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Solving a proportion with an unknown variable (example) 7th grade Khan Academy.mp3 | Well, if we did that to the numerator, in order to have an equivalent fraction, you have to do the same thing to the denominator. You have to multiply it. You have to multiply it times 5 over 4. And so we could say this n, this thing that we just solved for, this n is going to be equal to 36 times 5 divided by 4. Or you could say that this is going to be equal to 36 times 5 divided by 4. And now, 36 divided by 4, we know what that is. We can divide both the numerator and the denominator by 4. |
Solving a proportion with an unknown variable (example) 7th grade Khan Academy.mp3 | And so we could say this n, this thing that we just solved for, this n is going to be equal to 36 times 5 divided by 4. Or you could say that this is going to be equal to 36 times 5 divided by 4. And now, 36 divided by 4, we know what that is. We can divide both the numerator and the denominator by 4. You divide the numerator by 4, you get 9. Divide the denominator by 4, you get 1. You get 45. |
Solving a proportion with an unknown variable (example) 7th grade Khan Academy.mp3 | We can divide both the numerator and the denominator by 4. You divide the numerator by 4, you get 9. Divide the denominator by 4, you get 1. You get 45. So that's one way to think about it. 8 over 36 is equal to 10 over 45. Another way to think about it is, what do we have to multiply 8 by to get its denominator? |
Solving a proportion with an unknown variable (example) 7th grade Khan Academy.mp3 | You get 45. So that's one way to think about it. 8 over 36 is equal to 10 over 45. Another way to think about it is, what do we have to multiply 8 by to get its denominator? How much larger is the denominator 36 than 8? Well, let's just divide 36 over 8. So 36 over 8 is the same thing as, so we can simplify dividing the numerator and the denominator by 4. |
Solving a proportion with an unknown variable (example) 7th grade Khan Academy.mp3 | Another way to think about it is, what do we have to multiply 8 by to get its denominator? How much larger is the denominator 36 than 8? Well, let's just divide 36 over 8. So 36 over 8 is the same thing as, so we can simplify dividing the numerator and the denominator by 4. That's their greatest common divisor. That's the same thing as 9 halves. So if you multiply the numerator by 9 halves, you get the denominator. |
Solving a proportion with an unknown variable (example) 7th grade Khan Academy.mp3 | So 36 over 8 is the same thing as, so we can simplify dividing the numerator and the denominator by 4. That's their greatest common divisor. That's the same thing as 9 halves. So if you multiply the numerator by 9 halves, you get the denominator. So we're multiplying by 9 halves to get the denominator over here. Well, then we have to do the same thing over here. If 36 is 9 halves times 8, let me write this. |
Solving a proportion with an unknown variable (example) 7th grade Khan Academy.mp3 | So if you multiply the numerator by 9 halves, you get the denominator. So we're multiplying by 9 halves to get the denominator over here. Well, then we have to do the same thing over here. If 36 is 9 halves times 8, let me write this. 36, or let me write 8 times 9 halves is equal to 36. That's how we go from the numerator to denominator. Then to figure out what the denominator here is, if we want the same fraction, we have to multiply by 9 halves again. |
Solving a proportion with an unknown variable (example) 7th grade Khan Academy.mp3 | If 36 is 9 halves times 8, let me write this. 36, or let me write 8 times 9 halves is equal to 36. That's how we go from the numerator to denominator. Then to figure out what the denominator here is, if we want the same fraction, we have to multiply by 9 halves again. So then we'll get 10 times 9 halves is going to be equal to n, is going to be equal to this denominator. And so this is the same thing as saying 10 times 9 over 2. Divide the numerator and the denominator by 2, you get 5 over 1, which is 45. |
Solving a proportion with an unknown variable (example) 7th grade Khan Academy.mp3 | Then to figure out what the denominator here is, if we want the same fraction, we have to multiply by 9 halves again. So then we'll get 10 times 9 halves is going to be equal to n, is going to be equal to this denominator. And so this is the same thing as saying 10 times 9 over 2. Divide the numerator and the denominator by 2, you get 5 over 1, which is 45. So 45 is equal to n. Once again, we got the same way. Completely legitimate way to solve it. Now sometimes when you see a proportion like this, sometimes people say, oh, you can cross multiply. |
Solving a proportion with an unknown variable (example) 7th grade Khan Academy.mp3 | Divide the numerator and the denominator by 2, you get 5 over 1, which is 45. So 45 is equal to n. Once again, we got the same way. Completely legitimate way to solve it. Now sometimes when you see a proportion like this, sometimes people say, oh, you can cross multiply. And you can cross multiply, and I'll teach you how to do that. And that's sometimes a quick way to do it. But I don't like teaching it the first time you look at proportions, because it's really just something mechanical. |
Solving a proportion with an unknown variable (example) 7th grade Khan Academy.mp3 | Now sometimes when you see a proportion like this, sometimes people say, oh, you can cross multiply. And you can cross multiply, and I'll teach you how to do that. And that's sometimes a quick way to do it. But I don't like teaching it the first time you look at proportions, because it's really just something mechanical. You really don't understand what you're doing. And it really comes out of a little bit of algebra. And I'll show you the algebra as well. |
Solving a proportion with an unknown variable (example) 7th grade Khan Academy.mp3 | But I don't like teaching it the first time you look at proportions, because it's really just something mechanical. You really don't understand what you're doing. And it really comes out of a little bit of algebra. And I'll show you the algebra as well. But if you don't understand it, if it doesn't make as much sense to you at this point, don't worry too much about it. So we have 8 over 36 is equal to 10 over n. When you cross multiply, you're saying that the numerator here times the denominator over here is going to be equal to, so 8 times n is going to be equal to the denominator over here. Let me do this in a different color. |
