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### Course: Algebra basics>Unit 3

Lesson 3: Multi-step equations

# Why we do the same thing to both sides: Variable on both sides

We will use a scale to try to achieve balance and help explain why we do the same thing to both sides of an equation, even when the variable appears on both sides. Created by Sal Khan.

## Want to join the conversation?

• Why did he not divide 3 first and how would it effect the problem
(40 votes)
• He elected to eliminate the variable on the right side using the subtraction property of equality first. He absolutely could have used the division property of equality first by dividing by 3... Most likely he choose to avoid doing this first because it would have resulted in fractions making all subsequent steps more time consuming... Regardless the outcome would have been the same.
(79 votes)
• if in a question such as 16−2t=5t+9 do you subtract 2t from both sides or add or is it subtracting 5t from both sides or add. Why?
(18 votes)
• Subtracting both sides by 2t or 5t would still make the equation true ,but if you're trying to solve the equation, then subtracting both sides by 2t is the way to go because subtracting both sides by 5t is useless when you're trying solve the equation.

Remember! You're trying isolate the variable on one side when trying to solve an equation with one variable
(8 votes)
• What if in the masses for both of the sides are uneven like 10 for one side and 9 for the other. how would you solve it then?
(11 votes)
• You would end up with a
3y+3 = 7 + z
3y = 4 + z
which you can't literally go on unless you know the mass of at least either 'y' or 'z'.
(22 votes)
• This is interesting and confusing...at the same time
(9 votes)
• I find that math should always be confusing and interesting. If it is not confusing, then you understand what you are working on and should move ahead. If math is not interesting, then maybe you are to far ahead and it is just frustrating.
(17 votes)
• i really need help with this ! can any one help me
(10 votes)
• Please be specific. Is there a step in the problem you are confused about? Is it just the ENTIRE problem? Otherwise, we don´t know how to help.
(4 votes)
• I don't under stand what he means at
(4 votes)
• He is adding like terms. So he has 1y+1y+1y and 1+1+1 which is 3y+3
(13 votes)
• at it says that people had to find mass when at the jewelry store why did they have to do that
(7 votes)
• because most jewelry is measured in karats and they need to know how much to pay the jeweler.
(7 votes)
• This is so easy! Anyone else agree?
(7 votes)
• Agreed. Algebra is easy. And Fun.
(2 votes)
• because There is some way of dealing with variables on both sides called "transferring the terms", used most widely in China. ... The term actually goes to the other side of equation because on the original side the two terms cancel out or multiply to 1. So it's the same thing.
(6 votes)
• Where do you get the 3Y from?
(4 votes)
• u count on the scale, there are 3 blocks of y which means 3y
(4 votes)

## Video transcript

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.