Linear equations with variables on both sides
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Why we do the same thing to both sides: Variable on both sides
Alright, now we have a very interesting situation. On both sides of the scale, we have our mystery mass and now I'm calling the mystery mass having a mass of Y. Just to show you that it doesn't always have to be X. It can be any symbol as long as you can keep track of that symbol. But all these have the same mass. That's why I wrote Y on all of them. And we also have a little 1 kilogram boxes on both sides of the scale. So the first thing I wanna do, we're gonna do step by step and try to figure out what this mystery mass is. But the first thing I wanna do is, is, is, have you think about, whether you can represent this algebraically? Whether with, with a little bit of mathematic symbolry, you can represent what's going on this scale. Over here, I have three Ys and three of these boxes and their total mass is equal to this one Y. And I think I have about let's see, I have 7 boxes right over here. So I'll give you a few seconds to do that. So let's think about the total mass over here. We have 3 boxes and a mass Y. So they're going to have a mass of 3Y, and then you have 3 boxes with a mass of 1 kilogram. So they're going to have a mass of 3 kilograms. Now over here, I have 1 box with a mass of Y kilograms. So that's going to be my 1Y right over there. I could've written 1Y but that's, I don't need to. A Y is the same thing as 1Y. So I have the Y kilograms right there. And I have seven of these, right? 1,2,3,4,5,6,7. Yup, seven of these. So I have Y plus 7 kilograms on the right-hand side. And once again, it's balanced. The scale is balanced. This mass, total mass is equal to this total mass. So we can write an equal sign, right over there. So that's a good starting point. We were able to represent this situation to this real-life simple situation. You know, back in the day, when people actually had to figure out the mass of things if you were to go the jewelry store, whatever. They actually did had problems like this. We were able to represent it mathematically. Now the next thing to do is, what are some reasonable next steps? How can we start to simplify this a little bit? You know, once again, I'll give you a few seconds to think about that. Well, the neat thing about algebra is there's actually multiple paths that you could go down. You might say, well why won't we remove 3 of these what, of these yellow blocks from both sides? That would be completely legitimate. You might say, well, why won't we remove 1 of these Ys from both sides? That also would be legitimate. And we could do it in either order. So let's just pick one of them. Let's say that we've first want to remove, let's say that we first want to remove the, a Y from either sides. Just so that we feel a little bit more comfortable with all of our Ys sitting on the 1 side. And so the best way, if we don't want all of our Ys to sit on the 1 side, we can remove, we can remove a Y from each side. Remember, if you removed a Y from only 1 side, that would unbalance the scale. The scale was already balanced whatever you have to do to the one side after you do the other. So I'm gonna remove a Y, I'm gonna remove Y mass from both sides. Now what will that look like algebraically? Well. I remove the Y from both sides. So I subtracted Y from the left-hand side and I subtracted Y from the right-hand side. That's exactly what I did. The mass, it had a massive Y. I don't know what that is but I did take it away. I lifted that block, that little, that little block. And so, on the left-hand side, on the left-hand side, what am I left with? That you can pick it up mathematically, you can even look up here, and look up here what you're left with. If I had 3 of something, and I take away 1 of them, if I take away 1 of them, I'm left with 2 of that something. So I'm left with 2Y, right over here, you see it. I had 3, I got rid of 1, so I'm left with 2. And I still have those 3 yellow blocks. So I still have those 3 yellow blocks. On the right-hand side, I had a Y, I took away the Y, and so now I have no Ys left. We see it visually right over here. And I still have 7 of the yellow blocks. So I still have 7 of the yellow blocks. And since I took the exact same mass from both sides of the scale, the scale is still going to be balanced. It was balanced before, I took away the same thing from both sides. And so the scale is still, is still balanced. So this is going to be equal to that. Now, now just trying to look a bit similar to what we saw in that last video. But I will ask you, what can we do from this point? What can we do from this point to simplify it further or so, even better, think of it so we could isolate the, these Ys on the left-hand side. And I'll give you a few seconds to think about that. Well, if we want to isolate these Ys on the left-hand side, these 2 Ys, the best way is to get rid of this 3, to get rid of these 3 blocks. So why won't we do that? Let's take 3 blocks from this side but we can't just take it from that side if we want to keep it balanced. We have to do it to this side too. We gotta take away, we gotta take away 3 blocks. So we're subtracting 3 from that side, and subtracting 3 from the right side. So on the left-hand side, on the left-hand side, we're gonna be left with just these 2 blocks of mass Y. So our total mass is now going to be 2Y. These 3 minus 3 is 0, and you see that here. We're just left with 2Ys right over here, and on the right-hand side, we got rid of 3 of the blocks. So we only have 4 of them left. So you have 4 of them left. So you have 2 of these Y masses is equal to 4 kilograms. And because we did the same thing to both sides, the scale is still balanced. And now, well how do we solve this? And you might be able to solve this in your head. I have 2 times something is equal to 4. And you can kind of think about what that is, but if we want to stay true to what we've been doing before, let's think about it. I have 2 of something, is equal to something else. What if I multiply both sides by 2? Oh, sorry, what if I multiply both sides by 1/2 or another way is dividing both sides by 2. If I multiply this side by 1/2, if I essentially take away half of the mass, or I'll only leave half of the mass, Then, I'm only gonna have 1 block here, and if I take away half of the mass over here, I'm gonna have to take away 2 of these blocks right over there. And what I just did, you could say I multiplied both sides by 1/2 or just for a sake of a little change, You could say I divided both sides by 2. And on the left-hand side, I'm left with a mass of Y. And on the right-hand side, I'm left with a mass of 4 divided by 2 is 2. And once again I can still write this equal sign because the scale is balanced. I did the exact same thing to both sides. I left half to what was on the left-hand side, and half of it was on the right-hand side. It was balanced before, half of each side, so it's going to be balanced again. But there, we've done it. We've solved something that's actually not so easy to solve. Or it might not look so easy first. We figured out that our mystery mass Y is 2 kilograms. And you can verify this, this is the really fun thing about algebra. Is it, once you get to this point, you can go back and think about whether the original, the original problem you saw made sense. Let's do that. Let's think about whether the original problem made sense. And to do that, I want you to, I want you to calculate. Now that we know what the mass Y is it's 2 kilograms. What was the total mass on each side? Well, let's calculate it. We have 2 right, right over here, this is 2 kilograms, I'll do that in purple color,so this is a 2, this is a 2, this is a 2. So we had 6 kilograms plus these 3, we had 9 kilograms on the left-hand side. And on the right-hand side, I had these 7 plus 2 here, 7 plus 2 is 9 kilograms. That's why it was balanced. Our mystery mass, we had 9 kilograms, total on both sides.