Solving a proportion with an unknown variable (example) 7th grade Khan Academy.mp3 | And I'll show you the algebra as well. But if you don't understand it, if it doesn't make as much sense to you at this point, don't worry too much about it. So we have 8 over 36 is equal to 10 over n. When you cross multiply, you're saying that the numerator here times the denominator over here is going to be equal to, so 8 times n is going to be equal to the denominator over here. Let me do this in a different color. The denominator over here times the numerator over here. This is what it means to cross multiply. So this is going to be equal to 36 times 10. |
Solving a proportion with an unknown variable (example) 7th grade Khan Academy.mp3 | Let me do this in a different color. The denominator over here times the numerator over here. This is what it means to cross multiply. So this is going to be equal to 36 times 10. Or you could say, let me do this in a neutral color now. You could say that 8n is equal to 360. And so you're saying 8 times what is equal to 360? |
Solving a proportion with an unknown variable (example) 7th grade Khan Academy.mp3 | So this is going to be equal to 36 times 10. Or you could say, let me do this in a neutral color now. You could say that 8n is equal to 360. And so you're saying 8 times what is equal to 360? Or to figure out what that times what is, you divide 360 divided by 8. So we could divide, and this is a little bit of algebra here. We're dividing both sides of the equation by 8. |
Solving a proportion with an unknown variable (example) 7th grade Khan Academy.mp3 | And so you're saying 8 times what is equal to 360? Or to figure out what that times what is, you divide 360 divided by 8. So we could divide, and this is a little bit of algebra here. We're dividing both sides of the equation by 8. And we're getting n is equal to 360 divided by 8. And you can do that without thinking in strict algebraic terms. You can say 8 times what is 360? |
Solving a proportion with an unknown variable (example) 7th grade Khan Academy.mp3 | We're dividing both sides of the equation by 8. And we're getting n is equal to 360 divided by 8. And you can do that without thinking in strict algebraic terms. You can say 8 times what is 360? Well, 8 times 360 over 8. If I write 8 times question mark is equal to 360, well, question mark could definitely be 360 over 8. If I multiply these out, this guy and that guy cancel out, and it's definitely 360. |
Solving a proportion with an unknown variable (example) 7th grade Khan Academy.mp3 | You can say 8 times what is 360? Well, 8 times 360 over 8. If I write 8 times question mark is equal to 360, well, question mark could definitely be 360 over 8. If I multiply these out, this guy and that guy cancel out, and it's definitely 360. And that's why it's 360 over 8. But now we want to actually divide this to actually get our right answer or a simplified answer. 8 goes into 360. |
Solving a proportion with an unknown variable (example) 7th grade Khan Academy.mp3 | If I multiply these out, this guy and that guy cancel out, and it's definitely 360. And that's why it's 360 over 8. But now we want to actually divide this to actually get our right answer or a simplified answer. 8 goes into 360. 8 goes into 36 four times. 4 times 8 is 32. You have a remainder of 4. |
Solving a proportion with an unknown variable (example) 7th grade Khan Academy.mp3 | 8 goes into 360. 8 goes into 36 four times. 4 times 8 is 32. You have a remainder of 4. Bring down the 0. 8 goes into 45 times. 5 times 8 is 40. |
Solving a proportion with an unknown variable (example) 7th grade Khan Academy.mp3 | You have a remainder of 4. Bring down the 0. 8 goes into 45 times. 5 times 8 is 40. And then you have no remainder. And you're done. Once again, we got n is equal to 45. |
Solving a proportion with an unknown variable (example) 7th grade Khan Academy.mp3 | 5 times 8 is 40. And then you have no remainder. And you're done. Once again, we got n is equal to 45. Now, the last thing I'm going to show you involves a little bit of algebra. If any of the ways before this work, that's fine. And where this is sitting in the playlist, you're not expected to know the algebra. |
Solving a proportion with an unknown variable (example) 7th grade Khan Academy.mp3 | Once again, we got n is equal to 45. Now, the last thing I'm going to show you involves a little bit of algebra. If any of the ways before this work, that's fine. And where this is sitting in the playlist, you're not expected to know the algebra. But I want to show you the algebra just because I wanted to show you that this cross multiplication isn't some magic, that using algebra, we will get this exact same thing. But you could stop watching this if you find this part confusing. So let's rewrite our proportion. |
Solving a proportion with an unknown variable (example) 7th grade Khan Academy.mp3 | And where this is sitting in the playlist, you're not expected to know the algebra. But I want to show you the algebra just because I wanted to show you that this cross multiplication isn't some magic, that using algebra, we will get this exact same thing. But you could stop watching this if you find this part confusing. So let's rewrite our proportion. 8 over 36 is equal to 10 over n. And we want to solve for n. Well, the easiest way to solve for n is maybe multiply both. This thing on the left is equal to this thing on the right. So we can multiply them both by the same thing. |
Solving a proportion with an unknown variable (example) 7th grade Khan Academy.mp3 | So let's rewrite our proportion. 8 over 36 is equal to 10 over n. And we want to solve for n. Well, the easiest way to solve for n is maybe multiply both. This thing on the left is equal to this thing on the right. So we can multiply them both by the same thing. And the equality will still hold. So we can multiply both of them by n. On the right-hand side, the n's cancel out. On the left-hand side, we have 836 times n is equal to 10. |
