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More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3 | Let me show you that. Negative 9 plus negative 8, that is equal to negative 17. Close, but no cigar. So what other ones are? We have 6 and 12. That actually seems pretty good. If we have negative 6 plus negative 12, that is equal to negative 18. |
More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3 | So what other ones are? We have 6 and 12. That actually seems pretty good. If we have negative 6 plus negative 12, that is equal to negative 18. Notice, it's a bit of an art. You have to try the different factors here. So this will become negative 1, don't want to forget that, times x minus 6 times x minus 12. |
Negative exponent intuition Pre-Algebra Khan Academy.mp3 | I don't know, the inventor of mathematics wasn't one person. It was a convention that arose. But they defined this, and they defined this for the reasons that I'm going to show you. Well, what I'm going to show you is one of the reasons. And we'll see that this is a good definition, because once you learned exponent rules, all of the other exponent rules stay consistent for negative exponents and when you raise something to the 0th power. So let's take the positive exponents. Those are pretty intuitive, I think. |
Negative exponent intuition Pre-Algebra Khan Academy.mp3 | Well, what I'm going to show you is one of the reasons. And we'll see that this is a good definition, because once you learned exponent rules, all of the other exponent rules stay consistent for negative exponents and when you raise something to the 0th power. So let's take the positive exponents. Those are pretty intuitive, I think. So the positive exponents, so you have a to the 1, a squared, a cubed, a to the 4th. What's a to the 1? a to the 1, we said, was a. |
Negative exponent intuition Pre-Algebra Khan Academy.mp3 | Those are pretty intuitive, I think. So the positive exponents, so you have a to the 1, a squared, a cubed, a to the 4th. What's a to the 1? a to the 1, we said, was a. And then to get to a squared, what did we do? We multiplied by a, right? a squared is just a times a. |
Negative exponent intuition Pre-Algebra Khan Academy.mp3 | a to the 1, we said, was a. And then to get to a squared, what did we do? We multiplied by a, right? a squared is just a times a. And then to get to a cubed, what did we do? We multiplied by a again. And then to get to a to the 4th, what did we do? |
Negative exponent intuition Pre-Algebra Khan Academy.mp3 | a squared is just a times a. And then to get to a cubed, what did we do? We multiplied by a again. And then to get to a to the 4th, what did we do? We multiplied by a again. Or the other way you could imagine is when you decrease the exponent, what are we doing? We are multiplying by 1 over a, or dividing by a. |
Negative exponent intuition Pre-Algebra Khan Academy.mp3 | And then to get to a to the 4th, what did we do? We multiplied by a again. Or the other way you could imagine is when you decrease the exponent, what are we doing? We are multiplying by 1 over a, or dividing by a. Similarly, decrease again, you're dividing by a. And go from a squared to a to the 1st, you're dividing by a. So let's use this progression to figure out what a to the 0 is. |
Negative exponent intuition Pre-Algebra Khan Academy.mp3 | We are multiplying by 1 over a, or dividing by a. Similarly, decrease again, you're dividing by a. And go from a squared to a to the 1st, you're dividing by a. So let's use this progression to figure out what a to the 0 is. So this is the first hard one. So a to the 0. So you're the inventor, the founding mother of mathematics, and you need to define what a to the 0 is. |
Negative exponent intuition Pre-Algebra Khan Academy.mp3 | So let's use this progression to figure out what a to the 0 is. So this is the first hard one. So a to the 0. So you're the inventor, the founding mother of mathematics, and you need to define what a to the 0 is. And maybe it's 17, maybe it's pi, I don't know. It's up to you to decide what a to the 0 is. But wouldn't it be nice if a to the 0 retained this pattern? |
Negative exponent intuition Pre-Algebra Khan Academy.mp3 | So you're the inventor, the founding mother of mathematics, and you need to define what a to the 0 is. And maybe it's 17, maybe it's pi, I don't know. It's up to you to decide what a to the 0 is. But wouldn't it be nice if a to the 0 retained this pattern? That every time you decrease the exponent, you're dividing by a, right? So if you're going from a to the 1st to a to the 0, wouldn't it be nice if we just divided by a? So let's do that. |
Negative exponent intuition Pre-Algebra Khan Academy.mp3 | But wouldn't it be nice if a to the 0 retained this pattern? That every time you decrease the exponent, you're dividing by a, right? So if you're going from a to the 1st to a to the 0, wouldn't it be nice if we just divided by a? So let's do that. So if we go from a to the 1st, which is just a, and divide by a, so we're just going to divide it by a. What is a divided by a? Well, it's just 1. |
Negative exponent intuition Pre-Algebra Khan Academy.mp3 | So let's do that. So if we go from a to the 1st, which is just a, and divide by a, so we're just going to divide it by a. What is a divided by a? Well, it's just 1. So that's where the definition, or that's one of the intuitions behind why something to the 0th power is equal to 1. Because when you take that number and you divide it by itself one more time, you just get 1. So that's pretty reasonable. |
