Video Transcript: Graphing Systems of Equations

Systems of equations: trolls, tolls (1 of 2) 

You are traveling in some type of a strange fantasy land. And you're trying to get to the castle  up here to save the princess or the prince or whomever you're trying to save. But to get there,  you have to cross this river. You can't swim across it. It's a very rough river. So you have to cross  this bridge. And so as you approach the bridge, this troll shows up. That's the troll. And he says,  well, I'm a reasonable troll. You just have to pay $5. And when you look a little bit more  carefully, you see that there actually was a sign there that says $5 toll to cross the bridge. Now,  unfortunately for you, you do not have any money in your pocket. And so the troll says, well,  you can't cross. But you say, I need to really, really get to that castle. And so the troll says, well,  I'll take some pity on you. Instead of paying the $5, I will give you a riddle. And the riddle is this.  And now, I'm speaking as the troll. I am a rich troll because I get to charge from everyone who  crosses the bridge. And actually, I only accept $5 or $10 bills. It's a bit of a riddle why they  accept American currency in this fantasy land. But let's just take that as a given for now. So I  only take $5 or $10 bills. I'm being the troll. Obviously, if you give me a $10, I'll give you $5 back.  And I know, because I count my money on a daily basis. I like to save my money as the troll. I  know that I have a total of 900 bills. So let me write that down. I have a total of 900 bills, a total  of 900 $5 and $10 bills. And he says, because I'm very sympathetic, I'll give you another piece of information. He says, if you were to add up the value of all of my money, which is all in $5 and  $10 bills that I have, I, speaking as the troll about dollars bills, is $5,500. This is a rich troll. And  so the riddle is-- exactly, exactly. And if you give the wrong answer and if you're not able to  solve it in 10 minutes, he's just going to push you into the river or do something horrible to you.  He says, exactly how many 5's and 10's do I, the troll, have? So the first thing I'm going to have  you think about is, is this even a solvable problem? Because if it's not a solvable problem, you  should probably run as fast as you can in the other direction. So now, I will tell you, yes, it is a  solvable problem. And let's start thinking about it a little bit algebraically. And to do that, let's  just set some variables. And I will set the variables to be what we're really trying to solve for.  We're trying to solve for the number of $5 bills we have and the number of $10 bills that we  have. So let's just define some variables. I'll say f for 5. Let's let f equal the number of $5 bills  that we have. And I'll use the same idea. Let's let t is equal to the number $10 bills that we have. Now, given this information, and now I'm not sure if I'm speaking as-- well, let's say I'm still  speaking as the troll. I'm a very sympathetic troll, and I'm going to give you hints. Given this  information and setting these variables in this way, can I represent the clues in the riddle  mathematically? So let's focus on the first clue. Can I represent this clue that the total of 900 $5  and $10 bills, or can I represent that mathematically, that I have a total of 900 $5 and $10 bills?  Well, what's going to be our total of bills? It's going to be the number of 5's that we have, which  is f. The number of 5's that we have is f. And then the number of 10s that we have is t. The total  number of 5's plus the total number of 10s, that's our total number of bills. So that's going to be equal to 900. So this statement, this first clue in our riddle, can be written mathematically like  this if we defined the variables like that. And I just said f for 5, because f for 5 in t for 10. Now,  let's look at the second clue. Can we represent this one mathematically given these variable  definitions that we created? Well, let's think separately about the value of the $5 bills and the  value of the $10 bills. What's the value of all of the $5 bills? Well, each $5 bill is worth $5. So it's  going to be 5 times the number of $5 bills that we have. So if I have one $5 bill, it will be $5. If I 

have 100 $5 bills, then it's going to be $500. How ever many $5 bills, I just multiply it by 5. That's the value of the $5 bills. Let me write that down. Value of the $5 bills. Now, same logic. What's  the value of the $10 bills? Well, the value of the $10 bills is just going to be 10 times however  many bills I have-- value of the $10 bills. So what's going to be the total value of my bills? Well,  it's going to be the value of the $5 bills plus the value of the $10 bills? And he tells me what that  total value is. It's $5,500. So if I add these two things, they're going to add up to be $5,500. So  this second statement we can represent mathematically with this second equation right over  here. And what we essentially have right over here, we have two equations. Each of them have  two unknowns. And just using one of these equations, we can't really figure out what f and t  are. You can pick a bunch of different combinations that add up to 900 here. You could pick a  bunch of different combinations, where if you work out all the math, you get $5,500. So  independently, these equations, you don't know what f and t are. But what we will see over the  next several videos is that if you use both of this information, if you say that there's an f and a t  that has to satisfy both of these equations, then you can find a solution. And this is called a  system of equations. Let me write that down-- system of equations. 

