Two Variable Systems of Linear Equations - Video Tutorials & Practice Problems

Introduction to Systems of Linear Equations

Hey, everyone. Welcome back. So earlier in this course, we learned how to plot and solve a single equation on a graph, something like why was eight X minus four. But now you might start to see problems that have multiple equations in them. You'll be asked to solve these types of problems. And this is what your book referred to as a system of equations. Now, in this video, we're only gonna focus on linear equations. So basically, whenever you see a system of equations, it really just means that there are multiple equations like we have in our problem. So I'm gonna explain the basic difference between these types of problems. And then later on, we'll learn a bunch of different ways to solve these types of problems. Let's go ahead and get started. So remember that the solution to a single equation is you're really just trying to find xy pairs that satisfy this equation. So for example, if we were asked to determine if each point was a solution to this equation, these three points over here, we'd all, all we have to do is we could plug them into our equation and figure out if to make true statements, we can also just graph them in see if they're on the line graphically just means if these points are solutions, that means they're on the line. So for example, something like negative two comma zero. So if you plot that point that's gonna be over here. Is this a solution to this equation? Well, it's not on the line. So it's not, if you were to plug this into this equation, it would at least lead to a false statement at this 0.0 comma negative four. And that's gonna be over here, it's on the line. So therefore, this is a solution to the system of uh sorry to this specific equation. If you plug these numbers into this equation, you'll get a true statement. Now, what about the 0.1 comma four? If you plot that, what you'll see is that this is right over here. So is this, well, this actually is on the line? So it also is a solution to this equation. So it's pretty straightforward if it's on the line, it satisfies this equation. Now, let's take a look at this situation over here. We have multiple equations and I've actually already plotted them for you. If you plot them in slope intercept form, you'll see that the graph indeed actually does look like this. We're asked to figure out if a, if a point is a solution to the system of equations, really what you're doing here is to solve these types of problems, you're gonna find coordinate pairs or XY pairs that satisfy not just one equation but all of the equations. So for example, if you were to look at these points over here, zero comma negative four, what we'll see here is that this only is on the blue line, but this is not a solution to the system of equations because it's not on the red line, it doesn't satisfy both of them at the same time. So this is not a point, this is not a solution to the system because this only satisfies the blue equation only. Let's take a look at the next one. The next one over here is the 0.2 comma negative two comma five. If you plot that out, that's gonna be right over here. So is this a solution to the system of equations? Well, this is kind of like the opposite of a where it only satisfies the red equation, right. So it's on the red line but notice how it's not on the blue line. So this satisfies the red only but not the blue. And so that's why it's not a solution to both. Let's take a look at our last point which is one comma four. So the 0.1 comma four you'll see is actually right over here. And if you take a look at this is this a solution to the system, it satisfies the red line, but it also satisfies the blue line. So it turns out that this is a solution to the system of equations because it satisfies both the lines. It's on both of the lines. All right. So what we can see here is that graphically the solution to the system of equations is actually just the place where the lines intersect or where they meet. All right. So you could plug in these points into both of these equations and you'll get true statements for both of them. But graphically the solution is actually just the point where the lines cross. All right. And that's really all there is to it. So I actually point out the big difference between these types of problems because what we saw is that for a single equation, the solution is that you actually had many points that satisfy just one line. So for example, we could pick these two points, but really any point along this line will satisfy just this one equation. Whereas over here, what we saw is that there's only really just one solution that satisfies all the lines, there's lots of points that satisfy one or the other, there's only one that satisfies all of them, right? And that's really the main difference between a single equation and a system of equations. Now, one of the things I want to just mention is that this is true for most problems where you'll have just one solution, but there aren't other types of solutions and we'll deal with them as we sort of go along anyway. So that's an introduction to a system of equations. Hopefully, that makes sense. Thanks for watching.