Negative exponent intuition Pre-Algebra Khan Academy.mp3 | Well, it's just 1. So that's where the definition, or that's one of the intuitions behind why something to the 0th power is equal to 1. Because when you take that number and you divide it by itself one more time, you just get 1. So that's pretty reasonable. But now let's go into the negative domain. So what should a to the negative 1 equal? a to the negative 1. |
Negative exponent intuition Pre-Algebra Khan Academy.mp3 | So that's pretty reasonable. But now let's go into the negative domain. So what should a to the negative 1 equal? a to the negative 1. Well, once again, it's nice if we can retain this pattern, where every time we decrease the exponent, we're dividing by a. So let's divide by a again. So 1 over a. |
Negative exponent intuition Pre-Algebra Khan Academy.mp3 | a to the negative 1. Well, once again, it's nice if we can retain this pattern, where every time we decrease the exponent, we're dividing by a. So let's divide by a again. So 1 over a. So we're going to take a to the 0 and divide it by a. a to the 0 is 1. So what's 1 divided by a? It's 1 over a. |
Negative exponent intuition Pre-Algebra Khan Academy.mp3 | So 1 over a. So we're going to take a to the 0 and divide it by a. a to the 0 is 1. So what's 1 divided by a? It's 1 over a. And let's do it one more time, and then I think you're going to get the pattern. Well, I think you probably already got the pattern. What's a to the minus 2? |
Negative exponent intuition Pre-Algebra Khan Academy.mp3 | It's 1 over a. And let's do it one more time, and then I think you're going to get the pattern. Well, I think you probably already got the pattern. What's a to the minus 2? Well, it'd be silly now to change this pattern. Every time we decrease the exponent, we're dividing by a. So to go from a to the minus 1 to a to the minus 2, let's just divide by a again. |
Negative exponent intuition Pre-Algebra Khan Academy.mp3 | What's a to the minus 2? Well, it'd be silly now to change this pattern. Every time we decrease the exponent, we're dividing by a. So to go from a to the minus 1 to a to the minus 2, let's just divide by a again. And what do we get? We get, if you take 1 over a and divide by a, you get 1 over a squared. And you could just keep doing this pattern all the way to the left, and you would get a to the minus b is equal to 1 over a to the b. Hopefully that gave you a little intuition as to why, well, first of all, the big mystery is, well, something to a 0th power, why does that equal 1? |
Negative exponent intuition Pre-Algebra Khan Academy.mp3 | So to go from a to the minus 1 to a to the minus 2, let's just divide by a again. And what do we get? We get, if you take 1 over a and divide by a, you get 1 over a squared. And you could just keep doing this pattern all the way to the left, and you would get a to the minus b is equal to 1 over a to the b. Hopefully that gave you a little intuition as to why, well, first of all, the big mystery is, well, something to a 0th power, why does that equal 1? First, keep in mind that that's just a definition. Someone decided that it should be equal to 1, but they had a good reason. And their good reason was they wanted to keep this pattern going. |
Negative exponent intuition Pre-Algebra Khan Academy.mp3 | And you could just keep doing this pattern all the way to the left, and you would get a to the minus b is equal to 1 over a to the b. Hopefully that gave you a little intuition as to why, well, first of all, the big mystery is, well, something to a 0th power, why does that equal 1? First, keep in mind that that's just a definition. Someone decided that it should be equal to 1, but they had a good reason. And their good reason was they wanted to keep this pattern going. And that's the same reason why they defined negative exponents in this way. And what's extra cool about it is not only does it retain this pattern of when you decrease exponents, you're dividing by a, or when you're increasing exponents, you're multiplying by a. But as you'll see in the exponent rules videos, all of the exponent rules hold. |
Examples of simplifying expressions involving rational numbers 7th grade Khan Academy.mp3 | What I want to do in this video is get some practice simplifying expressions that have some hairier numbers involved, and these numbers are kind of hairy. And like always, try to pause this video and see if you can simplify this expression before I take a stab at it. Alright, I'm assuming you have attempted it. Now let's look at it. We have negative 5.55 minus 8.55c plus 4.35c. So the first thing I want to do is, can I combine these c terms? And I definitely can. |
Examples of simplifying expressions involving rational numbers 7th grade Khan Academy.mp3 | Now let's look at it. We have negative 5.55 minus 8.55c plus 4.35c. So the first thing I want to do is, can I combine these c terms? And I definitely can. This is if you, we can add negative 8.55c to 4.35c first, and then that would be, let's see, that would be negative 8.55 plus 4.35, I'm just adding the coefficients, times c, and of course we still have that negative 5.55 out front. Negative 5.55, and I'll just put a plus there. Now how do we calculate negative 8.55 plus 4.35? |
Examples of simplifying expressions involving rational numbers 7th grade Khan Academy.mp3 | And I definitely can. This is if you, we can add negative 8.55c to 4.35c first, and then that would be, let's see, that would be negative 8.55 plus 4.35, I'm just adding the coefficients, times c, and of course we still have that negative 5.55 out front. Negative 5.55, and I'll just put a plus there. Now how do we calculate negative 8.55 plus 4.35? Well there's a couple of ways to think about it or visualize it. One way is to say, well this is the same thing as the negative of 8.55 minus 4.35. And 8.55 minus 4.35, let's see, 8 minus 4 is gonna be the negative, 8 minus 4 is 4, 55 hundredths minus 35 hundredths is 20 hundredths, so I can write 4.20, which is really just the same thing as 4.2. |