Systems of equations trolls, tolls (2 of 2) 

Where we left off, we were trying our very best to get to the castle and save whomever we  were needing to save. But we had to cross the bridge and the troll gave us these clues because  we had no money in our pocket. And if we don't solve his riddle, he's going to push us into the  water. So we are under pressure. And at least we made some headway in the last video. We  were able to represent his clues mathematically as a system of equations. What I want to do in  this video is think about whether we can solve for this system of equations. And you'll see that  there are many ways of solving a system of equations. But this time I want to do it visually.  Because at least in my mind, it helps really get the intuition of what these things are saying. So  let's draw some axes over here. Let's draw an f-axis. That's the number of fives that I have. And  let's draw a t-axis. That is the number of tens I have. And let's say that this right over here is 500 tens. That is 1,000 tens. And let's say this is-- oh, sorry, that's 500 fives. That's 1,000 fives. This is 500 tens, And this is 1,000 tens. So let's think about all of the combinations of f's and t's that  satisfy this first equation. If we have no tens, then we're going to have 900 fives. So that looks  like it's right about there. So that's the point 0 tens, 900 fives. But what if went the other way?  If we have no fives, we're going to have 900 tens. So that's going to be the point 900 tens, 0  fives. So all the combinations of f's and t's that satisfy this are going to be on this line right over  there. And I'll just draw a dotted line just because it's easier for me to draw it straight. So that  represents all the f's and t's that satisfy the first constraint. Obviously, there's a bunch of them,  so we don't know which is the one that is actually what the troll has. But lucky for us, we have a  second constraint-- this one right over here. So let's do the same thing. In this constraint, what  happens if we have no tens? If tens are 0, then we have 5f is equal to 5,500. Let me do a little  table here, because this is a little bit more involved. So for the second equation, tens and fives.  If I have no tens, I have 5f is equal to 5,500, f will be 1,100. I must have 1,100 fives. If I have no  fives, then this is 0, and I have 10t is equal to 5,500, that means I have 550 tens. So let's plot  both at those point. t equals 0, f is 11. That's right about there. So that is 0. 1,100 is on the line  that represents this equation. And that when f is 0, t is 550. So let's see, this is about-- this 

would be 6, 7, 8, 9, so 550 is going to be right over here. So that is the point 550 comma 0. And  all of these points-- let me try to draw a straight line again. I could do a better job than that. So  all of these points are the points-- let me try one more time. We want to get this right. We don't  want to get pushed into the water by the troll. So there you go. That looks pretty good. So  every point on this blue line represents an ft combination that satisfies the second constraint.  So what is an f and t, or number of fives and number of tens that satisfy both constraints? Well,  it would be a point that is sitting on both of the lines. And what is a point that is sitting on both  of the lines? Well, that's where they intersect. This point right over here is clearly on the blue  line and is clearly on the yellow line. And what we can do is, if we drew this graph really, really  precisely, we could see how many fives that is and how many tens that is. And if you look at it,  if you look at very precisely, and actually I encourage you to graph it very precisely and come up with how many fives and how many tens that is. Well, when we do it right over here, I'm going  to eyeball it. If we look at it right over here, it looks like we have about 700 fives, and it looks  like we have about 200 tens. And this is based on my really rough graph. But let's see if that  worked. 700 plus 200 is equal to 900. And if I have 700 fives-- let me write this down. 5 times  700 is going to be the value of the fives, which is $3,500. And then 10 plus 10 times 200, which is $2,000, $2,000 is the value of the 10s. And if you add up the two values, you indeed get to $  5,500 So this looks right. And so we can tell the troll-- Troll! I know! I know how many $5 and  $10 bills you. You have 700 $5 bills, and you have 200 $10 bills. The troll is impressed, and he lets you cross the bridge to be the hero or heroine of this fantasy adventure. 

