Algebra (all content)
Here's how we solve a two step equation. It begins with the concept of equality: what we do to one side of the equation must be done to the other. Created by Sal Khan.
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- Do the rules of order of operations also count, but then in reverse?
So first you do:
3x + 2 = 14
3x = 12
x = 4
3x + 2 = 14
x + 2 = 4 6/9
x = 2 6/9(306 votes)
- You can change the order of operations. Your example has a flaw. You'll need to divide the 2 also by 3 on the left side.
3x + 2 = 14
x + 2/3 = 14/3
x = 12/3
x = 4(338 votes)
- Cant sal just / 3 (divide by three) than x 1/3 (multipy by 1/3)?(34 votes)
- My teacher usually gives my class a challenge problem every week, each is supposed to be more challenging than the last, I was wondering how to solve for f(13)
I think I know how to solve, but I'm not sure(11 votes)
- Wouldn't it also work to think, 3*4=12, therefore, 3x=12 is the same thing as 3*4=12, leaving x=4?(10 votes)
- Does anyone see those two blocks he missed or is it just me?
And did he draw that scale? If he did props to him.(8 votes)
- what if the x is on both sides(6 votes)
- If the unknown number is present on both sides then you can replace x by a number, such as 2 or 3 on both sides(2 votes)
- When using the railroad track method, couldn't you just say:
3x +2 = 14
-2 = -2
3x = 12
/3 = /3
x = 4
Would that work instead of multiplying 1/3 of each side?(5 votes)
Now we mixed up things a little bit more: on the left side of the scale, not only do we have these identical unknown masses with mass X, these three blue things, we also have some of the 1kg masses over here, actually, we have two of them. Now, we are going to figure out what X is. But before we even do that, I want you to think about a mathematical equation that can represent what is going on; that equates what we have on the left hand, with what we have on the right side of the scale. I will give you a few seconds to think about it... So let's think about what we have on the left side: we have 3 masses with mass X, so you can say we have 3x and then we have 2 masses of 1 kilogram, so in total we have 2 kg. So + 2. So one way to think about the total mass on the left-hand side is 3x + 2. Three masses with mass X, plus two kilograms. That is what we have on the left-hand side. Now, let us think about what we have on the right-hand side. We can simply count them: [counts to 14] Fourteen blocks, each has a mass of 1 kg, so the total mass will be 14 kg. And we see that the scale is balanced, not tilting down or upwards. So this mass over here must be equal to this total mass. The scale is balanced, so we can write an 'equal'-sign. (let me do that in a white coulour, I do not like that brown) Now, what I want you to think about, and you can think about it either through the symbols or through the scales, is: how would you go about -- let us think about a few things: how would you first go about at least getting rid of these little 1kg blocks? I will give you a second to think about that... Well, the simplest thing is: you can take these 1kg blocks off of the left-hand side, but remember, if you just took these blocks off of the left-hand side, and it was balanced before, now the left-hand side will be lighter and it will move up. But we want to keep it balanced so we can keep saying 'equal'. That this mass is equal to that mass. So, if we remove 2 block from the left-hand side, we need to remove 2 from the right-hand side. So, we can remove two there, and then we can remove two over there. Mathematically, what we are doing is: we are subtracting 2 kilograms from each side. We are subtracting 2 from this side, So on the left-hand side we now have 3x + 2, minus 2 we are left with just 3x, and on the right-hand side we had 14 and we took away 2 (let me write this:) we took away 2, so we are going to be left with 12 blocks. And you see that there, the ones that I have not crossed out, there are 12 left, and here you have 3 of those X-blocks. Since we removed the same amount from both sides, our scale is still balanced. And our equation: 3x is now equal to 12. Now, this turns into a problem very similar to what we saw in the last video, so now I ask you: what can we do to isolate one x, to only have one 'X' on the left-hand side of the scale, while keeping the scale balanced? The easiest way to think about it is: If I want one X on this left-hand side, that is a third of the total X's here. So what if I were to multiply the left-hand side by one-third -- -- but if I want to keep the scale balanced, I have to multiply the right-hand side by one-third. If we can do that mathematically, Over here I can multiply the left-hand side by 1/3, and if I want to keep my scale balanced I also have to multiply the right-hand side by 1/3. Multiplying it physically literally means: just keeping a third of what we had originally We would get rid of two of these. If we want to keep a third of what we had here originally, -- there are 12 blocks left over after removing those first two -- so, 1/3 of 12: we are only going to have four of these little 1kg boxes left. Let me remove all but four. (so, remove those, and those...) And I have left [counts them] 4 here. And so, what you are left with, the only thing you have left, is this 'X' - I will shade it in to show this is the one we actually have left - and then we have these 1 kilogram boxes. You see it mathematically over here: 1/3 * 3x -- or you could have said 3x divided by 3 -- either way, that gives us -- these threes cancel out, so that would give you an 'X' and on the right-hand side: 12 * 1/3 - which is the same as 12/3, is equal to 4. And, since we did the same thing to both sides, the scale is still balanced. So you see that the mass of this thing must be the same as the mass of these 4 left-over blocks. x must be equal to 4 kilograms.