Examples of simplifying expressions involving rational numbers 7th grade Khan Academy.mp3 | Now how do we calculate negative 8.55 plus 4.35? Well there's a couple of ways to think about it or visualize it. One way is to say, well this is the same thing as the negative of 8.55 minus 4.35. And 8.55 minus 4.35, let's see, 8 minus 4 is gonna be the negative, 8 minus 4 is 4, 55 hundredths minus 35 hundredths is 20 hundredths, so I can write 4.20, which is really just the same thing as 4.2. So all of this, all of this can be replaced with a negative 4.2. So my entire expression has simplified to negative 5.55, and instead of saying plus negative 4.2c, I can just write it as minus 4.2, 4.2c, and we're done. We can't simplify this anymore. |
Examples of simplifying expressions involving rational numbers 7th grade Khan Academy.mp3 | And 8.55 minus 4.35, let's see, 8 minus 4 is gonna be the negative, 8 minus 4 is 4, 55 hundredths minus 35 hundredths is 20 hundredths, so I can write 4.20, which is really just the same thing as 4.2. So all of this, all of this can be replaced with a negative 4.2. So my entire expression has simplified to negative 5.55, and instead of saying plus negative 4.2c, I can just write it as minus 4.2, 4.2c, and we're done. We can't simplify this anymore. We can't add this term that doesn't involve the variable to this term that does involve the variable. So this is about as simple as we're gonna get. So let's do another example. |
Examples of simplifying expressions involving rational numbers 7th grade Khan Academy.mp3 | We can't simplify this anymore. We can't add this term that doesn't involve the variable to this term that does involve the variable. So this is about as simple as we're gonna get. So let's do another example. So here I have these, I have some more hairy numbers involved, these are all expressed as fractions. And so let's see, I have 2 fifths m, minus 4 fifths, minus 3 fifths m. So how can I simplify? Well I could add all the m terms together. |
Examples of simplifying expressions involving rational numbers 7th grade Khan Academy.mp3 | So let's do another example. So here I have these, I have some more hairy numbers involved, these are all expressed as fractions. And so let's see, I have 2 fifths m, minus 4 fifths, minus 3 fifths m. So how can I simplify? Well I could add all the m terms together. So let me just change the order, I could rewrite this as 2 fifths m, minus 3 fifths m, minus 4 fifths. All I did is I changed the order, and we can see that I have these 2 m terms, I can add those two together. So this is going to be 2 fifths, minus 3 fifths, times m, and then I have the minus 4 fifths still on the right hand side. |
Examples of simplifying expressions involving rational numbers 7th grade Khan Academy.mp3 | Well I could add all the m terms together. So let me just change the order, I could rewrite this as 2 fifths m, minus 3 fifths m, minus 4 fifths. All I did is I changed the order, and we can see that I have these 2 m terms, I can add those two together. So this is going to be 2 fifths, minus 3 fifths, times m, and then I have the minus 4 fifths still on the right hand side. Now what's 2 fifths minus 3 fifths? Well that's gonna be negative 1 fifth. It's gonna be negative 1 fifth. |
Examples of simplifying expressions involving rational numbers 7th grade Khan Academy.mp3 | So this is going to be 2 fifths, minus 3 fifths, times m, and then I have the minus 4 fifths still on the right hand side. Now what's 2 fifths minus 3 fifths? Well that's gonna be negative 1 fifth. It's gonna be negative 1 fifth. So I have negative 1 fifth m, minus 4 fifths. Minus 4 fifths. And once again I'm done, I can't simplify it anymore, I can't add this term that involves m somehow to this negative 4 fifths. |
Examples of simplifying expressions involving rational numbers 7th grade Khan Academy.mp3 | It's gonna be negative 1 fifth. So I have negative 1 fifth m, minus 4 fifths. Minus 4 fifths. And once again I'm done, I can't simplify it anymore, I can't add this term that involves m somehow to this negative 4 fifths. So we are done here. Let's do one more example. So here, this is interesting, I have a parentheses and all the rest. |
Examples of simplifying expressions involving rational numbers 7th grade Khan Academy.mp3 | And once again I'm done, I can't simplify it anymore, I can't add this term that involves m somehow to this negative 4 fifths. So we are done here. Let's do one more example. So here, this is interesting, I have a parentheses and all the rest. And like always, pause the video, see if you can simplify this. Alright, let's work through it together. Now the first thing that I want to do is, let's distribute this 2 so that we just have three terms that are just being added and subtracted. |
Examples of simplifying expressions involving rational numbers 7th grade Khan Academy.mp3 | So here, this is interesting, I have a parentheses and all the rest. And like always, pause the video, see if you can simplify this. Alright, let's work through it together. Now the first thing that I want to do is, let's distribute this 2 so that we just have three terms that are just being added and subtracted. So if we distribute this 2, we're gonna get 2 times 1 fifth m is 2 fifths m. Let me make sure you see that m, m is right here. 2 times negative 2 fifths is negative 4 fifths, and then I have plus 3 fifths. Now how can we simplify this more? |