Testing a solution to a system of equations 

Is negative 1 comma 7 a solution for the system of linear equations below? And they give us the  first equation is x plus 2y is equal to 13. Second equation is 3x minus y is equal to negative 11. In  order for negative 1 comma 7 to be a solution for the system, it needs to satisfy both equations. Or another way of thinking about it, x equals 7, and y-- sorry, x is equal to negative 1. This is the  x coordinate. X equals negative 1, and y is equal to 7, need to satisfy both of these equations in  

order for it to be a Solution. So let's try it out. Let's try it out with the first equation. So we have  x plus 2y is equal to 13. So if we're thinking about that, we're testing to see if when x is equal to  negative 1, and y is equal to 7, will x plus 2y equals 13? So we have negative 1 plus 2 times 7-- y  should be 7-- this needs to be equal to 13. And I'll put a question mark there because we don't  know whether it does. So this is the same thing as negative 1 plus 2 times 7 plus 14. That does,  indeed, equal 13. Negative 1 plus 14, this is 13. So 13 does definitely equal 13. So this point it  does, at least, satisfy this first equation. This point does sit on the graph of this first equation, or on the line of this first equation. Now let's look at the second equation. I'll do that one in blue.  We have 3 times negative 1 minus y, so minus 7, needs to be equal to negative 11. I'll put a  question mark here because we don't know whether it's true or not. So let's see, we have 3  times negative 1 is negative 3. And then we have minus 7 needs to be equal to negative 11-- I  put the question mark there. Negative 3 minus 7, that's negative 10. So we get negative 10  equaling negative 11. No, negative 10 does not equal a negative 11. So x equaling negative 1,  and y equaling 7 does not satisfy the second equation. So it does not sit on its graph. So this  over here is not a solution for the system. So the answer is no. It satisfies the first equation, but  it doesn't satisfy the second. In order to be a solution for the system, it has to satisfy both 

equations. 

Systems of equations with graphing: y=7/5x-5 & y=3/5x-1 

Just in case we encounter any more trolls who want us to figure out what types of money they  have in their pockets, we have devised an exercise for you to practice with. And this is to solve  systems of equations visually. So they say right over here, graph this system of equations and  

solve. And they give us two equations. This first one in blue, y is equal to 7/5x minus 5, and then  this one in green, y is equal to 3/5x minus 1. So let's graph each of these, and we'll do it in the  corresponding color. So first let's graph this first equation. So the first thing I see is its y intercept is negative 5. Or another way to think about it, when x is equal to 0, y is going to be  negative 5. So let's try this out. So when x is equal to 0, y is going to be equal to negative 5. So  that makes sense. And then we see its slope is 7/5. This was conveniently placed in slope intercept form for us. So it's rise over run. So for every time it moves 5 to the right it's, going to  move seven up. So if it moves 1, 2, 3, 4, 5 to the right, it's going to move 7 up. 1, 2, 3, 4, 5, 6, 7.  So it'll get right over there. Another way you could have done it is you could have just tested  out some values. You could have said, oh, when x is equal to 0, y is equal to negative 5. When x  is equal to 5, 7/5 times 5 is 7 minus 5 is 2. So I think we've properly graphed this top one. Let's  try this bottom one right over here. So we have when x is equal to 0, y is equal to negative 1. So  when x is equal to 0, y is equal to negative 1. And the slope is 3/5. So if we move over 5 to the  right, we will move up 3. So we will go right over there, and it looks like they intersect right at  that point, right at the point x is equal to 5, y is equal to 2. So we'll type in x is equal to 5, y is  equal to 2. And you could even verify by substituting those values into both equations, to show  that it does satisfy both constraints. So let's check our answer. And it worked. 