Example 1

Hey, everyone. Let's get started with this example problem here. So we're gonna graph this system of equations that's given to us below. And because we're trying to identify what the intersection point is. So basically, we have to figure out where these lines will cross each other in order to do that, we have to actually graph them. So let's go ahead and get started here. So Y equals two X plus three. Notice how both of these equations are already in slope intercept form. We have Y equals MX plus by equals something, right? So two X plus three, what does that look like? Well, it passes through the 0.0 comma three and it has a slope of positive two, positive two means it's gonna go up two and then over one. So I'm gonna put a point here or we can go down to and to the left one. So if you connect these dots with, with a line which you'll see is that this graph should look something like this. All right. So that's the equation or that's the graph Y equals to X plus three. Let's do the same thing. Now for the red equation, which should be a little bit simpler because it's just X plus four. That means that it has a uh win intercept of zero comma four. And instead of a slope of two, it has a slope of one. So what I'm gonna do now is I'm gonna slope of one means up 1/1, down, 1/1. So it's gonna look a little bit different. And if you sort of draw the line that connects these points, it should look something like that to look something like that. So that's the equation Y equals X plus four. So we have to identify with the intersection point is, it's just the point where the lines cross. And clearly we can see these lines cross right at this point over here. The coordinates to this point are one comma five. All right. So with these two lines over here, the intersection, it's just gonna be the point one comma five. Are we done yet? Well, not quite because the rest of the problem says that we have to take this intersection point and we have to verify that it's a solution to both equations. So what does that mean? But when we did this for one single equation, we would plug in the X and Y values, we'd take those and we would plug them into the equation to figure out if we got a true statement. It's the same idea here. We're just gonna take these X and Y coordinates and we're gonna plug them into both of the equations. We have to get a true statement for both of them. Then we can say it's a solution to both equations. All right. So it's pretty straightforward what I'm gonna do here is I'm just gonna rewrite Y equals two X plus three and remember what I'm gonna do here and I'm just gonna take these X and Y values the five And I'm gonna plug that in for X or sorry for Y, I'll take the one and I'll replace that in for the X and then we'll solve this equation here. All right. So we'll see what happens. We get Y is equal to two times one plus three. Right now. You can see pretty clearly here that the left and right sides will be the same because you'll just get a statement that just says five equals five on the left and right sides. And that is a true statement. Five does equal five. So this is a solution. Now, we'll just do the same exact thing with the red equation. Y equals X plus four, do the same thing. So we're just gonna replace with five and X with one and you'll get that five is equal to one plus four. And just like above, we'll just get a true statement. Five does equal five. So because this point over here one comma five, when you plug it into both equations, you just get true statements for both of them, five equals five. And we can say conclusively that this point is a solution to both equations. All right. Thanks for watching.

Example 2

Everyone. So let's take a look at this example, probably working out together. We've got these three graphs that are showing these three pairs of lines. And we want to take these graphs and match them to their system of equations that's shown here below. So really, I'm gonna just try to match equations to graphs. Let's take a look at the first one over here because I've got Y equals three X plus five and Y equals negative to X plus 10. Notice how both of these are in slope intercept form. So it's gonna be a little bit easier to sort of sort of try to match the equation with the graph. Then for example, this equation which is a little bit messier. All right. So let's take a look at this three X plus five. I'm looking for something that crosses the Y axis here at positive five. So let's take a look at this sort of first graph over here. I'm looking for if anything crosses the Y equals five points and it actually doesn't. So neither of these two graphs do. What about this one? In fact, both of these actually cross at Y equals 10 or Y intercept of 10. And over here, what I see is that the red equation does actually cross the axis at Y intercept of five. All right. But let's just double check the next equation. This says Y equals negative two X plus 10. So now I'm looking for a Y intercept of positive 10. Remember this is like Y equals MX plus B, I'm looking for the B term which is 10. And if you look here, the blue equation does actually cross here at Y equals 10. So that means that this is probably going to be equations or, or the graph C over here that I'm looking at but just to sort of triple check what we're gonna do here is we're gonna look at this intersection point and we're gonna plug it into both of these equations here. So remember we're just gonna take a look here and we're gonna plug these X and Y values into their equations and see if we get true statements for both. All right. So if you bring this equation down, what this says is that eight is equal to three times one plus five and eight does in fact equal three plus five. So you just get eight equals eight, which is a true statement if you do it for the bottom equation, for the blue equation, which you'll see is you're gonna get eight equals negative two times one plus 10 and eight does. In fact, equal negative two plus 10 because that just equals eight as well. So this is definitely gonna be a true statement. All right. So let's take a look at now, the second equation or a second pair of equations, I've got Y equals four X plus eight. So if you look through my graphs, what you're looking for is remember this is gonna be in slope intercept form. So I'm looking for something that crosses the axis at a positive eight. So does anything cross the axis at positive eight? It actually does over here. So that might be actually be the answer. But let's take a look at the second graph over here just to kind of make, make sure in fact what we actually said here for graph B is that both of the lines cross at a Y intercept of 10. So because of that, this definitely can't be the right answer. And in fact, this one is going to be a all right. Now, you can go ahead and pause the video if you want to and just double check that this intersection point is gonna work for both these equations. But you will see that if you plug in the values of like for example, negative one and four, if you plug them into both of these equations, you will get true statements for both. Feel free to pause and sure sure to check that yourself. And that just means that by default, this equation B over here or sorry, this equation three is gonna line up with graph B because both of these things have a Y intercept of 10. And in fact, what you can actually see here is if I sort of rearrange this equation, this actually becomes Y equals negative three, X plus 10. And this equation over here becomes Y equals X plus 10. Notice how both of them have Y intercept of 10. So that lines up with these two graphs over here, these are the red and blue lines. All right. So that's the answer to this problem. Hopefully, that makes sense. Thanks for watching.