Solving a proportion with an unknown variable (example) 7th grade Khan Academy.mp3 | So we can multiply them both by the same thing. And the equality will still hold. So we can multiply both of them by n. On the right-hand side, the n's cancel out. On the left-hand side, we have 836 times n is equal to 10. Now if we want to solve for n, we can literally multiply. If we want just an n here, we'd want to multiply this side times 36. Do that in a different color. |
Solving a proportion with an unknown variable (example) 7th grade Khan Academy.mp3 | On the left-hand side, we have 836 times n is equal to 10. Now if we want to solve for n, we can literally multiply. If we want just an n here, we'd want to multiply this side times 36. Do that in a different color. We'd want to multiply this side times 36 times 8. Because if you multiply these guys out, you get 1. And you just have an n. But since we're doing it to the left-hand side, we also have to do it to the right-hand side. |
Solving a proportion with an unknown variable (example) 7th grade Khan Academy.mp3 | Do that in a different color. We'd want to multiply this side times 36 times 8. Because if you multiply these guys out, you get 1. And you just have an n. But since we're doing it to the left-hand side, we also have to do it to the right-hand side. So 10 times 36 over 8. These guys cancel out. And we're left with n is equal to 10 times 36 is 360 over 8. |
Solving a proportion with an unknown variable (example) 7th grade Khan Academy.mp3 | And you just have an n. But since we're doing it to the left-hand side, we also have to do it to the right-hand side. So 10 times 36 over 8. These guys cancel out. And we're left with n is equal to 10 times 36 is 360 over 8. And notice, we're getting the exact same value that we got with cross-multiplying. And with cross-multiplying, you're actually doing two steps. Actually, you're doing an extra step here. |
Solving a proportion with an unknown variable (example) 7th grade Khan Academy.mp3 | And we're left with n is equal to 10 times 36 is 360 over 8. And notice, we're getting the exact same value that we got with cross-multiplying. And with cross-multiplying, you're actually doing two steps. Actually, you're doing an extra step here. You're multiplying both sides by n so that you get your 8n. And then you're multiplying both sides by 36 so that you get your 36 on both sides. And you get this value here. |
Solving a proportion with an unknown variable (example) 7th grade Khan Academy.mp3 | Actually, you're doing an extra step here. You're multiplying both sides by n so that you get your 8n. And then you're multiplying both sides by 36 so that you get your 36 on both sides. And you get this value here. But at the end, when you simplify it, you'll get the exact same answer. So those are all different ways to solve this proportion. Probably the most obvious way, or the easiest way to do it in your head, was either just looking at what you have to multiply the numerator by, and then doing the same thing to the denominator, or maybe by cross-multiplication. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | 7x equals 14, this is the exact same thing as saying 7 times x, 7 times x is equal to 14. Now you might be able to do this in your head, you could literally go through the 7 times tables and say 7 times 1 is equal to 7, so that won't work, 7 times 2 is equal to 14, so 2 works here. So you would immediately be able to solve it, you would be able to just by trying different numbers out say hey, that's going to be a 2. But what we're going to do in this video is think about how to solve this systematically, because what we're going to find is as these equations get more and more complicated, you're not going to be able to just think about it and do it in your head. So it's really important that 1, you understand how to manipulate these equations, but even more important, understand what they actually represent. This literally just says 7 times x is equal to 14. In algebra, we don't write the times there, when you write two numbers next to each other, or a number next to a variable like this, it just means that you are multiplying, it's just a shorthand, a shorthand notation. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | But what we're going to do in this video is think about how to solve this systematically, because what we're going to find is as these equations get more and more complicated, you're not going to be able to just think about it and do it in your head. So it's really important that 1, you understand how to manipulate these equations, but even more important, understand what they actually represent. This literally just says 7 times x is equal to 14. In algebra, we don't write the times there, when you write two numbers next to each other, or a number next to a variable like this, it just means that you are multiplying, it's just a shorthand, a shorthand notation. And in general we don't use the multiplication sign, because it's confusing, because x is the most common variable used in algebra, and if I were to write 7 times x is equal to 14, and I write my times sign or my x a little bit strange, it might look like xx or times times, so in general when you're dealing with equations, especially when one of the variables is an x, you wouldn't use the traditional multiplication sign. You might use something like this. You might use dot to represent multiplication, so you might have 7 times x is equal to 14, but this is still a little unusual. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | In algebra, we don't write the times there, when you write two numbers next to each other, or a number next to a variable like this, it just means that you are multiplying, it's just a shorthand, a shorthand notation. And in general we don't use the multiplication sign, because it's confusing, because x is the most common variable used in algebra, and if I were to write 7 times x is equal to 14, and I write my times sign or my x a little bit strange, it might look like xx or times times, so in general when you're dealing with equations, especially when one of the variables is an x, you wouldn't use the traditional multiplication sign. You might use something like this. You might use dot to represent multiplication, so you might have 7 times x is equal to 14, but this is still a little unusual. If you have something multiplying by a variable, you'll just write 7x. That literally means 7 times x. Now, to understand how you can manipulate this equation to solve it, let's visualize this. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | You might use dot to represent multiplication, so you might have 7 times x is equal to 14, but this is still a little unusual. If you have something multiplying by a variable, you'll just write 7x. That literally means 7 times x. Now, to understand how you can manipulate this equation to solve it, let's visualize this. So 7 times x, what is that? That's the same thing. So I'm just going to rewrite this equation, but I'm going to rewrite it in visual form. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | Now, to understand how you can manipulate this equation to solve it, let's visualize this. So 7 times x, what is that? That's the same thing. So I'm just going to rewrite this equation, but I'm going to rewrite it in visual form. So 7 times x. So that literally means x added to itself 7 times. That's the definition of multiplication. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | So I'm just going to rewrite this equation, but I'm going to rewrite it in visual form. So 7 times x. So that literally means x added to itself 7 times. That's the definition of multiplication. So it's literally x plus x plus x plus x plus x. Let's see, that's 5x's plus x plus x. So that right there is literally 7x's. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | That's the definition of multiplication. So it's literally x plus x plus x plus x plus x. Let's see, that's 5x's plus x plus x. So that right there is literally 7x's. This is 7x right there. Let me rewrite it down. This right here is 7x. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | So that right there is literally 7x's. This is 7x right there. Let me rewrite it down. This right here is 7x. Now, this equation tells us that 7x is equal to 14. So it's saying that this is equal to 14. Now let me just draw 14 objects here. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | This right here is 7x. Now, this equation tells us that 7x is equal to 14. So it's saying that this is equal to 14. Now let me just draw 14 objects here. So let's say I have 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14. So literally we're saying 7x is equal to 14 things. These are equivalent statements. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | Now let me just draw 14 objects here. So let's say I have 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14. So literally we're saying 7x is equal to 14 things. These are equivalent statements. Now, the reason why I drew it out this way is so that you really understand what we're going to do when we divide both sides by 7. So let me erase this right here. So the standard step whenever... Oh, I didn't want to do that. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | These are equivalent statements. Now, the reason why I drew it out this way is so that you really understand what we're going to do when we divide both sides by 7. So let me erase this right here. So the standard step whenever... Oh, I didn't want to do that. Let me do this. Let me draw that last circle. So in general, whenever you simplify an equation down to a coefficient is just the number multiplying the variable. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | So the standard step whenever... Oh, I didn't want to do that. Let me do this. Let me draw that last circle. So in general, whenever you simplify an equation down to a coefficient is just the number multiplying the variable. So some number multiplying the variable, or we could call that a coefficient times a variable, equal to something else. What you want to do is just divide both sides by 7 in this case, or divide both sides by the coefficient. So if you divide both sides by 7, what do you get? |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | So in general, whenever you simplify an equation down to a coefficient is just the number multiplying the variable. So some number multiplying the variable, or we could call that a coefficient times a variable, equal to something else. What you want to do is just divide both sides by 7 in this case, or divide both sides by the coefficient. So if you divide both sides by 7, what do you get? 7 times something divided by 7 is just going to be that original something. 7's cancelled out, and 14 divided by 7 is 2. So your solution is going to be x is equal to 2. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | So if you divide both sides by 7, what do you get? 7 times something divided by 7 is just going to be that original something. 7's cancelled out, and 14 divided by 7 is 2. So your solution is going to be x is equal to 2. But just to make it very tangible in your head, what's going on here is when we're dividing both sides of the equation by 7, we're literally dividing both sides by 7. This is an equation. It's saying that this is equal to that. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | So your solution is going to be x is equal to 2. But just to make it very tangible in your head, what's going on here is when we're dividing both sides of the equation by 7, we're literally dividing both sides by 7. This is an equation. It's saying that this is equal to that. Anything I do to the left-hand side, I have to do to the right. If they start off being equal, I can't just do an operation to one side and have it still be equal. They were the same thing. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | It's saying that this is equal to that. Anything I do to the left-hand side, I have to do to the right. If they start off being equal, I can't just do an operation to one side and have it still be equal. They were the same thing. So if I divide the left-hand side by 7, so let me divide it into 7 groups. So there are 7 x's here, so that's 1, 2, 3, 4, 5, 6, 7. So it's 1, 2, 3, 4, 5, 6, 7 groups. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | They were the same thing. So if I divide the left-hand side by 7, so let me divide it into 7 groups. So there are 7 x's here, so that's 1, 2, 3, 4, 5, 6, 7. So it's 1, 2, 3, 4, 5, 6, 7 groups. Now if I divide that into 7 groups, I'll also want to divide the right-hand side into 7 groups. 1, 2, 3, 4, 5, 6, 7. So if this whole thing is equal to this whole thing, then each of these little chunks that we broke into, these 7 chunks, are going to be equivalent. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | So it's 1, 2, 3, 4, 5, 6, 7 groups. Now if I divide that into 7 groups, I'll also want to divide the right-hand side into 7 groups. 1, 2, 3, 4, 5, 6, 7. So if this whole thing is equal to this whole thing, then each of these little chunks that we broke into, these 7 chunks, are going to be equivalent. So this chunk, you could say, is equal to that chunk. This chunk is equal to this chunk. They're all equivalent chunks. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | So if this whole thing is equal to this whole thing, then each of these little chunks that we broke into, these 7 chunks, are going to be equivalent. So this chunk, you could say, is equal to that chunk. This chunk is equal to this chunk. They're all equivalent chunks. There are 7 chunks here, 7 chunks here. So each x must be equal to 2 of these objects. So we get x is equal to, in this case, we had the objects drawn out, where there's 2 of them. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | They're all equivalent chunks. There are 7 chunks here, 7 chunks here. So each x must be equal to 2 of these objects. So we get x is equal to, in this case, we had the objects drawn out, where there's 2 of them. x is equal to 2. Now, let's just do a couple more examples here, just so it really gets in your mind that we're dealing with an equation, and any operation that you do to one side of the equation, you should do to the other. So let me scroll down a little bit. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | So we get x is equal to, in this case, we had the objects drawn out, where there's 2 of them. x is equal to 2. Now, let's just do a couple more examples here, just so it really gets in your mind that we're dealing with an equation, and any operation that you do to one side of the equation, you should do to the other. So let me scroll down a little bit. So let's say I have 3x is equal to 15. Now, once again, you might be able to do this in your head. This is saying 3 times some number is equal to 15. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | So let me scroll down a little bit. So let's say I have 3x is equal to 15. Now, once again, you might be able to do this in your head. This is saying 3 times some number is equal to 15. You could go through your 3 times tables and figure it out. But if you just wanted to do this systematically, and it is good to understand it systematically, you say, OK, this thing on the left is equal to this thing on the right. What do I have to do to this thing on the left to have just an x there? |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | This is saying 3 times some number is equal to 15. You could go through your 3 times tables and figure it out. But if you just wanted to do this systematically, and it is good to understand it systematically, you say, OK, this thing on the left is equal to this thing on the right. What do I have to do to this thing on the left to have just an x there? Well, to have just an x there, I want to divide it by 3. And my whole motivation for doing that is to have 3 times something divided by 3. The 3's will cancel out, and I'm just going to be left with an x. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | What do I have to do to this thing on the left to have just an x there? Well, to have just an x there, I want to divide it by 3. And my whole motivation for doing that is to have 3 times something divided by 3. The 3's will cancel out, and I'm just going to be left with an x. Now, 3x was equal to 15. If I'm dividing the left side by 3, in order for the equality to still hold, I also have to divide the right side by 3. And what does that give us? |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | The 3's will cancel out, and I'm just going to be left with an x. Now, 3x was equal to 15. If I'm dividing the left side by 3, in order for the equality to still hold, I also have to divide the right side by 3. And what does that give us? Well, the left-hand side, we're just going to be left with an x. So it's just going to be an x. And then the right-hand side, what is 15 divided by 3? |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | And what does that give us? Well, the left-hand side, we're just going to be left with an x. So it's just going to be an x. And then the right-hand side, what is 15 divided by 3? Well, it is just 5. Now, you could have also done this equation in a slightly different way, although they are really equivalent. If I start with 3x is equal to 15, you might say, hey, Sal, instead of dividing by 3, I could also get rid of this 3. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | And then the right-hand side, what is 15 divided by 3? Well, it is just 5. Now, you could have also done this equation in a slightly different way, although they are really equivalent. If I start with 3x is equal to 15, you might say, hey, Sal, instead of dividing by 3, I could also get rid of this 3. I could just be left with an x. If I multiply both sides of this equation by 1 3rd. So if I multiply both sides of this equation by 1 3rd, that should also work. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | If I start with 3x is equal to 15, you might say, hey, Sal, instead of dividing by 3, I could also get rid of this 3. I could just be left with an x. If I multiply both sides of this equation by 1 3rd. So if I multiply both sides of this equation by 1 3rd, that should also work. You say, look, 1 3rd of 3 is 1. When you just multiply this part right here, 1 3rd times 3, that is just 1. 1x is equal to 15 times 1 3rd is equal to 5. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | So if I multiply both sides of this equation by 1 3rd, that should also work. You say, look, 1 3rd of 3 is 1. When you just multiply this part right here, 1 3rd times 3, that is just 1. 1x is equal to 15 times 1 3rd is equal to 5. And 1 times x is the same thing as just x. So this is the same thing as x is equal to 5. And these are actually equivalent ways of doing it. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | 1x is equal to 15 times 1 3rd is equal to 5. And 1 times x is the same thing as just x. So this is the same thing as x is equal to 5. And these are actually equivalent ways of doing it. If you divide both sides by 3, that is equivalent to multiplying both sides of the equation by 1 3rd. Now, let's do one more, and I'm going to make it a little bit more complicated. So let's say, and I'm going to change the variable a little bit. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | And these are actually equivalent ways of doing it. If you divide both sides by 3, that is equivalent to multiplying both sides of the equation by 1 3rd. Now, let's do one more, and I'm going to make it a little bit more complicated. So let's say, and I'm going to change the variable a little bit. So let's say I have 2y plus 4y is equal to 18. Now all of a sudden, it's a little harder to do in your head. We're saying 2 times something plus 4 times that same something is going to be equal to 18. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | So let's say, and I'm going to change the variable a little bit. So let's say I have 2y plus 4y is equal to 18. Now all of a sudden, it's a little harder to do in your head. We're saying 2 times something plus 4 times that same something is going to be equal to 18. So here it's harder to think about what number that is. You could try it. You might say, well, if y was 1, it would be 2 times 1 plus 4 times 1. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | We're saying 2 times something plus 4 times that same something is going to be equal to 18. So here it's harder to think about what number that is. You could try it. You might say, well, if y was 1, it would be 2 times 1 plus 4 times 1. Well, that doesn't work. But let's think about how to do it systematically. You could keep guessing, and you might eventually get the answer. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | You might say, well, if y was 1, it would be 2 times 1 plus 4 times 1. Well, that doesn't work. But let's think about how to do it systematically. You could keep guessing, and you might eventually get the answer. But how do you do this systematically? Let's visualize it. So if I have 2 y's, what does that mean? |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | You could keep guessing, and you might eventually get the answer. But how do you do this systematically? Let's visualize it. So if I have 2 y's, what does that mean? It literally means I have 2 y's added to each other. So it's literally y plus y. And then to that, I'm adding 4 y's, which are literally 4 y's added to each other. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | So if I have 2 y's, what does that mean? It literally means I have 2 y's added to each other. So it's literally y plus y. And then to that, I'm adding 4 y's, which are literally 4 y's added to each other. So it's y plus y plus y plus y. And that has got to be equal to 18. So that is equal to 18. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | And then to that, I'm adding 4 y's, which are literally 4 y's added to each other. So it's y plus y plus y plus y. And that has got to be equal to 18. So that is equal to 18. Now, how many y's do I have here on the left-hand side? How many y's do I have? I have 1, 2, 3, 4, 5, 6 y's. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | So that is equal to 18. Now, how many y's do I have here on the left-hand side? How many y's do I have? I have 1, 2, 3, 4, 5, 6 y's. So you could simplify this as 6 y's equal to 18. And if you think about it, it makes complete sense. So this is the whole thing right here, the 2 y plus the 4 y is 6 y. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | I have 1, 2, 3, 4, 5, 6 y's. So you could simplify this as 6 y's equal to 18. And if you think about it, it makes complete sense. So this is the whole thing right here, the 2 y plus the 4 y is 6 y. So 2 y plus 4 y is 6 y, which makes sense. If I have 2 apples plus 4 apples, I'm going to have 6 apples. If I have 2 y's plus 4 y's, I'm going to have 6 y's. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | So this is the whole thing right here, the 2 y plus the 4 y is 6 y. So 2 y plus 4 y is 6 y, which makes sense. If I have 2 apples plus 4 apples, I'm going to have 6 apples. If I have 2 y's plus 4 y's, I'm going to have 6 y's. Now, that's going to be equal to 18. That is going to be equal to 18. And now, hopefully, we understand how to do this. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | If I have 2 y's plus 4 y's, I'm going to have 6 y's. Now, that's going to be equal to 18. That is going to be equal to 18. And now, hopefully, we understand how to do this. If I have 6 times something is equal to 18, if I divide both sides of this equation by 6, I'll solve for the something. So divide the left-hand side by 6. And divide the right-hand side by 6. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | And now, hopefully, we understand how to do this. If I have 6 times something is equal to 18, if I divide both sides of this equation by 6, I'll solve for the something. So divide the left-hand side by 6. And divide the right-hand side by 6. And we are left with y is equal to 3. And you could try it out. That's what's cool about an equation. |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | And divide the right-hand side by 6. And we are left with y is equal to 3. And you could try it out. That's what's cool about an equation. You can always check to see if you got the right answer. Let's see if that works. 2 times 3 plus 4 times 3 is equal to what? |
How to solve equations of the form ax = b Linear equations Algebra I Khan Academy.mp3 | That's what's cool about an equation. You can always check to see if you got the right answer. Let's see if that works. 2 times 3 plus 4 times 3 is equal to what? 2 times 3, this right here, is 6. And then 4 times 3 is 12. 6 plus 12 is indeed equal to 18. |