Systems of equations with graphing: exact & approximate solutions The following two equations form a linear system. This is one equation; it has X and Y so it's  gonna define a line. And then I have another equation that involves X and Y, so it's gonna define another line. It says: "Graph the system of equations "and find its solution." So we're gonna try  to find it visually. So let's graph this first one. To graph this line, I have the little graphing tool  here. Notice if I can figure out two points, I can move those points around and it's going to  define our line for us. I'm gonna pick two X values and figure out the corresponding Y values  and then graph the line. So let's see how I could do this. So let's see; an easy one is what  happens when X is equal to zero? Well if X is equal to zero, everything I just shaded goes away  and we're left with -3y is equal to nine. So -3y equals nine. Y would be negative three. So when  X is equal to zero, Y would be negative three. So let me graph that. When X is equal to zero, X is  zero, Y is negative three. Now another easy point actually instead of trying another X value,  let's think about when Y is equal to zero 'cause these equations are in a standard form so it's  easy to just test. Well what are the X and Y intercepts? So when Y is equal to zero, this term  goes away, and you have negative X is equal to nine, or X would be equal to negative nine. So  when Y is zero, X is negative nine. So when Y is zero, X is negative nine, or when X is negative  nine, Y is zero. So I've just plotted this first equation. So now let's do the second one. We'll do  the same thing. What happens when X is equal to zero? When X is equal to zero, so this is going  to be our Y intercept now. When X is equal to zero, -6y is equal to negative six. Well Y would  have to be equal to one. So when X is zero, Y is equal to one. So when X is zero, Y is equal to 

one. Get one more point here. When Y is zero, when this term is zero, Y being zero would make  this entire term zero, then 6x is equal to negative six or X is equal to negative one. So when Y is  zero, X is negative one or when X is negative one, Y is zero. When X is negative one, Y is zero.  And so just like that, I've plotted the two lines. And the solution to the system are the X and Y  values that satisfy both equations; and if they satisfy both equations, that means they sit on  both lines. And so in order to be on both lines, they're going to be at the point of intersection.  And I see this point of intersection right over here, it looks pretty clear that this is the point X is  equal to negative three and Y is equal to negative two. So it's the point negative three comma  negative two. So let me write that down. Negative three comma negative two. And then I could check my answer; got it right. Let's do another. Let's do another one of these. Maybe of a  different type. So over here it says: "A system of two linear equations "is graphed below.  "Approximate the solution of the system." Alright so here I just have to just look at this carefully and think about where this point is. So let's think about first its X value. So its X value, it's about  right there in terms of its X value. It looks like, so this is negative one. This is negative two, so  negative 1.5 is gonna be right over here. It's a little bit to the left of negative 1.5, so it's even  more negative, I would say negative 1.6. And I'm approximating it, negative 1.6. Hopefully it  has a little leeway in how it checks the answer. What about the Y value? So if I look at the Y  value here, it looks like it's a little less than one and a half. One and a half would be halfway  between one and two. It looks like it's a little less than halfway between one and two, so I'd give it 1.4, positive 1.4. And let's check the answer, see how we're doing. Yep, we got it right. Let's  actually just do one more for good measure. So this is another system. They've just written the  equations in more of our slope intercept form. So let's see, Y is equal to negative seven, X plus  three. When X is equal to zero, we have our Y intercept. Y is equal to three. So when X is equal  to zero, Y is equal to three. And then we see that our slope is negative seven. When you  increase X by one, you decrease Y by seven. So when you increase X by one, you decrease Y by  one, two, three, four, five, six, and seven. When X goes from zero to one, Y went from three to  negative four, it went down by seven, so that's that first one. Now the second one: our Y  intercept. When X is equal to zero, Y is negative three, so let me graph that. When X is zero, Y is  equal to negative three. And then its slope is negative one. When X increases by one, Y  decreases by one. So the slope here is negative one. So when X increases by one, Y decreases  by one. And there you have it. You have your point of intersection. You have the X-Y pair that  satisfies both equations. That is the point of intersection. It's gonna sit on both lines which is  why it's the point of intersection. And that's the point X equals one, Y is equal to negative four.  So you have X equals one and Y is equal to negative four. And I can check my answer and we got it right. 