Solving Systems of Equations - Substitution

Welcome back everyone. So we saw in an earlier video how to solve a system of equations by graphing. So we would graph these two lines for example, and we would just find the place where they intersect and that would give us our X and Y values. But some problems like the one we're gonna work out down here will ask us to solve a system of equations without graphing. And they'll ask you specifically in some cases to solve it by using something called the substitution method. So that's what I wanna show you how to do in this video. I wanna show you this substitution method and I'm gonna show you it's actually a set of really straightforward steps here, these five steps. And we're gonna see a lot of stuff that we've already seen how to do before like plugging in numbers and for variables and expressions and things like that. And also just solving some simple equations for X and Y. Let's go ahead and get started. Now, first, I actually want to talk about why we even need this method in the first place. Remember the solution to a system of equations is really just a set of numbers. So for these two equations, there's just gonna be a two number X and Y equals something. And when I plug them into both of these equations, I'll get true statements. So if I can't graph these equations, can I just sit here guessing a bunch of numbers for X and Y? And you totally could. The problem is it's gonna take you a long time because you might find a pair that works for one of the equations or the other, but not both of them. And so this kind of guess and check method isn't gonna be useful. I'm gonna show you a much more straightforward and systematic way to solve this. The basic idea is that to solve these types of problems without graphing, we're just going to substitute one equation into another and that's gonna make our equations simpler. That's why we call it the substitution method. Let's go ahead and get started. The first thing you're gonna do here is you're gonna choose the easiest equation to isolate X or Y. And you're gonna call that equation A. So for example, this equation over here, Y equals seven, X minus 14 is already isolated for X or sorry, isolated for Y. So I'm just gonna call that my equation A and by default, this is gonna be equation B. That's the first step. The second thing you wanna do is actually just solve that equation for X or Y, whichever variable is the easiest to do it. It actually doesn't matter. You can solve for X or Y. It really won't matter. All right. So the first two steps are already more or less done for us. Now, let's take a look at the third step, which is the actual substitution. We're gonna take this equation A and we're going to substitute it into B what does that mean? We have an expression in this equation? It's, we say we see that Y is equal to seven X minus 14. So wherever I see why in equation B, I'm actually just now going to replace it with that expression over here. All right. So now what does this become? Well, equation B just becomes two X minus Y. But now, instead of Y, I'm actually just going to replace it with the expression that I have over here seven X minus 14. All right. Now, this is gonna equal four over here. And now all they have to do here is they just have to solve the notice how after the substitution, we've gone from two equations in which both of them have two variables that are unknown. And now in this equation, we see that we've eliminated, we've, we've sort of gotten rid of the Y because we've substituted it. And now we only have X's and we know exactly how to solve for that. So this really just becomes uh this is gonna become two X and now we remember we have to distribute the minus sign into this uh uh this into this parenthesis. So this is gonna be two X minus seven X plus 14 equals four. Now we can solve for this uh this really just becomes negative five X like this. And then over here we subtract 14 from both sides. What we're gonna see is that this is negative five X equals negative 10. And if we solve this equation, we're just gonna get that X is equal to two, right? So notice how, now we've actually gotten one of our answers. We've gotten X equals two. This is one of our numbers that we, if we plug it into one of the equations, uh or either one of the equations will get true answers or true statements. So now that we've already gotten of the first half of our solution, which is X equals two, how do we get the Y value? Well, now that we're done with step number three, we're gonna move on to step number four, we're gonna plug the value that we just got this X equals two back into either equation. It actually doesn't matter whether you plug it back into A or B and then you just have to solve. So for example, we're gonna take this X equals two and I'm just gonna go ahead and plug it back into equation A because it's already solved for Y. So what you'll see here is if you plug this back into a, this is that Y equals seven and then instead of X, I'm actually just gonna write a two because that's what X equals, right? It just equals two minus 14. So if you look at this, this actually just equals 14 minus 14. And what you'll see here is that Y is equal to zero. All right. So here is the other half of my solution X equals two and Y equals zero. So these are the two numbers that if I plug them into these two equations, I'll get true statements for both. And if you don't believe me, we can actually move back to step number five, which just says that you should check your answer by plugging the values into equations. So let's go ahead and do that step number five. All right. So this is gonna be step number five. So for the blue equation, this says that Y equals seven X minus 14. But now I'm actually just gonna replace the values with X and Y. So instead of Y I plug in zero and instead of X I plug in two, so does zero equal seven times two minus 14. And actually, we'll see that this does end up being a true statement, right? Because zero equals 14 minus 14. And so in other words, zero equals zero. That's a true statement. Now, that was equation A over here. So it was equation A, let's do the same exact thing for B. So for B, what we have is that two X except, and whatever I see X and plug into minus Y, which is zero, does that equal four? And we'll see that four minus zero does actually equal four, which is a true statement. So if you ever are just uncertain about the values they've gotten, you can always plug them back into your equation just to double check. So anyway, that's the substitution method. Thanks for watching. Let me know if you have any questions.

Use substitution to solve the following system of linear equations.

$4x+y=1$

$x-y=4$

• A. $\left(-1,3\right)$
• B. $\left(-3,1\right)$
• C. $\left(3,-1\right)$
• D. $\left(1,-3\right)$

Use substitution to solve the following system of linear equations.

