Reasoning with systems of equations | Equivalent systems of equations | Algebra I | Khan Academy

So let's say I had the equation (2x + y = 8). This is a single equation with two unknowns, and there are many different (xy) pairs that would satisfy this equation. Now let's add a second equation: (x + y = 5). Once again, if we only looked at this second equation, there are many different (xy) pairs that would satisfy it. You could have (x = 4), (y = 1); (x = 3), (y = 2); many, many, actually there's an infinite number that would satisfy this right over here.

But what's interesting about a system of equations is you're using both of these equations as constraints. You're saying, is there at least one (xy) pair that would satisfy both of these equations? And as we'll see in many future videos, this is a very useful thing to think about in many, not only fields of mathematics but in many fields of knowledge generally.

But the focus of this video is to think about why the operations we can perform on either or both of these equations, why they are valid and why they're intuitive. So the first series of operations is when we just try to manipulate one equation by themselves. For example, we could multiply both sides of this purple equation, this top equation, by negative one. You would get (-2x - y = -8).

Now all I did right over here, this is you could do this even if we weren't thinking about a system, is (2x + y) is truly equal to (8). We're assuming that that is true because they're telling us that's true. So if (2x + y) is equivalent to (8), then the negative of (2x + y) should be equivalent to (-8). Or another way to think about it: if both of these sides are equal, if I multiply the left side by something, in order for the equality to hold true, I have to multiply the right side by the same thing.

So this equation on the right, this purple equation, is an equivalent equation to our original one. It looks different, but the same (xy) pairs that satisfy this right equation are going to satisfy this left equation and vice versa.

Now another operation that sometimes feels a little bit less intuitive when you first learn it, when you're solving systems, is when you add two equations together. So for example, we have that purple equation, the one that we've now multiplied by negative one, and now we have our original, I guess that's teal or blue equation, so (x + y = 5).

And we learn when we're solving systems of equations that we can get a new equation by adding the two left sides and adding the two right sides. So you might have seen something like this. When we add the two left sides, let's see, (-2x + x) would be (-x), and then (-y + y), well, that's just going to cancel out, so we have no (y)'s left. And then that's going to be equal to (-8 + 5), which is equal to (-3).

And before I even go on to try to solve this, why were we able to do that? Pause this video and think about that. Well, let me give you an example. If we had started with (-2x - y = -8) just as a single equation, and if I added (5) to both sides of that, so if I added (5) on the left-hand side and I added (5) on the right-hand side, I think that would have made intuitive sense to you.

(-2x - y + 5 = -8 + 5). Hopefully, this is a little bit intuitive because, once again, and I'm really saying the same thing over and over again, the left side is truly equal to (-8). So if I add (5) to it, it's still going to be truly equal to (-8 + 5).

So hopefully this makes a little bit of intuitive sense. The key realization with what we did up here is we essentially added (5) to both sides. You might say, well no, we only added (5) to the right-hand side, but remember (x + y) we are saying is equal to (5). It’s just like adding the same thing to both sides of the equation.

And then when you do that, that's where we essentially were able to eliminate this (y) variable, and now we got one equation with one unknown. From there, you can just do a valid algebraic operation. You could say, "Okay, I just want (x) over here." What if I were to divide both sides by (-1)?

Once again, because (-x = -3), if I divide (-1), if I divide (-x) by (-1), I also have to divide (-3) by (-1) in order to maintain the equality. So then you're going to get (x = 3).

And so that would be the (x) value of that (xy) pair that satisfies both. And then to figure out the (y) value, you would say, "Alright, if (x = 3), I should be able to go back into either of these equations to find the corresponding (y) value."

And it's a little bit easier to go into that second one. You could say, "Alright, 3 + y = 5." (3 + y) must be equal to (5), and then of course, if you subtract (3) from both sides, because once again we're saying (3 + y) is literally equal to (5), then you're going to get (y = 2).

And so we found an (xy) pair that satisfies both equations. Really, everything that we wrote down over here, these are all equivalent statements. One of them is going to be true if and only if the other statements, the other equations are also true.

So (2x + y = 8) and (x + y = 5) if and only if (x = 3) and (y = 2) if and only if (-2x - y = -8) if and only if (-x = -3).