Equations with variables on both sides
Let's try to solve a more involved equation. So, let's say that we have 2x plus 3, 2x plus 3 is equal to is equal to 5x minus 2. So this might look a little daunting at first. We have x's on both sides of the equation. We're adding and subtracting numbers. How do you solve it? And we'll do it in a couple of different ways. The the important thing to remember is, we just want to isolate an x. Once you've isolated an x, you have x equals something. Or x equals something. You're done, you've solved the equation. You can actually go back and check whether that works, So what we're going to do is just do a bunch of operations on both sides of the equation, to eventually isolate the x. But while we do those, I actually want to visualize what's occurring. Because I don't want you just say, oh what are the rules or the steps of solving equations. And I forgot whether this is a allowed or that isn't allowed. If you visualize what's happening, it'll actually be common sense what's allowed. So let's visualize it. So we have 2x right here on the left-hand side. So that's literally, that's x plus x. And then you have plus 3. Plus 3, I'll do it like this. So that's equal to plus 1, plus 1, plus 1. That's the same thing as 3. I could've drawn 3 circles here as well. Let do the same color. Plus 3. And then that is equal to 5 x's. Do that in blue. That is equal to 5 x's. So, 1, 2, 3, 5, 6. And I want to make it clear. You would never actually have to do it this way when you're solving the problem. You would just have to do the algebraic steps. But I'm doing this for you so you can actually visualize what this equation is saying. the left-hand side is these two orange x's plus 3. The right-hand side is 5x minus 2. So minus 2, we could write as -- so let me do this in a different color, I'll do it in pink. So, minus 2, I'll do as minus 1 and minus 1. Now, we want to isolate the x's on the same side of the equation. So, how could we do that? Well, there's two ways of doing it. We could subtract these two x's from both sides of the equation. And that would be pretty reasonable. Because then you'd have 5 x's minus the 2 x's. You'd have a positive number of x's on the right-hand side. Or, you could actually subtract 5x from both sides. And that's what's neat about algebra. As long as you do legitimate operations, you will eventually get the right answer. So let's just start off subtracting 2x from both sides of the equation. And what I mean there, I mean we're going to remove 2 x's from the left-hand side. And if we were to move 2 x's on the left-hand side, we have to remove 2 x's the right-hand side. Just like that. So what does that give us? We're subtracting 2 x's. 2 x's from the left. And we're also going to subtract 2 x's from the right. Now, what does our left-hand side simplify to? We have 2x plus 3 minus 2x. The 2 x's cancel out. So you're just left with -- you're just left with the 3. And you see that over here. We took 2 of these x's away. We're just left with the plus 1, plus 1, plus 1. And then on the right-hand side, 5x minus 2x. We have it over here. We have 5 x's minus 2 x's. You only have 1, 2, 3, x's left over. 3 is equal to 3x. And then you have your minus 2 there. You have your minus 2. So, normally if you were to do the problem, you would just have to write what we have here on the left-hand side. So what can we do next? Remember, we want to isolate the x's. Well, we have all of our x's on the right-hand side right here. If we could get rid of this negative 2, off of the right-hand side, then the x's will be alone. They'll be isolated. So how can we get rid of this negative 2, if we visualize it over here. This negative 1, this negative 1. Well, we could add 2 to both sides of this equation. Think about what happens there. So, if we add 2, so I'm going to do it like this. Plus 1, plus 1. So you could literally see. We're adding 2. And then we're going to add 2 to the left-hand side. 1 plus, 1 plus. What happens? Let me do it over here as well. So we're going to add 2. We're going to add 2. So what happens to the left-hand side? 3 plus 2 is going to be equal to 5. And that is going to be equal to 3x minus 2 plus 2. These guys cancel out. And you're just left with 3x. And we see it over here. We have the left-hand side is 1 plus 1 plus 1 plus 1 plus 1. We have 5 1's, or 5. And the right-hand side, we have the 3 x's, right over there. And then we have the negative 1, negative 1. Plus 1, plus 1, negative 1, these cancel out. They get us to 0. They cancel out. So we're just left with 5 is equal to 3x. So we have 1, 2, 3, 4, 5 is equal to 3x. Let me clear everything that we've removed, so it looks a little bit cleaner. These are all of the things that we've removed. Let me clear that out. And then let me clear that out, like that. Edit. Clear. So now we are just left with 1, 2, 3, 4, 5. Actually, let me move this over. So I could just move this over right over here. We now have 1, 2, 3, 4, 5. These are the two that we added here, is equal to 3x. These guys canceled out. That's why we have nothing there. Now, to solve this, we just divide both sides of this equation by 3. And this is going to be a little hard to visualize over here. But if we divide over here both sides by 3, what do we get? We divide the left by 3. We divide the right by 3. The whole reason why we divided by 3 is because the x was being multiplied by 3. 3 is the coefficient on the x. Fancy word, it literally just means the number multiplying the variable. The number we're solving, the variable we're solving for. So these 3's cancel out. The right-hand side of the equation is just x. The left-hand side is 5/3. So 5/3, we could say is is equal to 5/3. And this is different than everything we've seen so far. I now have the x on the right-hand side, the value on the left-hand side. That's completely fine. This is the exact same thing as saying 5/3 is is equal to x is the same thing as saying x's equal to 5/3. Completely equivalent. Completely equivalent. We sometimes get more used to this one, but this is completely the same thing. Now, if we wanted to write this as a mixed number, if we want to write this as a mixed number, 3 goes into 5 one time with remainder 2. So it's going to be 1 2/3. So it's going to be 1 2/3. So we could also write that x is equal to 1 2/3. And I'll leave it up for you to actually substitute back into this original equation. And see that it works out. Now, to visualize it over here, you know, how did he get 1 2/3, let's think about it. Instead of doing 1, I'm going to do circles. I am going to do circles. Actually, even better, I'm going to do squares. So I'm going to have 5 squares on the left-hand side. I'll do it in this same yellow color right here. So I have 1, 2, 3, 4, 5. And that is going to be equal to the 3 x's. x plus x plus x. Now, we're dividing both sides of the equation by 3. We're dividing both sides of the equation by 3. Actually, that's where we did it up here, we divided both sides by 3. So how do you do that right-hand side's pretty straightforward. You want to divide these 3 x's into 3 groups. That's 1, 2, 3 groups. 1, 2, 3. Now, how do you divide 5 into 3? And they have to be thought through even groups. And the answer tells us. Each group is going to be 1 2/3. So, 1 2/3. So it's going to be 2/3 of this, the next one. And then we're going to have 1 2/3. So this is 1/3. We're going to need another. Another 1, so this is 1 1/3. We're going to need 1 more 1/3, so this is going to be right here. And then we're left with 2/3 and 1. So we've broken it up into 3 groups. This right here. Let me make it clear. Let me make it clear, this right here is 1 2/3. 1 2/3. And then this right here, this 1/3. That's another 1/3, so that's 2/3, and then that's 1 right there. So that's 1 2/3. And then finally this is 2/3 and this is 1, so this is 1 2/3. So when you divide both sides by 3 you get 1 2/3. Each section, each bucket, is 1 2/3 on the left-hand side. On the left-hand side, or 5/3. And on the right-hand side we just have an x. So it still works. A little bit harder to visualize with fractions.