Setting up a system of equations from context example (pet weights) In this video, we're gonna get some more practice setting up systems of equations. Not solving  them, but just setting them up. So we're told Sanjay's dog weighs five times as much as his cat.  His dog is also 20 kilograms heavier than his cat. Let c be the cat's weight and let d be the dog's  weight. So pause this video and see if you can set up a system of equation, two linear equations with two unknowns that we could use to solve for c and d, but we don't have to in this video. All right so let's do it together. So, what I like to do is usually there's a sentence or two that 

describes each of the equations we wanna set up. So this first one tells us Sanjay's dog weighs  five times as much as his cat. So how much does his dog weigh? So his dog weighs d, so we  know d is going to be equal to five times as much as his cat weighs. So his cat weighs c, so d is  going to be equal to five times as much as his cat weighs. So that's one linear equation using d  and c. And so what's another one? Well, then we are told his dog is also 20 kilograms heavier  than his cat. So we could say that the dog's weight is going to be equal to the cat's weight plus  what? Plus 20 kilograms. We're assuming everything's in kilograms, so I don't have to write the  units. But there you have it, I have just set up two equations in two unknowns, two linear  equations, based on the information given in this word problem, which we could then solve,  and I encourage you to do so if you're curious. But sometimes, the difficult part is just to find, is  to re-express the information that you're given in a mathematical form. But as you see, as you  get practice, it becomes somewhat intuitive. What we see in blue is just another way of writing  what we underline in blue and what we see in yellow is just another way of writing or  expressing wat we underlined in yellow up there. 

Setting up a system of linear equations example (weight and price) In this video, we're going to get a little bit of practice setting up systems of linear equations  based on a word problem. We're not actually going to end up solving it. You can do that if you  like just for kicks. But really we're going to just focus on setting it up. So here we're told Lauren  uses a blend of dark roast beans and light roast beans to make coffee at her cafe. She needs 80  kilograms of beans in total for her next order. Dark roast beans cost $3 per kilogram, light roast  beans cost $2 per kilogram, and someone wants to spend $220 total. Let D be the number of  kilograms of dark roast beans she buys and L be the number of kilograms of light roast beans  she buys. All right, so based on this information that we've been given, see if you can pause this video and set up a system of equations. And it's going to have two equations with two  unknowns, D and L, that in theory we could solve to figure out the right number of kilograms of dark roast beans and light roast beans that Karen should use. So pause this video and try to  work that out. All right, now let's do it together. And what I'm going to do is I'm going to  underline. So let's see, we know that D is dark roast beans and the L is the number of kilograms  of light roast beans. And then they tell us here, they say she needs 80 kilograms of beans in  total. So what we could say is, hey, the number of kilograms of dark roast beans plus the  number of kilograms of light roast beans needs to be equal to 80 kilograms. So the number of  kilograms of dark roast beans plus the number of kilograms of light roast beans. (laughs) I'm  having trouble saying light roast beans. Well, this, what I just underlined here, it says needs to  be 80 kilograms in total. So that needs to be 80. So this number of kilograms plus this number  of kilograms is going to be equal to your total number of kilograms. All right, so I have one  equation with two unknowns. Let's see if we can get another one. So next, they say dark roast  beans cost $3 per kilogram, light roast beans cost $2 per kilogram, and she wants to spend $220 total. So what I just underlined in this aquamarine color we can set up another equation with.  And if you haven't already set up your system of equations, see if you can now do that. See if  you can set up this second equation. Pause the video. All right, well, the way to think about it is  we just have to have an expression for how much did she spend on dark roast beans, how much did she spend on light roast beans, and then we need to add those two together, and that 

needs to be equal to $220 'cause that's how much she wants to spend in total. So how much  does she spend on dark roast beans? Well, it's going to be the number of kilograms of dark  roast beans that she buys, and it says that it cost $3 per kilogram, so we're gonna multiply it by  three. $3 per kilogram times the number of kilograms of dark roast beans. This is how much she spends on dark roast beans. And so how much is she going to spend on light roast beans? Well,  she buys L kilograms of light roast beans. They told us that there. And they cost $2 per  kilogram, so $2 per kilogram times the number of kilograms. This is how much she spends on  light roast beans. So you add how much she spends on dark roast to how much she spends on  light roast, and so this is going to be $220 in total. And there you have it. We have our two  equations with two unknowns, and so now we could go and solve it, but you can do that  outside of this video. But the whole point of this video is to understand how to construct these  based on the constraints, based on the information that we see in this. So typically when you're trying to set these up, there's often a sentence or two that will focus on one equation. So this  first one is saying, hey, the kilograms, let's add those up for the total number of kilograms. And  then there's another sentence or two that'll focus on, in this case, some other equation. In this  case, it's the price. So the price of the dark plus the price of the light is going to be equal to the  total amount she wants to spend. 

