Two Variable Systems of Linear Equations - Online Tutor, Practice Problems & Exam Prep

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 y=8x-4. But now, you might start to see problems that have multiple equations in them, and you'll be asked to solve these types of problems. This is what your book will refer to as a system of equations. In this video, we're only going to focus on linear equations. Basically, whenever you see a system of equations, it really just means that there are multiple equations like we have in our problem. I'm going to 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.

Remember that the solution to a single equation is really just trying to find x y pairs that satisfy this equation. For example, if we were asked to determine if each point was a solution to this equation, these three points over here, we could plug them into our equation and figure out if they make true statements, but we can also just graph them and see if they're on the line. Graphically, if these points are solutions, that means that they're on the line. For example, something like (−2, 0). If you plot that point, that's going to 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 lead to a false statement. How about this point, (0, −4)? Well, that's going to be over here. It's on the line, so, therefore, this is a solution to this specific equation. If you plug these numbers into this equation, you'll get a true statement. Now what about the point (1, 4)? If you plot that out, 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 where we have multiple equations, and I've 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. When we're asked to figure out if a point is a solution to the system of equations, what you're doing here is to solve these types of problems, you're going to find coordinate pairs or x y 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, (0, −4), 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 point (2, 5). If you plot that out, that's going to 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 (1, 4). So the point (1, 4), 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 of the lines. It's on both of these lines.

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. 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. And that's really all there is to it. I want to 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 are lots of points that satisfy one or the other, but there's only one that satisfies all of them. And that's really the main difference between a single equation and a system of equations. One of the things I want to mention is that this is true for most problems where you'll have just one solution, but there are other types of solutions, and we'll deal with them as we go along. Anyway, that's an introduction to a system of equations. Hopefully, that made sense. Thanks for watching.

Example 1

Everyone, let's get started with this example problem here. So, we're going to graph this system of equations that's given to us below, and because we're trying to identify what the intersection point is, 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=2x+3, notice how both of these equations are already in slope-intercept form. We have y=mx+b, y=something. Right? So 2x+3, what does that look like? Well, it passes through the point (0, 3) and it has a slope of positive 2. Remember, positive 2 means it's going to go up 2 and then over 1, so I'm going to put a point here, or we can go down 2 and to the left one. So if you connect these dots with a line, what you'll see is that this graph should look something like this.

Alright? So that's the equation, or that's the graph, y=2x+3. Let's do the same thing now for the red equation, which should be a little simpler because it's just xx+4. That means that it has a y-intercept of (0, 4). And instead of a slope of 2, it has a slope of 1. So what I'm going to do now is a slope of 1 means up 1 over 1, down 1 over 1. So, it's going to look a little bit different. And if you sort of draw the line that connects these points, it should look something like that. It should look something like that.

So that's the equation y=x+4. So we have to identify what 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 (1, 5). Alright? So with these two lines over here, the intersection is just going to be the point (1, 5).

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? Well, 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 going to take these x and y coordinates, and we're going to 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.

Alright? So it's pretty straightforward. What I'm going to do here is I'm just going to rewrite y=2x+3. And remember, what I'm going to do here is I'm just going to take these x and y values, the 5, and I'm going to plug that in for y. I'll take the 1 and replace that in for x, and then we'll solve this equation here.

Alright? So we get yisequalto2times1plus3. Alright? Now you can see pretty clearly here that the left and right sides will be the same, You'll just get a statement that just says 5 equals 5 on the left and right sides, and that is a true statement. 5 does equal 5. So this is a solution. Now we'll just do the same exact thing with the red equation, y=x+4. Do the same thing, so we're just going to replace y with 5 and x with 1, and you'll get that 5 is equal to 1+4. And just like above, we'll just get a true statement. 5 does equal 5. So because this point over here, (1, 5), when you plug it into both equations, you just get true statements for both of them, 5 equals 5, and we can say conclusively that this point is a solution to both equations. Alright? Thanks for watching.