Examples of simplifying expressions involving rational numbers 7th grade Khan Academy.mp3 | Now the first thing that I want to do is, let's distribute this 2 so that we just have three terms that are just being added and subtracted. So if we distribute this 2, we're gonna get 2 times 1 fifth m is 2 fifths m. Let me make sure you see that m, m is right here. 2 times negative 2 fifths is negative 4 fifths, and then I have plus 3 fifths. Now how can we simplify this more? Well I have these two terms here that don't involve the variable, those are just numbers, I can add them to each other. So I have negative 4 fifths plus 3 fifths. So what's negative 4 plus 3? |
Examples of simplifying expressions involving rational numbers 7th grade Khan Academy.mp3 | Now how can we simplify this more? Well I have these two terms here that don't involve the variable, those are just numbers, I can add them to each other. So I have negative 4 fifths plus 3 fifths. So what's negative 4 plus 3? That's gonna be negative 1. So this is gonna be negative 1 fifth. What we have in yellow here, and I still have the 2 over 5 m. 2 fifths m minus 1 fifth. |
Why we do the same thing to both sides basic systems Linear equations Algebra I Khan Academy.mp3 | On the left-hand side of the scale, I have two different types of unknown masses. One are these x masses, and we know that they have the same identical mass, and we'll call that identical. Each of them have a mass of x. But then we have this other, this blue thing, and that has a mass of y, which isn't necessarily going to be the same as a mass of x. But when I have two of these x's and a y, it seems like their total mass, y is the case, that their total mass balances out these 8 kilograms right over here. Each of these are 1-kilogram blocks. It balances them out. |
Why we do the same thing to both sides basic systems Linear equations Algebra I Khan Academy.mp3 | But then we have this other, this blue thing, and that has a mass of y, which isn't necessarily going to be the same as a mass of x. But when I have two of these x's and a y, it seems like their total mass, y is the case, that their total mass balances out these 8 kilograms right over here. Each of these are 1-kilogram blocks. It balances them out. So the first question I'm going to ask you is, can you express this mathematically? Can you express what we're seeing here, the fact that this total mass balances out with this total mass? Can you express that mathematically? |
Why we do the same thing to both sides basic systems Linear equations Algebra I Khan Academy.mp3 | It balances them out. So the first question I'm going to ask you is, can you express this mathematically? Can you express what we're seeing here, the fact that this total mass balances out with this total mass? Can you express that mathematically? Well, let's just think about our total mass on this side. We have two masses of mass x, so those two are going to total at 2x. And then you have a mass of y, so then you're going to have another y. |
Why we do the same thing to both sides basic systems Linear equations Algebra I Khan Academy.mp3 | Can you express that mathematically? Well, let's just think about our total mass on this side. We have two masses of mass x, so those two are going to total at 2x. And then you have a mass of y, so then you're going to have another y. So the total mass on the left-hand side, and let me write it a little bit closer to the center, just so it doesn't get too spread out. On the left-hand side, I have 2x plus a mass of y. That's the total mass. |
Why we do the same thing to both sides basic systems Linear equations Algebra I Khan Academy.mp3 | And then you have a mass of y, so then you're going to have another y. So the total mass on the left-hand side, and let me write it a little bit closer to the center, just so it doesn't get too spread out. On the left-hand side, I have 2x plus a mass of y. That's the total mass. The total mass on the left-hand side is 2x plus y. The total mass on the right-hand side is just 8. 1, 2, 3, 4, 5, 6, 7, 8. |
Why we do the same thing to both sides basic systems Linear equations Algebra I Khan Academy.mp3 | That's the total mass. The total mass on the left-hand side is 2x plus y. The total mass on the right-hand side is just 8. 1, 2, 3, 4, 5, 6, 7, 8. It is equal to 8. And since we see that the scale is balanced, this total mass must be equal to this total mass, so we can write an equal sign there. Now my question to you. |
Why we do the same thing to both sides basic systems Linear equations Algebra I Khan Academy.mp3 | 1, 2, 3, 4, 5, 6, 7, 8. It is equal to 8. And since we see that the scale is balanced, this total mass must be equal to this total mass, so we can write an equal sign there. Now my question to you. Is there anything we can do, just based on the information that we have here, to solve for either the mass x or for the mass y? Is there anything that we can do? Well, the simple answer is, just with this information here, there is actually very little. |
Why we do the same thing to both sides basic systems Linear equations Algebra I Khan Academy.mp3 | Now my question to you. Is there anything we can do, just based on the information that we have here, to solve for either the mass x or for the mass y? Is there anything that we can do? Well, the simple answer is, just with this information here, there is actually very little. You might say, well, let me take a y from both sides. You might take this y block up. But if you take this y block up, you have to take away y from this side, and you don't know what y is, and if you think about it algebraically, you might get rid of the y here, subtracting y, but then you're going to have to subtract y from this side too. |