Solving the troll riddle visually Algebra II Khan Academy.mp3 | But we had to cross the bridge, and the troll gave us these clues because we had no money in our pocket. And if we don't solve his riddle, he's gonna push us into the water. So we are under pressure. And at least we made some headway in the last video. We were able to represent his clues mathematically as a system of equations. What I wanna do in this video is think about whether we can solve for the system of equations. And you will see that there are many ways of solving a system of equations. |
Solving the troll riddle visually Algebra II Khan Academy.mp3 | And at least we made some headway in the last video. We were able to represent his clues mathematically as a system of equations. What I wanna do in this video is think about whether we can solve for the system of equations. And you will see that there are many ways of solving a system of equations. But this time I wanna do it visually, because at least in my mind, it helps really kind of get the intuition of what these things are saying. So let's draw some axes over here. Let's draw an F axis. |
Solving the troll riddle visually Algebra II Khan Academy.mp3 | And you will see that there are many ways of solving a system of equations. But this time I wanna do it visually, because at least in my mind, it helps really kind of get the intuition of what these things are saying. So let's draw some axes over here. Let's draw an F axis. That's the number of fives that I have. And let's draw a T axis. That is the number of tens I have. |
Solving the troll riddle visually Algebra II Khan Academy.mp3 | Let's draw an F axis. That's the number of fives that I have. And let's draw a T axis. That is the number of tens I have. And let's say that this right over here, this right over here is 500 tens. That is 1,000 tens. And let's say that this is, sorry, that's 500 fives. |
Solving the troll riddle visually Algebra II Khan Academy.mp3 | That is the number of tens I have. And let's say that this right over here, this right over here is 500 tens. That is 1,000 tens. And let's say that this is, sorry, that's 500 fives. That's 1,000 fives. This is 500 tens. 500 tens. |
Solving the troll riddle visually Algebra II Khan Academy.mp3 | And let's say that this is, sorry, that's 500 fives. That's 1,000 fives. This is 500 tens. 500 tens. And this is 1,000 tens. So let's think about all of the combinations of Fs and Ts that satisfy this first equation. If T, if we have no tens, then we're gonna have 900 fives. |
Solving the troll riddle visually Algebra II Khan Academy.mp3 | 500 tens. And this is 1,000 tens. So let's think about all of the combinations of Fs and Ts that satisfy this first equation. If T, if we have no tens, then we're gonna have 900 fives. So if we have no tens, we're gonna have 900 fives. So that looks like it's right about there. So that's the point zero tens, 900, 900 fives. |
Solving the troll riddle visually Algebra II Khan Academy.mp3 | If T, if we have no tens, then we're gonna have 900 fives. So if we have no tens, we're gonna have 900 fives. So that looks like it's right about there. So that's the point zero tens, 900, 900 fives. But what if it went the other way? If we have no fives, we're gonna have 900 tens. So that's gonna be the point 900 tens, zero fives. |
Solving the troll riddle visually Algebra II Khan Academy.mp3 | So that's the point zero tens, 900, 900 fives. But what if it went the other way? If we have no fives, we're gonna have 900 tens. So that's gonna be the point 900 tens, zero fives. So all the combinations of Fs and Ts that satisfy this are going to be on this line, are going to be on that line right over there. And I just draw a dotted line just because it's easier for me to draw it straight. So that represents all the Fs and Ts that satisfy the first constraint. |
Solving the troll riddle visually Algebra II Khan Academy.mp3 | So that's gonna be the point 900 tens, zero fives. So all the combinations of Fs and Ts that satisfy this are going to be on this line, are going to be on that line right over there. And I just draw a dotted line just because it's easier for me to draw it straight. So that represents all the Fs and Ts that satisfy the first constraint. Obviously, there's a bunch of them. So we don't know which is the one that is actually what the troll has. But lucky for us, we have a second constraint, this one right over here. |
Solving the troll riddle visually Algebra II Khan Academy.mp3 | So that represents all the Fs and Ts that satisfy the first constraint. Obviously, there's a bunch of them. So we don't know which is the one that is actually what the troll has. But lucky for us, we have a second constraint, this one right over here. So let's do the same thing. In this constraint, what happens if we have no tens? If tens are zero, then we have five F is equal to 5,500. |
Solving the troll riddle visually Algebra II Khan Academy.mp3 | But lucky for us, we have a second constraint, this one right over here. So let's do the same thing. In this constraint, what happens if we have no tens? If tens are zero, then we have five F is equal to 5,500. Let me do a little table here because this is a little bit more involved. It's a little bit more involved. So for the second equation, tens and fives. |
Solving the troll riddle visually Algebra II Khan Academy.mp3 | If tens are zero, then we have five F is equal to 5,500. Let me do a little table here because this is a little bit more involved. It's a little bit more involved. So for the second equation, tens and fives. If I have no tens, if I have no tens, I have five F is equal to 5,500, F will be 1,100. F is, I must have 1,100 fives. If I have no fives, if I have no fives, and this is zero, and I have 10 T is equal to 5,500, that means I have 550, 550 tens. |
Solving the troll riddle visually Algebra II Khan Academy.mp3 | So for the second equation, tens and fives. If I have no tens, if I have no tens, I have five F is equal to 5,500, F will be 1,100. F is, I must have 1,100 fives. If I have no fives, if I have no fives, and this is zero, and I have 10 T is equal to 5,500, that means I have 550, 550 tens. So let's plot both of those points. T equals zero, F is 1,100, that's right about there. So that is zero, 1,100 is on the line that represents this equation. |