Example 2

Everyone. So let's take a look at this example problem we're 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 shown here below. So, really, I'm just going to try to match equations to graphs. Let's take a look at the first one over here because I've got y=3x+5 and y=-2x+10. Notice how both of these are in slope-intercept form. So it's going to be a little bit easier to sort of try to match the equation with the graph than, for example, this equation, which is a little bit messier. Alright? So let's take a look at 3x+5. I'm looking for something that crosses the y-axis here at positive 5. So let's take a look at this first graph over here. I'm looking for if anything crosses at y=5 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=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 5. Alright? But let's just double-check the next equation, where this says y=-2x+10. So now I'm looking for a y-intercept of positive 10. Remember, this is like y=mx+b. Looking for the b term, which is 10. And if you look here, the blue equation does actually cross here at y=10. So that means that this is probably going to be equations for the graph c over here that I'm looking at. But just to sort of triple check, what we're going to do here is we're going to look at this intersection point, and we're going to plug it into both of these equations here. So remember, we're just going to take a look here, and we're going to plug these x and y values into their equations and see if we get true statements for both of them. Alright? So if you bring this equation down, what this says is that 8=3×1+5, and 8 does in fact equal 3+5, so you just get 8=8, which is a true statement. If you do it for the bottom equation, for the blue equation, what you'll see is you're going to get 8=-2×1+10, and 8 does in fact equal -2+10 because that just equals 8 as well. So this is definitely going to be a true statement. Alright? So let's take a look at now the second equation or second pair of equations. I've got y=4x+8. So if you look through my graphs, what you're looking for is remember, this is going to be in slope-intercept form. So I'm looking for something that crosses the axis at a positive 8. So does anything cross the axis at positive 8? It actually does over here, so that might be actually be the answer.

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 want to show you how to do in this video. I want to show you the substitution method. I'm gonna show you it's actually a set of really straightforward steps here, these 5 steps. And we're gonna see a lot of stuff that we've already seen how to do before, like plugging in numbers 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 going to be a two numbers, 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 = 7x - 14 , is already 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 want to 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.

Now let's take a look at the 3rd step, which is the actual substitution. We're going to take this equation, a, and we're going to substitute it into b. What does that mean? We have an expression in this equation. We see that y is equal to 7x - 14. So wherever I see y in equation b, I'm actually just now going to replace it with that expression over here. So now what does this become? Well, equation b just becomes 2x - y. But now instead of y, I'm actually just going to replace it with the expression that I have over here, 7x - 14. And now this is gonna equal 4 over here.

Now all I have to do here is I just have to solve b. Notice how after the substitution, we've gone from 2 equations in which both of them have 2 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 this is gonna become 2x - 7x + 14 = 4. Now we can solve for this. This really just becomes -5x, like this. And then over here, if we subtract 14 from both sides, what we're gonna see is that this is -5x = -10. And if we solve for this equation, we're just gonna get that x is equal to 2. So notice how now we've actually gotten one of our answers. We've gotten x equals 2. This is one of our numbers that if we plug it into one of the equations or either one of the equations, we'll get true answers or true statements.

So now that we've already gotten sort of the first half of our solution, which is x equals 2, how do we get the y value? Well, now that we're done with step number 3, we're gonna move on to step number 4. We're gonna plug the value that we just got, this x equals 2, 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 2, 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 says that y equals 7, and then instead of x, I'm actually just gonna write a 2 because that's what x equals. Right? It just equals 2 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 0.

