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 three 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 I 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 go to 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 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 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 negative one comma 3. Now it makes perfect sense because those are the answers that we got. We got x is equal to negative one, y is equal to three. 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. Alright?

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're going to be left with some weird equation over here where 0x+0y=0, or in other words, 0=0. What does that mean? Well, I really just sort of 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 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 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 a true 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 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 third 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 ca...