Why we do the same thing to both sides basic systems Linear equations Algebra I Khan Academy.mp3 | Well, the simple answer is, just with this information here, there is actually very little. You might say, well, let me take a y from both sides. You might take this y block up. But if you take this y block up, you have to take away y from this side, and you don't know what y is, and if you think about it algebraically, you might get rid of the y here, subtracting y, but then you're going to have to subtract y from this side too. So you're not going to get rid of the y. Same thing with the x's. You actually don't have enough information. |
Why we do the same thing to both sides basic systems Linear equations Algebra I Khan Academy.mp3 | But if you take this y block up, you have to take away y from this side, and you don't know what y is, and if you think about it algebraically, you might get rid of the y here, subtracting y, but then you're going to have to subtract y from this side too. So you're not going to get rid of the y. Same thing with the x's. You actually don't have enough information. y depends on what x is, and x depends on what y is. Lucky for us, however, we do have some more of these blocks laying around. And what we do is, we take one of these x blocks, and I stick it over here, and I also take one of the y blocks, and I stick them over here, and I stick it right over there. |
Why we do the same thing to both sides basic systems Linear equations Algebra I Khan Academy.mp3 | You actually don't have enough information. y depends on what x is, and x depends on what y is. Lucky for us, however, we do have some more of these blocks laying around. And what we do is, we take one of these x blocks, and I stick it over here, and I also take one of the y blocks, and I stick them over here, and I stick it right over there. And then I keep adding these ones until I balance this thing out. So I keep adding these ones. So obviously, if I just place this, this will go down, because there's nothing on that side. |
Why we do the same thing to both sides basic systems Linear equations Algebra I Khan Academy.mp3 | And what we do is, we take one of these x blocks, and I stick it over here, and I also take one of the y blocks, and I stick them over here, and I stick it right over there. And then I keep adding these ones until I balance this thing out. So I keep adding these ones. So obviously, if I just place this, this will go down, because there's nothing on that side. But I keep adding these blocks until it all balances out. And I find that my scale balances once I have 5 kilograms on the right-hand side. So once again, let me ask you this information, that now having an x and a y on the left-hand side, and a 5 kilograms on the right-hand side, and the fact that they're balanced, how can we represent that mathematically? |
Why we do the same thing to both sides basic systems Linear equations Algebra I Khan Academy.mp3 | So obviously, if I just place this, this will go down, because there's nothing on that side. But I keep adding these blocks until it all balances out. And I find that my scale balances once I have 5 kilograms on the right-hand side. So once again, let me ask you this information, that now having an x and a y on the left-hand side, and a 5 kilograms on the right-hand side, and the fact that they're balanced, how can we represent that mathematically? Well, our total mass on the left-hand side is x plus y, and our total mass on the right-hand side is x plus y, and on the right-hand side I have 5 kilograms. And we know this is what actually balanced the scale. So these total masses must be equal to each other. |
Why we do the same thing to both sides basic systems Linear equations Algebra I Khan Academy.mp3 | So once again, let me ask you this information, that now having an x and a y on the left-hand side, and a 5 kilograms on the right-hand side, and the fact that they're balanced, how can we represent that mathematically? Well, our total mass on the left-hand side is x plus y, and our total mass on the right-hand side is x plus y, and on the right-hand side I have 5 kilograms. And we know this is what actually balanced the scale. So these total masses must be equal to each other. And this information by itself, once again, there's nothing I can do with it. I don't know what x and y are. If y is 4, maybe x is 1. |
Why we do the same thing to both sides basic systems Linear equations Algebra I Khan Academy.mp3 | So these total masses must be equal to each other. And this information by itself, once again, there's nothing I can do with it. I don't know what x and y are. If y is 4, maybe x is 1. Maybe x is 4, y is 1. Who knows what these are? The interesting thing is we can actually use both of this information to figure out what x and y actually is. And I'll give you a few seconds to think about how we can approach this situation. |
Why we do the same thing to both sides basic systems Linear equations Algebra I Khan Academy.mp3 | If y is 4, maybe x is 1. Maybe x is 4, y is 1. Who knows what these are? The interesting thing is we can actually use both of this information to figure out what x and y actually is. And I'll give you a few seconds to think about how we can approach this situation. Well, think about it this way. We know that x plus y is equal to 5. So if we were to get rid of an x and a y on this side, on the left-hand side of the equation, if we get rid of an x and a y on the left-hand side of the scale, what would we have to get rid of on the right-hand side of the scale to take away the same mass? |
Why we do the same thing to both sides basic systems Linear equations Algebra I Khan Academy.mp3 | And I'll give you a few seconds to think about how we can approach this situation. Well, think about it this way. We know that x plus y is equal to 5. So if we were to get rid of an x and a y on this side, on the left-hand side of the equation, if we get rid of an x and a y on the left-hand side of the scale, what would we have to get rid of on the right-hand side of the scale to take away the same mass? Well, if we take away an x and a y on the left-hand side, we know that an x plus a y is 5 kilograms. So we would just have to take 5 kilograms from the right-hand side. So let's think about what that would do. |