So here is the other half of my solution, x equals 2 and y equals 0. 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 5, which just says that you should check your answer by plugging the values into both equations. So let's go ahead and do that, step number 5. So for the blue equation, this says that y = 7x - 14, but now I'm actually just gonna replace the values with x and y. So instead of y, I plug in 0, and instead of x, I plug in 2. So does 0 equal 7 \times 2 - 14? And actually, we'll see that this does end up being a true statement because 0 = 14 - 14, and so in other words, 0 = 0. That's a true statement. Now that was equation A over here. Let's do the same exact thing for B. So for B, what we have is that 2x - y, except wherever I see x I'm going to plug 2 into - y, which is 0, does that equal 4? And we'll see that 4 minus 0 does actually equal 4, which is a true statement. So if you ever are just uncertain about the values you'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 isolate it like x equals 2, 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=5×2−3, which we can solve. This just ends up being y=7, and then we would solve it that way,x=2,y=7. In this video, I'm going to show you a different method called the elimination method of getting rid of one of the variables to make your equations simpler. Alright? So it involves a few different steps. I'm going to break it down for you and 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 going to 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 x's and y's and the numbers just straight down. So you just add everything top to bottom. What you'll see here is that 1 and negative one actually will cancel out. It'll get rid of that x variable just like we got rid of it in the substitution method.

The y's will combine. We'll get 2y, and then 1 and 5 becomes 6. Now this equation is much easier to solve. It's just y equals 3. And then from here, we can 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 going to use this method. 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. Alright? Now problems won't always be this simple, so I'm actually going to 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 example over here. We've got 3x+2y=1 and −x+y=3. Let's take a look at the first step here. The first step is you're going to 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 x's are on top of x's, y's on y's, and then constants on constants. So this first step is already done for us because we have these equations in standard form. So the next thing we want to do is just start the process of adding these things together. Now if I try to do this right now, we're going to see is if you try to add these equations together, you're going to get 3x−x=2x, 2y+y=3y, and 1+3=4. So unlike what happened above, we didn't get rid of one of the variables. And that's because we didn't have x's that canceled out or y's that canceled out or anything like that. Alright? So we can't automatically just go ahead and start adding these equations.

What we're going to have to do here is we're going to 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 x's and y's to cancel out. And the way that they do that...

How to Multiply Equations in Elimination Method

Welcome back, everyone. So in an earlier video, I mentioned that step 2 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, 1 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 a more 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 4 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 2 and 3 so that you can do the rest later on. 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 signs, 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, \(7x + 13y = 12\), \(-7x + 2y = 18\), you don't have to do anything because these x's will just cancel already. So you'll eliminate the x's. The sum will become \(15y = 30\). And then you'll solve for this, and you'll see that \(y = 2\). And then you can just solve the rest from here.

Let's take a look at 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 \(5x+7y = 17\) and \(6x+7y = 12\), 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 equation by negative one. It really doesn't matter which one you do it to because either way, you'll get now equal and opposite signs.

What I'm gonna do is I'm just gonna multiply this top equation over here by negative one. So if I drop this down, what happens is basically all the coefficients will flip sign. \(-5x - 7y = -17\). Now, over here I've got \(6x + 7y = 12\). 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 y's will cancel, the \(-5x + 6x\) becomes \(x\), \(-17 + 12\) becomes \(-5\). And then from here, you have one of your answers, and you can already just go ahead and plug it and solve the other one.

Now let's move to the third situation here. But you may actually see some situations in which the coefficients of x or y will be factors of each other. Basically, it just means that they're evenly divisible by each other. Let's take a look at the situation over here. We have \(12x - 5y = 24\), \(3x - 2y = 6\). Now if you look at these 2 coefficients, they are not evenly divisible, but the \(12\) and the \(3\) are. So what are you supposed to do in this situation? Well, you'll multiply the equation with the smaller coefficient by whatever is the quotient between these coefficients, which is \(12 \div 3 = 4\).

So all I have to do is I have to multiply this equation here by \(-4\) because then you're gonna get your coefficients to cancel. So what this becomes over here is \(-12x + 8y = -24\). Now what happens is this basically just turns into the other situations. We'll see that the x's will cancel, leaving you only with the y's. Here we have \(8y - 5y = 3y\). And now over here, we have the \(24\) and \(-24\) that will cancel. So this just becomes \(0\), \(y = 0\), and then you can just finish the rest of the problem from there.