$4x+2y=7$

$x+5y=4$

• A. $\left(\frac12,\frac32\right)$
• B. $\left(\frac32,\frac12\right)$
• C. $\left(-\frac23,\frac13\right)$
• D. $\left(-\frac13,\frac23\right)$

Solving Systems of Equations - Elimination

Hey, everyone. Welcome back. So in another video, we looked at how to use the substitution method to solve a system of linear equations. The idea was that we would take a variable and isolated but like X equals two and we would plug it into another equation to get rid of one of the variables and make the equation simpler. This just becomes Y equals five times two minus three which we can solve. This just ends up being Y equals seven. And then we would solve it that way X equals two Y equals in this video. I'm gonna show you a different method called the elimination method of getting rid of one of the variables to make your equations simpler. All right. So it involves a few different steps. I'm gonna break it down for you. I'm show you that it's really not so bad. But the gist of it is that if I wanted to solve a system of equations like this, then I could use the substitution method. I absolutely could. But I'm gonna show you a different but much faster way to solve these kinds of problems. The basic gist of it is that instead of substituting, we're actually going to add the equations together. What do I mean by that? So we've added variables and we've added like expressions before adding equations is actually pretty straightforward. You line up the equations and then you would actually just add the coefficients of the XS and Ys and the numbers just straight down. So you just add everything top to bottom. You'll see here is that one and negative one actually will cancel out. It'll get rid of that X variable just like we got rid of it. In substitution method, the Ys will combine, we'll get two Y and then one in five becomes six. Now this equation is much easier to solve. It's just Y equals three. And then from here, we could actually solve the rest of the problem by plugging it back into one of the other equations. So the whole idea is that we want to add the equations together because we want to eliminate one of the variables. That's why we call it the elimination method. So I just want to point out here again, you're gonna use this method. Uh It's generally good to use this method whenever your equations are already in standard form because you can do stuff like adding the equations together or if they have large coefficients right now, problems won't always be the simple. So I'm actually gonna break it down for you step by step to show you how to do this let's go ahead and get started. So we have another uh example over here, we've got three X plus two, Y equals one and negative X plus Y equals three. Let's take a look at the first step here. First step is you're gonna want to write both of the equations in standard form in case they aren't already. And you want to align the coefficients vertically on top of each other. Basically what that means here is you want to line up the equation so that XS are on top of Xs Ys on Ys and then constant on constants. So this first step is already done for us because we have these equations in standard form. So the next thing we wanna do is just start the process of adding these things together. Now, if I try to do this right now, we're gonna see is if you try to add these equations together, you're gonna get three X and negative X becomes two, X, two Y and Y becomes three Y and one and three becomes four. So unlike what happened above, we didn't get rid of one of the variables. And that's because we didn't have Xs that canceled out or why is it canceled or anything like that? All right. So we can't automatically just go ahead and start adding these equations. So we're gonna have to do here is we're gonna have to multiply one of the equations or both of them by some number, it could be positive or negative. And the whole goal here is that we do want those Xs and Ys to cancel out. And the way that they, they do that is they have to be equal and opposite. It's kind of like up there in the other example that we did, we had one X and negative one X, those were equal and opposite. We're gonna have to do that here. So let's go ahead and just focus on this three X and negative three X, right? There's actually different ways to do this. But let's just focus on the three X and negative X. I want those things to the equal and opposite. So I'm gonna have to multiply one of the equations by some number in order to get there. So do I multiply this equation? Well, this one has the bigger coefficient. So I don't want to do that. I want to focus on the smaller equation. So this one, so can I multiply this equation by some number in order to get those coefficients to be the same? What if I legislate say for example, multiply by negative two? Well, if you multiply by negative two, what happens is I'm gonna get three X plus two Y equals one. And then over here, I'm gonna get, when you multiply all these coefficients by two by uh two, you're gonna get negative two X up plus two Y equals six. So does this help? Well, if you try to add these things together. Now, what you're gonna see is that you still don't cancel out a variable three X and negative two, X is X two Y and two Y is four, Y one and six makes seven. So this doesn't work because we didn't get the variables um or the coefficients to be equal and opposite. So multiplying it by two won't help us. Let's try to just a different number. What if we try now multiplying it by three and see what happens here. So if we have, if we do this, we're gonna see that three X plus two Y is equal to one. And then we get negative three X plus three Y again multiply all the coefficients by three equals and then three times three equals nine. Now if you try to add these two equations together, what you'll see is that one of the variables will cancel. So that second step is really important. It's also the trickiest step here and we're gonna go into much more detail later on. That second step is done. We have equal and opposite signs. Now, all we have to do is we just have to add the equations vertically to eliminate one of the variables. Once we add these two equations, we'll see is that negative three X and three X will cancel out leaving you just zero. And then what we see here is that we have two Y and three Y added together become five Y. So this becomes five, Y one and nine becomes 10. And so we'll see that Y equals two. And so we have one of our answers over here. Remember the answer to this system of equations is gonna be an X and Y pair that you plug into both of these equations. To get a true statement. We have one of those numbers already Y equals two. And now from here, we're really actually just gonna do the same exact thing that we did for substitution. All we have to do is just plug the value that we got back into either equation and then solve. In fact, these are exactly the same steps that we did for substitution. So really, we're all actually kind of already done with all the new stuff. All right. So we're gonna take this Y equals two and then plug it back into either one of these original equations. It really just doesn't matter which one you pick. I'm just gonna go ahead and pick the simpler one, which is this one over here? All right. So I'm gonna bring this second equation all the way down. And we're gonna see is that negative X plus Y. So we're not gonna plug in Y now because we know that Y is now equal to two and this is gonna equal uh three. And we just solve this equation subtract two from both sides, subtract two from both sides. And we're gonna see that negative X is equal to one and therefore X is equal to negative one. So we just flip the sign. And so that is the other variable that we need X is equal to negative one and Y is equal to two. All right. So that's step four. Now, I'm actually not gonna go ahead and solve for this uh this fifth do this fifth step. You can actually double check it yourself. I highly suggest that you pause the video and just double check that these numbers work for both of these equations. You should find that there's true statements for both of them. All right. So that's it for the elimination method. Hopefully, this makes sense. Uh Let's get uh go ahead and get some more practice.

