers with mine, press play.

(Describer) Titles: Kelsey is working two part-time jobs this summer to save money to purchase a car. She works at a computer repair store where she earns 15 dollars per hour, and at a clothing store where she earns 12 dollars per hour. Kelsey worked a total of 60 hours last week and earned 825 dollars. How many hours did Kelsey work at each part-time job?

(Describer) Titles: Kelsey is working two part-time jobs this summer to save money to purchase a car. She works at a computer repair store where she earns 15 dollars per hour, and at a clothing store where she earns 12 dollars per hour. Kelsey worked a total of 60 hours last week and earned 825 dollars. How many hours did Kelsey work at each part-time job?

(Describer) Titles: Kelsey is working two part-time jobs this summer to save money to purchase a car. She works at a computer repair store where she earns 15 dollars per hour, and at a clothing store where she earns 12 dollars per hour. Kelsey worked a total of 60 hours last week and earned 825 dollars. How many hours did Kelsey work at each part-time job?

(female narrator) Kelsey is working two part-time jobs this summer to save money to purchase a car. She works at a computer repair store where she earns $15 per hour, and at a clothing store where she earns $12 per hour. Kelsey worked 60 hours last week and earned $825. How many hours did Kelsey work at each part-time job? All right, let's see how you did. I do believe I have the answer here. I'll show you what I got and then do the work. Let's get this out of the way. So Kelsey worked 35 hours at the computer repair store, and 25 hours at the clothing store. If you want to see how I got that answer, watch me do this. I do believe-- nope, I didn't. Let's read it first, and then go back and highlight all of our key information. "Kelsey is working two part-time jobs this summer "to save money to purchase a car. "She works at a computer repair store "where she earns $15 per hour, and at a clothing store "where she earns $12 per hour. "Kelsey worked 60 hours total last week and earned $825. How many hours did Kelsey work at each part-time job?" We read through it once, now let's go back and let's highlight all that important information. Let me switch to my highlighter. So Kelsey is working two part-time jobs this summer to save money to purchase a car. She works at a computer repair store where she earns $15 an hour. So at the computer repair store, $15 per hour. And at a clothing store where she earns $12 per hour. So clothing store, $12 per hour. "Kelsey worked a total of 60 hours last week "and she earned $825. How many hours did Kelsey work at each part-time job?" So let's go through this. Computer repair store, $15 an hour. Clothing store, $12 an hour. She worked 60 hours, she earned $825. We're trying to figure out-- I forgot to highlight that-- how many hours she worked at each job. Let's switch to the pen and see if you took these steps to solve this. I'm going to let X represent the hours at the computer store and let Y represent the hours at the clothing store. You could've reversed that; either way is fine. Let's get that answer out of my way so I can work down there. So we're going to let X... Switch to the pen, there we go. X is the computer store-- I'll abbreviate that for computer-- and Y is the clothing store. We're doing a lot of scrolling on this one. And Y is the clothing store. So scroll back up. At the computer repair store she earns $15 per hour, and at the clothing store she earns $12 per hour. So I'm going to say 15X... plus 12Y equals 825. That'll handle the money that she's earning. And she worked a total of 60 hours last week. So the hours at the computer store plus the hours at the clothing store were 60. That should've been your system. Your system should've resembled something like this.

(Describer) 15x plus 12y equals 825, and x plus y equals 60.

(Describer) 15x plus 12y equals 825, and x plus y equals 60.

(Describer) 15x plus 12y equals 825, and x plus y equals 60.

Let's get some more workspace, do a little maneuvering up here. We can actually keep that right there. Step 1: I'm going to keep using the substitution method on these. So I'm going to take that second equation and solve it for Y again. So X plus Y equals 60. Subtract X from both sides. So Y equals negative X plus 60. So I have that second equation solved for Y, which will help me to be able to substitute back into that first equation, okay? All right, I'm going to get right underneath this one here. So that first equation is 15X plus 12Y equals 825. So 15 X plus 12Y equals 825, okay? And we'd already solved that second one, so we're going to substitute that negative X plus 60 into this equation for Y. So 15X plus 12 times negative X plus 60 equals 825. Let's get some more space up here. Let's clean this up. Let's bring down the 15X. 12 times negative X. That's negative 12X. 12 times 60. Just to be safe, let's get the calculator on that one. So 12 times 60. 720. Let's go back to our work here. Let's move that out of the way. So plus 720 equals 825. Let's get some more workspace. We can combine these two like terms in the front, So 15X minus 12X, that's 3X. Plus 720 equals 825. Okay? We're trying to get that 3X by itself, so let's subtract 720 from both sides. Let's get a little more workspace. Getting ahead of myself here. So that's going to cancel out, 720 minus 720. There we go. So let's bring down the 3X. 825 minus 720. That's 105. Then our last step: divide each side by three. I'll use calculator for that: 105 divided by 3. Just to make sure that's right. So 105 divided by 3. 35. Let's go back to our work. And we have X equals 35. Let's go back up to the problem to see what meaning that has in relation to the word problem. So if X is 35, and we said that X represented the number of hours she worked at the computer store, then we know that she worked 35 hours at the computer store. Let's write that down. Kind of crowded up here so let's go back to the bottom and write that down. So she worked 35 hours... at the computer store. And if we go back to the problem, I believe it said that she worked a total of 60 hours. Right, so she worked a total of 60 hours. Let's see how many hours are left over for her job at the clothing store. Let's scroll back down. So 60 minus 35. Let's get a little more space here. And that's actually 25. So she worked 35 hours at the computer store, and she worked 25 hours at the clothing store. Your work should have been something similar to that when you were solving that real-world problem. I hope you're feeling confident about how to solve real-world problems involving systems of linear equations. See you back here soon for some more Algebra 1. Bye!

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In this program, students will learn about concepts involving inequalities and systems of equations. These can be used to solve linear programming application problems. Part of the "Welcome to Algebra I" series.

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Runtime: 34 minutes

Welcome to Algebra I
Episode 1
31 minutes
Grade Level: 7 - 12
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Episode 2
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Episode 3
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Episode 4
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Episode 5
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Episode 10
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