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 \(6x + 2y = 10\), and then we have \(-4x + 3y = 15\). 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 to cancel out the x's. I'm gonna multiply this equation by \(4\), and I'm gonna multiply this equation over here by \(6\).

So what we're gonna see here is that now the first equation becomes \(24x + 8y = 40\). And then this over here will be \(-24x - 18y = -90\). Now what you'll see is that basically just turns into the other situations; the x's will cancel, leaving you with just the y's. And what we'll see here is you have \(8 - 18y = -10y\), and then \(40 + 90 = 50\). What you're gonna see here is that \(y = -5\), and then you could solve the rest of the problem from there. So really, this situation will actually always work, but it's always just a good idea to look for any of these other three apply. It's also just, you know, sometimes you may be able to look at the y's 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 going to get some more practice with what to multiply your equation or equations by in order to eliminate one of your variables. But the whole idea is you could take these 2 equations here. We're going to have to multiply 1 or both of them by some number because we want either the x or the y coefficients to cancel out when we add them. Alright? 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 are actually many different correct answers. So, without further ado, let's go ahead and get started and break it down here. We've got these two equations, 2x+3y=1 and x-y=3. Now remember here, we're just going to 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 2x and I've got 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 going to do this. Are they equal with the same sign? They're not either, so we're not going to do any of this. But they are factors of each other. So in other words, you know, 2 is a factor or, sorry. 1 is always a factor of anything else. So what you could do is you could 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 going to have to multiply by the difference or, sorry, the quotient between these coefficients. 21, which means we're going to multiply it by just 2. Now one of the things you might notice here is if you multiply by 2, what you're going to get is 2x on the bottom, and notice how both of those are going to 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 the signs to be opposite. So we're not going to multiply by 2, we're going to multiply by negative 2. So let's go ahead and do that. If I bring the system of equations down, I'm going to rewrite this 2x+3y=1. And then what happens is the negative two times one becomes -2x. The negative two into the negative y is going to become a positive 2y, it has to flip sign. And the negative 2 going into the 3 is going to be negative 6. So everything gets multiplied by 2 and then flips the sign. Alright? 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 x's now will cancel. And then the 3y plus the 2y will actually become 5y. And then your one and negative six will actually just become negative 5. Alright? So again, what you're going to do here is this is you know, we don't have to fully solve this problem, 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 going to 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 2. But I'm actually going to show you that there is another possibility. You could have looked at this problem a little bit differently. So I'm going to go ahead and rewrite this here and show you another very common possibility that you could have seen. So 2x+3y=1, and then I've got x-y=3. Now you could have looked at these two equations here, and you could have looked at the fact that 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? They're not. Are they equal with the same signs? They're not, but they are factors of each other because, again, we have a situation where we have a coefficient of 1. 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 quotient, and the quotient 31 is just 3. Notice how with this equation we don't have to multiply by negative 3 because these coefficients are already opposite signs. Alright? So what happens when we just multiply this equation by 3? We're going to get 2x+3y=1, and then the 3 goes into the x just to become 3x, and then the 3 goes into the negative one y to become negative three y, and then 3 goes into 3 to make 9. Now what happens is if you line these up and you add these together, what you'll see is that 2x and 3x become 5x, but then the 3y and negative three y will end up canceling out. And then finally, what you'll see here is that the 1 and the 9 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 2. 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 x y equals negative 1. So in fact, this kind of is actually the solution to the whole problem. So I just wanted to show you some different possibilities. Thanks for watching.