How to Multiply Equations in Elimination Method

Welcome back everyone. So in an earlier video, I mentioned that step two of the elimination method was the trickiest step. This is where you had to look at your equations and figure out what to multiply them by one or both of them by some number positive or negative. So that your X's and Y coefficients would cancel out. This is a little bit of a trial and error method that we saw here. I'm gonna give you a little bit more information and give you a little bit more of a systematic way to figure out how to multiply your equations in this method. And it really just comes down to looking at the coefficients of each equation. So what I'm gonna do is actually just gonna walk you through the four different possible scenarios that you'll see. We'll walk through each one of these examples and I'm not gonna fully solve each one of them, but we're just gonna do steps two and three so that you can do the rest later on. Let's get, let's go ahead and get started here. So it really just comes down to the coefficients. Let's take a look at the first situation here, if your coefficients of X or Y are equal with opposite sign, then you actually just have to do nothing. You don't have to do anything and you can just go ahead and start the addition process. So really, if you have something like these two equations over here, seven X plus 13, Y equals 12, negative seven X equals two, Y equals 18, you don't have to do anything because these X's will just cancel already. So you'll just eliminate the X's, the 13 and two will become 15, Y 18 and 12 will become 30. And then you'll solve for this and you'll see that Y equals two and then after this, you can just solve the rest here. All right, let's stick to the second one here. What if the coefficients of X or Y are equal with the same sign? Well, what happens here is if you have something like five X plus seven, Y equals 17 and six X plus seven, Y equals 12, these things are the same, but they're the same sign as well. So if you add them, they won't cancel out. All you have to do in these situations is you have to multiply either by negative one. It really doesn't matter which one you do it to because either way you'll get now uh equal and opposite sign. What I'm gonna do is I'm just gonna multiply this top equation over here by negative one. So I'm just gonna multiply this um by yeah, negative one over here. So this just be careful. This doesn't mean X minus one. All right. So this just means multiply by negative one. OK. So if I sort of drop this down, what happens is basically all the coefficients will flip sign negative five X minus seven Y equals negative 17. Now, over here, I've got six X plus seven Y equals 12. Now, if you take a look here, we've basically just gotten back to this situation over here where we have equal and opposite signs. So we're done. All we have to do is just add them. What you'll see is that the Ys will cancel the negative five X and six X becomes X negative 17 and 12 becomes negative five. And then from here you have one of your answers and you can already just go ahead and plug it in. So the other one. All right. Now let's move to the second or third situation here. But you may actually see some situations in which um the coefficients of X or Y will be factors of each other. Basically just means that they're evenly divisible by each other. Let's take a look at the situation over here. We have 12, X minus five, Y equals 24 3, X minus two, Y equals six. Now, if you look at these two coefficients, five and two are not evenly divisible, but these are the 12 and the three. So what are you supposed to do in this situation? Well, you'll multiply the one, the equation with the smaller coefficients. And what do you multiply them by? Well, whatever is the quotient between these coefficients? 12, divided by three is four. So all I have to do is I have to multiply this uh equation here by four. And now if you do this, what you're gonna see if you multiply this by four, you're gonna get 12 and 12. So you may have to also insert a negative sign in somewhere. So what I'm gonna do is I'm gonna multiply this by negative four because then you're, you're gonna get your coefficients to cancel. So what this becomes over here is this becomes 12 X minus five Y equals 24. And then I have negative 12 X. So I basically just repeated this negative 12 X and then negative four times this two Y will actually become eight Y and then negative four times this over here will actually just become negative 24. Now, what happens is this basically just turns into the other situations, we'll see that the X's will cancel, leaving you just only with the Ys. So here we have eight Y and negative five Y becomes three Y. And now over here we have the 24 and negative 24 that will cancel. All right. So this just becomes zero, Y equals zero and, and, and then you can just finish the rest of the problem from there. All right. Now, if any of these three situations don't apply. So in other words, if your coefficients of X and Y are anything else, then one surefire method that will always work is you can always just multiply each equation by the other's coefficients. And you may exactly, and you may have to put a plus or minus sign in there. So for example, let's say we have something like six X, two Y equals 10. And then we have negative four X and three Y equals 15. So we can see that none of these things are evenly divisible into each other or factors. What we can do here is we can just multiply this equation over here just to cancel out the XS. I'm gonna multiply this equation by four. So I'm gonna multiply by four and I'm gonna multiply this equation over here by six. All right. So you kind of just like you almost just sort of like cross uh multiply each one of the equations by the other coefficients. OK. So what we're gonna see here is that now the first equation becomes 24 X plus you multiply 24 X, this becomes eight Y and then this over here will become negative 40. And then this over here will be negative 24 X, then this becomes a negative 18 X or ne negative 18 Y. And then this over here six and 15 becomes 90. Now, what you'll see is that they basically just turns into the other situations that the XS will cancel, leaving you with just the Ys. And what we'll see here is you have eight and negative 18 Y which is negative 10 and then the 40 sorry, negative 4090 become 50. What you're gonna see here is that Y is equal to negative five and then you could solve the rest of the problem using uh from there. All right. So really, if this situation will actually always work, but it's always just a good idea to look for any of any of these other three apply. It's also just, you know, sometimes you may be able to look at the Ys instead of the X's that may be easier. So there's a bunch of different ways to do this. So hopefully, this makes sense, folks, let me know if you have any questions and I'll see you in the next video.