Why we do the same thing to both sides basic systems Linear equations Algebra I Khan Academy.mp3 | So if we were to get rid of an x and a y on this side, on the left-hand side of the equation, if we get rid of an x and a y on the left-hand side of the scale, what would we have to get rid of on the right-hand side of the scale to take away the same mass? Well, if we take away an x and a y on the left-hand side, we know that an x plus a y is 5 kilograms. So we would just have to take 5 kilograms from the right-hand side. So let's think about what that would do. Well, then I would just have an x left over here, and I would just have some of these masses left over there, and I would know what x is. Let's think about how we can represent that algebraically. Essentially, if we're taking an x and a y from the left-hand side, if I'm taking an x and a y from the left-hand side, I'm subtracting an x, I'm subtracting an x, and I am subtracting, actually, let me think of it this way. |
Why we do the same thing to both sides basic systems Linear equations Algebra I Khan Academy.mp3 | So let's think about what that would do. Well, then I would just have an x left over here, and I would just have some of these masses left over there, and I would know what x is. Let's think about how we can represent that algebraically. Essentially, if we're taking an x and a y from the left-hand side, if I'm taking an x and a y from the left-hand side, I'm subtracting an x, I'm subtracting an x, and I am subtracting, actually, let me think of it this way. I am subtracting, I am subtracting an x, I am subtracting an x plus y. I'm subtracting an x and a y from the left-hand side, but then what am I gonna have to do on the right-hand side? Well, an x and a y we know has a mass of five, so we can subtract five from the right-hand side. And the only way I'm able to do this is because of the information that we got from the second scale. |
Why we do the same thing to both sides basic systems Linear equations Algebra I Khan Academy.mp3 | Essentially, if we're taking an x and a y from the left-hand side, if I'm taking an x and a y from the left-hand side, I'm subtracting an x, I'm subtracting an x, and I am subtracting, actually, let me think of it this way. I am subtracting, I am subtracting an x, I am subtracting an x plus y. I'm subtracting an x and a y from the left-hand side, but then what am I gonna have to do on the right-hand side? Well, an x and a y we know has a mass of five, so we can subtract five from the right-hand side. And the only way I'm able to do this is because of the information that we got from the second scale. And so I can take away five, so this is going to be equal to taking away five. Taking away an x and a y is the same thing as taking away five, and we know that because an x and a y is equal to five kilograms. And so if we take away an x and a y on the left-hand side, what are we left with? |
Why we do the same thing to both sides basic systems Linear equations Algebra I Khan Academy.mp3 | And the only way I'm able to do this is because of the information that we got from the second scale. And so I can take away five, so this is going to be equal to taking away five. Taking away an x and a y is the same thing as taking away five, and we know that because an x and a y is equal to five kilograms. And so if we take away an x and a y on the left-hand side, what are we left with? Well, this is going to be the same thing. Let me rewrite, let me rewrite this part. This taking away an x and a y is the same thing, is the same thing if you distribute the negative sign as taking away an x, taking away an x and taking away a y, and taking away a y. |
Why we do the same thing to both sides basic systems Linear equations Algebra I Khan Academy.mp3 | And so if we take away an x and a y on the left-hand side, what are we left with? Well, this is going to be the same thing. Let me rewrite, let me rewrite this part. This taking away an x and a y is the same thing, is the same thing if you distribute the negative sign as taking away an x, taking away an x and taking away a y, and taking away a y. And so on the left-hand side, we are left with, on the left-hand side, we are left with just two x, and we take away one of the x's. We're left with just an x. And we had a y and we took away one of the y's, so we're left with no y's left. |
Why we do the same thing to both sides basic systems Linear equations Algebra I Khan Academy.mp3 | This taking away an x and a y is the same thing, is the same thing if you distribute the negative sign as taking away an x, taking away an x and taking away a y, and taking away a y. And so on the left-hand side, we are left with, on the left-hand side, we are left with just two x, and we take away one of the x's. We're left with just an x. And we had a y and we took away one of the y's, so we're left with no y's left. And we see that here visually. We have just an x here. And what do we have on the right-hand side? |
Why we do the same thing to both sides basic systems Linear equations Algebra I Khan Academy.mp3 | And we had a y and we took away one of the y's, so we're left with no y's left. And we see that here visually. We have just an x here. And what do we have on the right-hand side? We had eight, and we knew that an x and a y is equal to five so we took away five, to keep the scale balanced. And so eight minus five is going to be three. Eight minus five is equal to three. |