Example 4

Alright, 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, 5x + 3y = 10. And then I've got -7x + 5y = 15. Let's go through our different possibilities and figure out what to multiply these equations by. Alright? 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 5 and negative 7 and then 35, so definitely not. 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 5 and negative seven factors of each other? No. Are 35 factors of each other? Also, no. We've got no situation here where the factors or their 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 3, then remember, kind of by default, we can just 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 5 and negative 7. Notice how these two things are already opposite signs. So what I can do is I can multiply this top equation by 7, and then I can multiply this bottom equation by 5. Because notice what's going to happen is that 7 times 5 will give you a number, and 5 times 7 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 other's coefficients, and this is what happens here. So when I multiply this top equation by 7, I'm gonna multiply all the coefficients and 7 \times 5 becomes 35x, 7 \times 3 will become 21y, and then 7 \times 10 will become 70. And on the bottom here, what you'll get is when you do 5 \times (-7), you'll get -35x. See how now we've gotten the same numbers, but opposite signs? 5 \times 5 becomes 25y, and then 5 \times 15 will become 75. Alright? 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 we have to do is just add the rest. So 21y + 25y will become 46y, and then 70+75 will become 145. Now again, the math is going to be kind of ugly here, but you'll actually notice here that you can solve for this variable. So you just divide by 46 from both sides, and what you would get is you would end up getting that y = \frac{145}{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. Alright? 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. Alright? But this is what you would multiply these equations by in order to get one of your 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 going to 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 3 categories, and it really just comes down to the number of solutions that we get. So what I'm going to 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 going to go ahead and just talk about the names. Let's go ahead and get started. We're just going to jump right into the problems here. We're going to solve each type of or 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.

In this first equation, we've got y=3, and then we've got x+y=2. In fact, we actually have the same equation throughout the 3 examples, and that graph is already shown to us right over here. Alright? So how do I solve this? Well, if I can use substitution or elimination, which one is going to be better here? Well, in this case, I already have one of the variables that's already isolated, so it's going to be easy to substitute it into the other equation. So I'm going to use substitution here. And really all that means is I'm going to just replace the y inside of the red equation with 3. And so we've seen how to do this before. This is just x+3=2. Notice how we've gotten rid of one of the variables. Now we just solve this like a regular equation. Subtract 3 from both sides, and we'll get that x=−1. And those are our answers. So we get that x=−1 and y=3. Let's graph this and see if that makes sense. We already have the graph of the red equation, and the blue equation is y=3. Remember, that's just going to be a horizontal line that has a y value of 3. So if you look at this, remember that the solution to a system of equations is the place where they intersect. And if you look at this, this is exactly the point −1, 3. Now it makes perfect sense because those are the answers that we got. We got x=−1, y=3. Alright? 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.

Let's take a look now at the second one. The second one has −x−y=−2 as the blue equation. So remember, I can go ahead and solve for this using elimination or substitution. Which is better here? Well, in this case, now what I have is I actually have the 2 equations in standard form, and so I'm going to use the elimination method. So this is going to be elimination. And so what I'm going to do here is look at the 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. What you'll see here is that the x's will cancel out, the y's will cancel out, and, also, strangely, the twos will cancel out. So what are you left with? You'd be left with some weird equation over here where 0x + 0y equals 0, or in other words, 0 equals 0. What does that mean? Well, I really just like to sort of 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 0 equal 0? And, actually, it does. It is a true statement. So this is kind of a weird one where you've 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 −2 over to the other side. Those things will flip signs. What you'll see here is that this is −x+2=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=−x+2. 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, then you'll get true statements 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 problem that you'll see, where you actually get numbers for x and y. When you have a dependent system or where 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 3rd situation here and see what's different about that. So here we have y=−x+3. We already have something solved for y. So just like the first problem, I'm going to use substitution. So I'm going to plug this expression in for y and see what happens. So this is x+−x+3=2. Notice now what happens is that when you solve for this, the x will cancel out with a −x, so the variables will just cancel out. And all you'll be left with is the statement 3 equals 2. So if you put a question mark around this, does this equal a true statement? Does 3 equal 2? Well, clearly, it doesn't. That's a false statement. So just like the 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, −x + 3, what you'll see is that this is actually going to be a line that is, basically, that has a slope of −1, but it goes through the y intercept of 3. Whereas the red equation has a y intercept of 2, 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 y's 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. Alright? And then finally, when you have an 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. That really that's already all of the different sort of categories that you'll see. These obviously are the most common. You sometimes will see these, and these are actually pretty rare. Anyway, so 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