Example 3

Everyone. Let's get started here. So we're gonna get some more practice with what to multiply your equation or equations by in order to eliminate one of your variables. And the whole idea is you take these two equations here, we're gonna have to multiply one or both of them by some number because we want either the X or the Y coefficients to cancel out when we add them. All right. So I just want to mention here that we're actually not going to fully solve the problem. And I also want to mention that there isn't actually one correct way to do this, there's actually many different correct answers. So without further ado let's go ahead and get started and break a desk down here. We've got these two equations, two X plus three Y equals one and X minus Y equals three. Now, remember here, we're just gonna have to figure out what to multiply these variables by to get them to cancel. So if you look at this, we could look at the X coefficients right. I've got two X and I've got one X. So notice how this is kind of like an implied one. Over here, let's just go through the different sort of possibilities. Are they equal with opposite signs? No, they're not. So we're not just gonna do this. Are there this equal with the same side? They're not either. So we're not gonna do any of this, but they are factors of each other. So in other words, you know, two is a factor or, or sorry, one is always a factor of anything else. Um So what you could do is you can multiply this equation with the smaller coefficients by the quotient. What does that mean? It means the equation with the smaller coefficients over here, we're gonna have to multiply by the difference or sorry, the quotient between these coefficients 2/1, which means we're gonna multiply it by just two. Now, one of the things you might notice here is if you multiply by two, which you're gonna get is two X on the bottom and notice how the both of those are gonna have the same sign. So whenever you're doing this, you could always just multiply by a negative number so that you get kind of get the, the signs to be opposite. So we're not gonna multiply by two, we're gonna multiply by negative two. So let's go ahead and do that. I bring the system of equations down. I'm gonna rewrite this two X plus three Y equals one. And then what happens is the negative two times one becomes negative two X, the negative two into the negative Y is going to become a positive two Y. It has to flip sign and the negative two going into the three is gonna be negative six. So everything gets multiplied by two and then flips the sign. All right. So now that we've done this, notice how when you sort of set these equations up and compare their coefficients. When you add these things together, the XS now will cancel and then the three Y plus the two Y will actually become five Y and then your one and negative six will actually just become negative five. All right. So again, uh what you're gonna do here is this is, you know, we don't have to fully solve this problem. Uh But what you'll see here is that you'll, you'll see that Y is equal to negative one. Now, once you get this, remember, you can always just plug these back into the other equations to figure out X. But we're not gonna do that because we don't want to actually fully solve the problem. So this is definitely one of the things that you could have done, you could have multiplied by negative two. But ma is gonna show you that there is another possibility you could have looked at this problem a little bit differently. So I'm gonna go ahead and rewrite this here and show you another sort of very common possibility that you could have seen. So two X plus three, Y equals uh one. And then I've got X minus Y equals three. Now, you could have looked at these two equations here and you could have looked at the fact the Y coefficients already have an opposite sign. And so you could have chosen to focus on those. And that would have been perfectly fine here. So again, let's go through the possibilities. Are they equal with opposite signs or not? Are they equal with the same sign? They're not, but they are factors of each other. Because again, we have a situation where we have a coefficient of one, right? There's kind of like an implied or hidden one that's over here. So what I can do is I can multiply the smaller coefficient by the quotients and the quotient 3/1 is just three. Notice how with this equation, we don't have to multiply by negative three because these, these coefficients are already opposite sign. All right. So what happens when we just multiply this equation by three, we're gonna get two X plus three Y equals one. And then the three goes into the one X just to become three X and then the three goes into the negative one Y to become negative three Y and then three goes into three to make nine. Now what happens is if you line these up and you add these together, which you'll see is that two X and three X become five X but then the three Y and negative three Y will end up canceling out. And then finally, what you'll see here is that the one and the nine become 10. So in this case, we chose to eliminate Y and solve for X. And what you would see here when you solve this is that X is equal to two. Now again, you could have just plugged this back into this equation over here to figure out the other variable. And you actually would have figured out that it was just xay equals negative one. So in fact, this kind of is actually the solution to the whole problem. Um So I just wanted to show you some different possibilities. Thanks for watching.