Why we do the same thing to both sides basic systems Linear equations Algebra I Khan Academy.mp3 | And what do we have on the right-hand side? We had eight, and we knew that an x and a y is equal to five so we took away five, to keep the scale balanced. And so eight minus five is going to be three. Eight minus five is equal to three. And just like that, using this extra information, we were able to figure out that the mass of x is equal to, the mass of x is equal to three. Now, one final question. We were able to figure out what the mass of x is, but can you now figure out what the mass of y is? |
Why we do the same thing to both sides basic systems Linear equations Algebra I Khan Academy.mp3 | Eight minus five is equal to three. And just like that, using this extra information, we were able to figure out that the mass of x is equal to, the mass of x is equal to three. Now, one final question. We were able to figure out what the mass of x is, but can you now figure out what the mass of y is? Well, we can go back to either one of these scales, but it'll probably be simpler to go back to this one. We know that the mass of x plus the mass of y is equal to five. So you could say, so one thing we know, that x is now equal to three. |
Why we do the same thing to both sides basic systems Linear equations Algebra I Khan Academy.mp3 | We were able to figure out what the mass of x is, but can you now figure out what the mass of y is? Well, we can go back to either one of these scales, but it'll probably be simpler to go back to this one. We know that the mass of x plus the mass of y is equal to five. So you could say, so one thing we know, that x is now equal to three. We know that this is now a three kilogram mass. We can rewrite this as three, three plus y, three plus y is equal to, is equal to five. Well now we say, well, we could take three away from both sides. |
Why we do the same thing to both sides basic systems Linear equations Algebra I Khan Academy.mp3 | So you could say, so one thing we know, that x is now equal to three. We know that this is now a three kilogram mass. We can rewrite this as three, three plus y, three plus y is equal to, is equal to five. Well now we say, well, we could take three away from both sides. If I take three away from the left-hand side, I just have to take three away from the right-hand side to keep my scale balanced, and I'll be left with that the mass of y is balanced with a mass of two, or y is equal to two. That's analogous to taking three from both sides of this equation. And on the left-hand side, I'm just left with a y. |
Why we do the same thing to both sides basic systems Linear equations Algebra I Khan Academy.mp3 | Well now we say, well, we could take three away from both sides. If I take three away from the left-hand side, I just have to take three away from the right-hand side to keep my scale balanced, and I'll be left with that the mass of y is balanced with a mass of two, or y is equal to two. That's analogous to taking three from both sides of this equation. And on the left-hand side, I'm just left with a y. And on the right-hand side, I am just left with a two. So x is equal to three kilograms, and y is equal to two kilograms. And what I encourage you to do is verify that it made sense right up here. |
Combining like terms, but more complicated Introduction to algebra Algebra I Khan Academy.mp3 | Now we have a very, very, very hairy expression. And once again, I'm going to see if you can simplify this. And I'll give you a little time to do it. So this one is even crazier than the last few we've looked at. We've got y's and xy's and x squared's and x's and, well, more just xy's and y squared's and on and on and on. And there will be a temptation, because you see a y here and a y here, to say, oh, maybe I can add this negative 3y plus this 4xy somehow, since I see a y and a y. But the important thing to realize here is that a y is different than an xy. |
Combining like terms, but more complicated Introduction to algebra Algebra I Khan Academy.mp3 | So this one is even crazier than the last few we've looked at. We've got y's and xy's and x squared's and x's and, well, more just xy's and y squared's and on and on and on. And there will be a temptation, because you see a y here and a y here, to say, oh, maybe I can add this negative 3y plus this 4xy somehow, since I see a y and a y. But the important thing to realize here is that a y is different than an xy. Think about if they were numbers. If y was 3 and an x was a 2, then a y would be a 3, while an xy would have been a 6. And a y is very different than a y squared. |
Combining like terms, but more complicated Introduction to algebra Algebra I Khan Academy.mp3 | But the important thing to realize here is that a y is different than an xy. Think about if they were numbers. If y was 3 and an x was a 2, then a y would be a 3, while an xy would have been a 6. And a y is very different than a y squared. Once again, if the y took on the value 3, then the y squared would be the value 9. So even though you see the same letter here, they aren't the same. I guess you cannot add these two or subtract these two terms. |
Combining like terms, but more complicated Introduction to algebra Algebra I Khan Academy.mp3 | And a y is very different than a y squared. Once again, if the y took on the value 3, then the y squared would be the value 9. So even though you see the same letter here, they aren't the same. I guess you cannot add these two or subtract these two terms. A y is different than a y squared is different than an xy. Now with that said, let's see if we can, if there is anything that we can simplify. So first let's think about the y terms. |
