Reasoning with linear equations | Solving equations & inequalities | Algebra I | Khan Academy

In this video, we're going to try to solve the equation (3 \cdot x + 1 - x = 9). And like always, I encourage you to pause this video and try to work through this on your own. But the emphasis of this video is to not just get to the right answer, but to really understand what's happening when we do algebraically valid operations on either side of this equation. Alright, so let's begin.

So the first thing that my brain wants to do—and we've talked about this before—there are many different ways to often solve an equation, but my brain wants to simplify. The first thing that looks a little bit hairy here is I have (3 \cdot) the expression (x + 1). So I might be able to simplify that if I multiply (3) times (x) and (3) times (1).

And so, if I did that, I would get (3x + 3 - x = 9). The key here is that the left-hand side of the second equation is equivalent to the left-hand side of the top equation. If the left-hand side of the top equation is equal to (9), well, so the left-hand side of this bottom equation is going to be equal to (9) as well. Another way to think about it is, if the entire top equation is true, the entire bottom equation must be true, and vice versa.

But let's keep going. So what's the next thing that we could do here if we're at least just trying to simplify the left-hand side? Well, my brain immediately sees I have three (x)s here, and then I want to take away an (x) right over there. If I have three (x)s and then I take away an (x), how many (x)s do I have? Well, then I have two (x)s, and then I still have that (+ 3). (3 + 3 = 9).

Once again, this left expression is equivalent to this left expression, which is equivalent to this top left expression. If any one of these equations are true, that must mean that all of these equations are true. But we could keep going. Now, the next thing I like to do is isolate the (x) terms on one side. So I want to have a left side that just has this (2x) there.

Well, the way to do that is I would have to subtract (3) from the left side. But, as we know, anything that we add or subtract or multiply or divide by on one side of the equation, we have to do on the other side of the equation. So I'll subtract (3) right over there.

Why does that make sense? Well, if (2x + 3) is truly equal to (9), then if I take (3) away from (2x + 3), if I just take (3) away, it's not necessarily going to be equal to (9) anymore. So I have to take (3) away from (9) in order for the equivalence of both sides of the equation to be true.

And so, what do I get? Well, then (3) and (-3) cancel out, and on the left-hand side, I'm just left with (2x), and on the right-hand side, I have (9 - 3 = 6). Now, I keep emphasizing it because really that's the point of this video: that (2x = 6) is an equivalent equation to (3 \cdot x + 1 - x = 9). Because we've done these algebraically valid operations, we'd be able to maintain the equality.

We've been able to say, "Look, if (2x) is equal to (6), then we know that (3 \cdot x + 1 - x) is equal to (9), and if this top equation is true, then we also know that this last equation is true." But when we're in the home stretch, what can we do to solve for (x)?

Well, if we just want to isolate an (x) here on the left-hand side, as you can imagine, we can divide the left-hand side by (2). But if we divide the left-hand side by (2), in order to maintain this equality, we have to divide the right-hand side by (2). Remember, an equation has the word equality in it, or at least the first part of the word equality.

If we multiply, divide, add, or subtract from one side, then we have to do it on the other side. So we are left with (x = \frac{6}{2}). Thus, (x = 3), and we have solved it.

Really, (x = 3) continues to be an equation, and so the equation (x = 3) is going to be true if this top equation is true, and this top equation is going to be true if the equation (x = 3) is going to be true.

Now let me finish off with a little bit of an interesting challenge for you. If I have (5x = 6x), one temptation might be—it kind of looks like this last step we had over here—so why don't we divide both sides by a common factor?

Maybe we could divide both sides here by (x), because this is (5) times (x) and this is (6) times (x). What would we get if we did that? Let's see, we would get (5 = 6), which it clearly is not equal to (6). But what just happened here? Is it the fact that (5x) can never be equal to (6x), and we did algebraically valid operations and we got (5 = 6), or did we do something wrong?

Pause this video and think about it. Well, some of you might have realized that if (x) is an arbitrary non-zero number, then you know that you could divide by (x). But what if (x) is zero? If (x) is zero, you can't divide by zero.

That's actually what's going on here, because if you do algebraically valid operations you will actually see that this is a case where (x) is going to have to be equal to zero. So let me rewrite it.

This is (5x = 6x). Even though it looks a lot like this last step that we had in the first equation and you're tempted to immediately kind of knee-jerk to divide, you have to realize that I have an (x) term on both sides.

Let me combine them. So I could subtract (5x) from both sides, or I could subtract (6x) from both sides. Let me subtract (5x) from both sides and then see what happens.

We are going to get on the left-hand side (5x - 5x = 0) and then that's equal to (6x - 5x = x). So we get the solution for that original equation: (5x = 6x) is indeed (x = 0).

So the big takeaway here is to appreciate the equivalence of these equations. If you do algebraically valid operations—hopefully, it's trying to make some intuitive sense why certain operations are valid and why other operations are not valid—but if you do valid operations, it's really saying that each step of your solution, each of those equations, are equivalent to the equations before it.

If one of them is true, then the others are true. But if you do an algebraically invalid operation, like you're dividing by (x) and (x) could be equal to zero, then you can start running into problems.