Example 4

All right folks. So let's get some more practice here figuring out what to multiply our equations by I've got these two equations over here. Five X plus three Y equals 10. And then I've got negative seven X plus five, Y equals 15. Let's go through a different possibilities and figure out what to multiply these equations by. All right. So are they equal with opposite signs? Are any of the coefficients, either the X or Y equal with opposite signs? Let's see, I got five and negative seven and then three and five. So definitely not. And they're also not equal with the same sign either. I've got all these different coefficients. Do I have any coefficients that are factors of each other? So for example, are five and negative seven factors of each other. No, are three and five factors of each other also, no, we've got no situation here where the factors where there are factors of each other. So we just have to multiply the smaller equation or the smaller coefficients by the quotient. So if it's none of these three, then remember kind of by default, we can just look, look at this one if it's anything else, then one situation that will always work no matter what is, we can multiply each equation by the other coefficients. So here's what I mean by this, I've got five and negative seven. Notice how these two things are already opposite signs. So what I can do is I can multiply this top equation by seven. And then I can multiply this bottom equation by five because notice how it's gonna happen is that seven times five will give you a number and five times seven will give you the same number. But the negative sign will make them opposites. So all I have to do is kind of just multiply them by each others, the others coefficients. And this is what happens here. So when I multiply this top equation by, by seven, I'm gonna multiply all the coefficients and seven times five becomes 35 X seven times three will become 21 Y and then seven times 10 will become 70. And on the bottom here, which you'll get is when you do five times negative seven, you'll get negative 35. See how now we've gotten the same numbers but opposite signs five times positive five becomes 20 five Y and then five times 15 will become 75. All right. Now when you go ahead and you add these two equations together, remember this will always work no matter what situation you have. Obviously, the math gets a little sort of tedious and you get, you might get some big numbers, but notice how the X coefficients will cancel as promised. And then all you have to do is just add the rest. So 21 and 25 will become 46 Y and then 7075 will become 100 and 45. Now again, the math is gonna be kind of ugly here, but you'll actually notice here that you can, can solve for this variable. So you just divide by 46 from both sides and which you would get is you would end up getting that Y is equal to 145 divided by 46. Now, that's not a clean number, but it actually doesn't matter because it still is going to be the solution to your Y variable. All right. So now you can take this Y variable and you can plug it into the other equations and you'd solve for the X variable. All right. But this is what you would multiply these equations by in order to get one of your uh coefficients to cancel. Thanks for watching. And I'll see you in the next one.

Use the elimination method to solve the following system of linear equations.

$2x+y=1$

$3x-y=4$

• A. $\left(-1,1\right)$
• B. $\left(1,1\right)$
• C. $\left(1,-1\right)$
• D. $\left(-1,-1\right)$

Use the elimination method to solve the following system of linear equations.

$10x-4y=5$

$5x-4y=1$

• A. $x=12,y=16$
• B. $x=16,y=12$
• C. $x=\frac34,y=\frac45$
• D. $x=\frac45,y=\frac34$