Solving the troll riddle visually Algebra II Khan Academy.mp3 | If I have no fives, if I have no fives, and this is zero, and I have 10 T is equal to 5,500, that means I have 550, 550 tens. So let's plot both of those points. T equals zero, F is 1,100, that's right about there. So that is zero, 1,100 is on the line that represents this equation. And that when F is zero, T is 550. So let's see, this is about, see this would be six, seven, eight, nine. So 550 is gonna be right over here. |
Solving the troll riddle visually Algebra II Khan Academy.mp3 | So that is zero, 1,100 is on the line that represents this equation. And that when F is zero, T is 550. So let's see, this is about, see this would be six, seven, eight, nine. So 550 is gonna be right over here. So that is the point 550, 550 comma zero. And all of these points, let me try to draw a straight line again. I can do a better job than that. |
Solving the troll riddle visually Algebra II Khan Academy.mp3 | So 550 is gonna be right over here. So that is the point 550, 550 comma zero. And all of these points, let me try to draw a straight line again. I can do a better job than that. I can do a better job than that. So all of these points are the points, let me try one more time, one more time. We wanna get this right. |
Solving the troll riddle visually Algebra II Khan Academy.mp3 | I can do a better job than that. I can do a better job than that. So all of these points are the points, let me try one more time, one more time. We wanna get this right. We don't wanna get pushed into the water by the troll. So there you go, that looks pretty good. So every point on this blue line represents an F, T combination that satisfies the second constraint. |
Solving the troll riddle visually Algebra II Khan Academy.mp3 | We wanna get this right. We don't wanna get pushed into the water by the troll. So there you go, that looks pretty good. So every point on this blue line represents an F, T combination that satisfies the second constraint. So what is an F and T, or number of fives and number of tens that satisfy both constraints? Well, it would be a point that is sitting on both of the lines. And what is a point that is sitting on both of the lines? |
Solving the troll riddle visually Algebra II Khan Academy.mp3 | So every point on this blue line represents an F, T combination that satisfies the second constraint. So what is an F and T, or number of fives and number of tens that satisfy both constraints? Well, it would be a point that is sitting on both of the lines. And what is a point that is sitting on both of the lines? Well, that's where they intersect. This point right over here is clearly on the blue line and it is clearly on the yellow line. And what we can do is if we drew this graph really, really, really precisely, we could see how many fives that is and how many tens that is. |
Solving the troll riddle visually Algebra II Khan Academy.mp3 | And what is a point that is sitting on both of the lines? Well, that's where they intersect. This point right over here is clearly on the blue line and it is clearly on the yellow line. And what we can do is if we drew this graph really, really, really precisely, we could see how many fives that is and how many tens that is. And if you look at it, if you look at it very precisely, and actually I encourage you to graph it very precisely and come up with how many fives and how many tens that is. Well, when we do it right over here, I'm going to eyeball it. If we look at it right over here, it looks like we have about 700 fives, 700 fives, and it looks like we have about 200 tens. |
Solving the troll riddle visually Algebra II Khan Academy.mp3 | And what we can do is if we drew this graph really, really, really precisely, we could see how many fives that is and how many tens that is. And if you look at it, if you look at it very precisely, and actually I encourage you to graph it very precisely and come up with how many fives and how many tens that is. Well, when we do it right over here, I'm going to eyeball it. If we look at it right over here, it looks like we have about 700 fives, 700 fives, and it looks like we have about 200 tens. And this is based on my really rough graph, but let's see if that worked. 700 plus 200, 700 plus 200 is equal to 900. And if I have 700 fives, 700, let me write this down. |
Solving the troll riddle visually Algebra II Khan Academy.mp3 | If we look at it right over here, it looks like we have about 700 fives, 700 fives, and it looks like we have about 200 tens. And this is based on my really rough graph, but let's see if that worked. 700 plus 200, 700 plus 200 is equal to 900. And if I have 700 fives, 700, let me write this down. Five times 700 is going to be the value of the fives, which is $3,500. And then 10 plus 10 times 200, 10 times 200, which is $2,000, $2,000 is the value of the tens. And if you add up the two values, you indeed get to $5,500. |
Solving the troll riddle visually Algebra II Khan Academy.mp3 | And if I have 700 fives, 700, let me write this down. Five times 700 is going to be the value of the fives, which is $3,500. And then 10 plus 10 times 200, 10 times 200, which is $2,000, $2,000 is the value of the tens. And if you add up the two values, you indeed get to $5,500. So this looks right. And so we can tell the troll, troll, I know, I know how many five and $10 bills you have. You have 700 $5 bills and you have 200 $10 bills. |
Example of adding fractions with unlike denominators word problem Pre-Algebra Khan Academy.mp3 | Cindy has 2 5ths of a gallon of red paint. Michael has got half a gallon of yellow paint. If they mix their paints together, will they have the one gallon they need? So let's think about that. We're going to add the 2 5ths of a gallon of red paint, and we're going to add that to half a gallon of yellow paint. And we want to see if this gets to being one whole gallon. So whenever we add fractions, right over here, we're not adding the same thing. |
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