Interpreting points in context of graphs of systems 

We're told that Lauren uses a blend of dark roast beans and light roast beans to make coffee at  her cafe. She needs 80 kilograms of beans in total for her next order. Dark roast beans cost $3  per kilogram, light roast beans cost $2 per kilogram, and she wants to spend $220 in total. And  they tell us here's a graph that shows a system of equations for this scenario where x is the  number of kilograms of dark roast beans she buys and y is the number of kilograms of light  roast beans she buys. All right, let me scroll down so we can take a look at this. And so sure  enough, so this blue line, and I'll write it out in blue, this x is the number of kilograms of dark  roast beans, y is the number of kilograms of light roast beans, and she wants to buy a total of  80 kilograms. That's what they told us up here. We can go back to look at that. She needs, I'll  underline this in blue. She needs 80 kilograms of beans. So that constraint that the sum of the  kilograms of dark and light is equal to 80, that's represented by this equation. And if we were to graph it, that is this blue line right over here. And then this other constraint, three x, well, the  dark roast beans cost $3 per kilogram, so three x is how much she spends on dark roast. Two y  is how much she spends on light roast 'cause it's $2 per kilogram. And 220 is the amount that  she spends in total. And they tell us that up here. Dark roast beans cost $3 per kilogram, light  roast beans cost $2 per kilogram, and she wants to spend $220. So this equation is another way  of expressing what I just underlined up here in green. And the green line shows all of the x y  combinations that would match these constraints. And so now let's do something interesting.  They've labeled some points here, point C, D, F, and E. And we're gonna think about what do  each of these points represent? So for example, this point C that is on the green line, but it sits  above the blue line, what does this tell us? What does this point C represent? Pause this video  and think about it. Well, if we're on the green line, that means that the amount that she spends  on dark roast plus the amount that she spends on light roast is adding up to exactly $220. So  she's definitely spending $220 at C, but how many total kilograms is she using? Well, the fact 

that for this given x, we're sitting above the line, that means that she's not using exactly 80  kilograms. And we can see that over here. She's using, looks like 10 kilograms of dark, and it  looks like something like 95 kilograms of light. If you were to add those two points together, it  looks like she's using something closer to 105 kilograms. So point C is a situation where she is  spending exactly $220, but she's using more than 80 kilograms 'cause it's not meeting this  second constraint. It's sitting above that line. Now let's think about point D. What does that  represent? Pause the video and try to figure that out. Well, because we sit on the blue line, that  means that we are meeting this constraint that the kilograms of dark and light combined is  equal to 80 kilograms. So she's using exactly 80 kilograms here, but what about her spending?  Well, because this point lies below the green line, that tells us that we are spending less than  $220. And we could even try it out. Three times 20 plus two times 60 is what? 60 plus 120 is  $180. And so this is a point where we're meeting this constraint, but we're not meeting this  constraint. We're underspending right over here. Now, what about point F? Well, point F sits  below both of these lines. So pause your video and think about what that means. Well, if we're  sitting below both of these lines, that means that neither are we spending $220, nor are we  using 80 kilograms. And you can see that if you actually look at the numbers. You don't have to  do this, but this is just to make you feel good about it. It looks like she is using 30 kilograms of  dark and 30 kilograms of light, so in total she is using, so this is a situation where she's using 60  kilograms in total, not 80. And so that's why we're not sitting on this blue line. And if you look  at how much she's spending, she has 30 kilograms of each, so three times 30 plus two times 30,  that's going to be 90 plus 60. That's also less than 220, and so that's why we see this point is  below these lines. And then last but not least, what does point E represent? Well, point E sits on both of these lines, so that means that it meets both of these constraints. This is a situation  where she is spending exactly $220, and the total number of kilograms she's using of dark and  light is exactly 80. And so if we wanted to say, hey, what combination of dark and light would  she need in order to meet both constraints, E represents that, the intersection of these two  lines.