Classifying Systems of Linear Equations

Hey, everyone, welcome back. So early in the course, when we studied how to solve the most basic types of linear equations, we saw that we could categorize them based on the number or types of solutions that we got. What I'm gonna show you in this video is that we can actually do the exact same thing with a system of equations. So we can put them into three categories and it really just comes down to the number of solutions that we get. So what I'm gonna do in this video is walk you through these three examples and we'll talk about the difference between these types of problems that you'll see. And then really, we're just gonna go ahead and just sort of talk about the names. Let's go ahead and get started. We're just gonna jump right into the problems here. We're gonna solve each type of each system of equations. Notice how it doesn't specify whether we use substitution or elimination, then we'll graph them and categorize them. Let's get started with the first one over here. So in this first equation, we've got Y equals three and then we've got X plus Y equals two. In fact, we actually have the same equation through throughout the three examples. And that graph is already shown to us right over here. All right. So how do I solve this? Well, if I can, I can use substitution or elimination, which one is gonna be better here? Well, in this case, I already have one of the variables that's already isolated. So it's gonna be easy to substitute it into the other equation. So I'm gonna use substitution here. And really all that means is I'm gonna just go to replace the Y inside of the red equation with three. And so we've seen how to do this before. This is just X plus three equals two. Notice how we've gotten rid of one of the variables. Now we just solve this like a regular equation subtract three from both sides. And we'll get that X is equal to negative one and those are our answers. So we get that X is equal to negative one and Y is equal to three. Let's graph this and see if that makes sense. We already have the graph of the red equation and the blue equation is uh Y equals three. Remember that's just gonna be a horizontal line that has a Y value of three. So if you look at this, remember the solution to a system of equations is the place where they intersect. And if you look at this, this is exactly the point negative one. Comma three. And that makes perfect sense because those are the answers that we got, we got X is equal to negative one, Y is equal to three. All right. So we've seen how to solve these types of problems before now. They just get a name. This is a consistent system of equations. Whenever you think of consistent, you can just think of it makes sense and you get an answer out of it and this is called an independent system. All right, let's take a look at now. The second one, the second one has negative X minus Y equals negative two as the blue equation. So remember I can go ahead and solve this using elimination or substitution, which is better here. Well, in this case, now what I have is I actually have the two equations in standard form. And so I'm gonna use the elimination method. So this is gonna be elimination. And so what I'm gonna do here is look at the uh equations and do I have coefficients that are equal and opposite? I do? So in fact, I actually don't have to multiply these equations by anything. I can just go ahead and start adding them, which you'll see here is that the XS will cancel out, the Ys will cancel out. And also strangely, the twos will cancel out. So what are you left with? You're gonna left, you be left with some weird uh equation over here where zero X plus zero, Y equals zero, or in other words, zero equals zero. What does that mean? Well, I really just sort of like to sort of a picture that there's like a question mark here. And what you're left with is kind of like a statement like a numerical statement does zero equals zero. And actually it does, it is a true statement. So this is kind of a weird one where you've sort of seen that all of the variables have canceled out of your problem. And all you're left with is just numbers and those numbers actually do reflect a true statement. So let's go ahead and graph this equation and see what's going on here. So this blue equation over here, I can transform into slope intercept form. So just move the Y over to the other side and the negative two over to the other side, those things will flip signs, which you'll see here is that this is negative X plus two equals Y. So let's graph that in fact, what they actually see is that this is exactly the same line as the red line. So these things are exactly the same line because if you were to write the red equation in slope intercept form, you would see that this is Y equals negative X plus two. It's the same exact equation just flipped sort of backwards. So the reason this is a true statement is because these things are actually the same line. So there is actually an infinite number of possibilities of X's and Y values no matter what you pick where you plug them into these two equations. So you'll get true statements for both because they're the same exact line. So this is also a consistent system of equations because you actually get an answer for this. But this is called a dependent system. So here's the difference for a consistent system of equations, you'll get an answer. But in this case where you have an independent system, it means that the lines are intersecting. So think independent means intersecting. And it's the sort of most normal type of, of problem that you'll see where you actually get numbers for X and Y when you have a dependent system or will you ever have a problem with which the variables cancel? And you're just left with a true statement. Those lines are exactly the same. Let's take a look now at our third situation here and see what's different about that. So here we have Y equals negative X plus three, we already have something solved for Y. So just like the first problem I'm gonna use substitution. So I'm gonna plug this expression in for Y and see what happens. So this is X plus and then I have negative X plus three and this is gonna equal two. Notice now what happens is that when you solve for this, the X will cancel out with a negative X. So the variables just can just cancel out and all you'll be left with is the statement three equals two. So if you put a question mark around this, does this equal a true statement? Does three equal two? Well, clearly, it doesn't that's a false statement. So just like this second situation over here, the variables canceled out of our problem and you're left with a false statement. Why is that? Let's take a look at the graph here. If you try to graph this equation negative X plus three, what you'll see is that this is actually gonna be a line that is the, that is uh basically that has a slope of negative one, but it goes through the Y intercept of three, whereas the red equation has a Y intercept of two, but it's the same slope. So these lines are actually parallel and that's why you'll never find a pair of X and Ys that will satisfy both of the equations at the same time. So whenever you have this sort of situation where you end up with a false statement, what that means graphically is that the lines are parallel. So let's talk about and quickly summarize the difference between these three situations. When the lines are intersecting you clearly just get one solution right? One answer right here at the intersection point when the lines are the same in a dependent system, that means that you have an infinite number of solutions all of these solutions over here will satisfy this equation. All right. And then finally, when you have a uh an inconsistent one, an inconsistent system means that the answer doesn't make any sense and you won't get an answer for this. And that's because there are actually zero solutions, there are no solutions that will satisfy both of the equations at the same time. Really, that's really all of the different sort of categories that you'll see. This is obviously are the most common. You sometimes will see these and these are actually pretty rare. Um Anyway, so that's, that's it for this one. Thanks for watching and I'll see you in the next video.

Solve the following system of equations. Classify it as CONSISTENT (INDEPENDENT or DEPENDENT) or INCONSISTENT.

$y=5x-17$

$15x-3y=51$

• A. Consistent and Independent
• B. Consistent and Dependent
• C. Inconsistent

Solve the following system of equations. Classify it as CONSISTENT (INDEPENDENT or DEPENDENT) or INCONSISTENT.

$2x+8y=7$

$x+4y=19$

• A. Consistent and Independent
• B. Consistent and Dependent
• C. Inconsistent