Combining like terms, but more complicated Introduction to algebra Algebra I Khan Academy.mp3 | I guess you cannot add these two or subtract these two terms. A y is different than a y squared is different than an xy. Now with that said, let's see if we can, if there is anything that we can simplify. So first let's think about the y terms. So you have a negative 3y there. Do we have any more y terms? Yes, we do. |
Combining like terms, but more complicated Introduction to algebra Algebra I Khan Academy.mp3 | So first let's think about the y terms. So you have a negative 3y there. Do we have any more y terms? Yes, we do. We have this 2y right over there. So I'll just write it out. I'll just reorder it. |
Combining like terms, but more complicated Introduction to algebra Algebra I Khan Academy.mp3 | Yes, we do. We have this 2y right over there. So I'll just write it out. I'll just reorder it. So we have negative 3y plus 2y. Now let's think about, I'm just going in an arbitrary order, but since our next term is an xy term, let's think about all of the xy terms. So we have plus 4xy right over here. |
Combining like terms, but more complicated Introduction to algebra Algebra I Khan Academy.mp3 | I'll just reorder it. So we have negative 3y plus 2y. Now let's think about, I'm just going in an arbitrary order, but since our next term is an xy term, let's think about all of the xy terms. So we have plus 4xy right over here. So let me just write it down. I'm just reordering the whole expression. Plus 4xy and then I have minus 4xy right over here. |
Combining like terms, but more complicated Introduction to algebra Algebra I Khan Academy.mp3 | So we have plus 4xy right over here. So let me just write it down. I'm just reordering the whole expression. Plus 4xy and then I have minus 4xy right over here. So minus 4xy. Then let's go to the x squared terms. I have negative 2 times x squared or minus 2x squared. |
Combining like terms, but more complicated Introduction to algebra Algebra I Khan Academy.mp3 | Plus 4xy and then I have minus 4xy right over here. So minus 4xy. Then let's go to the x squared terms. I have negative 2 times x squared or minus 2x squared. So let's look at this. So I have minus 2x squared. Do I have any other x squareds? |
Combining like terms, but more complicated Introduction to algebra Algebra I Khan Academy.mp3 | I have negative 2 times x squared or minus 2x squared. So let's look at this. So I have minus 2x squared. Do I have any other x squareds? Yes, I do. I have this 3x squared right over there. So plus 3x squareds. |
Combining like terms, but more complicated Introduction to algebra Algebra I Khan Academy.mp3 | Do I have any other x squareds? Yes, I do. I have this 3x squared right over there. So plus 3x squareds. And then let's see. I have an x term right over here. And that actually looks like the only x term. |
Combining like terms, but more complicated Introduction to algebra Algebra I Khan Academy.mp3 | So plus 3x squareds. And then let's see. I have an x term right over here. And that actually looks like the only x term. So that's plus 2x. And then I only have one y squared term. I'll circle that in orange. |
Combining like terms, but more complicated Introduction to algebra Algebra I Khan Academy.mp3 | And that actually looks like the only x term. So that's plus 2x. And then I only have one y squared term. I'll circle that in orange. So plus y squared. So all I have done is I've reordered the statement and I've color coded it based on the type of term we have. And now it should be a little bit simpler. |
Combining like terms, but more complicated Introduction to algebra Algebra I Khan Academy.mp3 | I'll circle that in orange. So plus y squared. So all I have done is I've reordered the statement and I've color coded it based on the type of term we have. And now it should be a little bit simpler. So let's try it out. If I have negative 3 of something plus 2 of that something, what do I have? Or another way to say it, if I have 2 of something and I subtract 3 of that, what am I left with? |
Combining like terms, but more complicated Introduction to algebra Algebra I Khan Academy.mp3 | And now it should be a little bit simpler. So let's try it out. If I have negative 3 of something plus 2 of that something, what do I have? Or another way to say it, if I have 2 of something and I subtract 3 of that, what am I left with? I'm left with negative 1 of that something. So I could write negative 1y or I could just write negative y. And another way to think about it, but I like to think about it intuitively more, is what's the coefficient here? |
Combining like terms, but more complicated Introduction to algebra Algebra I Khan Academy.mp3 | Or another way to say it, if I have 2 of something and I subtract 3 of that, what am I left with? I'm left with negative 1 of that something. So I could write negative 1y or I could just write negative y. And another way to think about it, but I like to think about it intuitively more, is what's the coefficient here? It is negative 3. What's the coefficient here? It's 2. |
Combining like terms, but more complicated Introduction to algebra Algebra I Khan Academy.mp3 | And another way to think about it, but I like to think about it intuitively more, is what's the coefficient here? It is negative 3. What's the coefficient here? It's 2. We're obviously both dealing with y terms. Not xy terms, not y squared terms, just y. And so negative 3 plus 2 is negative 1. |
Combining like terms, but more complicated Introduction to algebra Algebra I Khan Academy.mp3 | It's 2. We're obviously both dealing with y terms. Not xy terms, not y squared terms, just y. And so negative 3 plus 2 is negative 1. Or negative 1y is the same thing as negative y. So those simplify to this right over here. Now let's look at the xy terms. |
Combining like terms, but more complicated Introduction to algebra Algebra I Khan Academy.mp3 | And so negative 3 plus 2 is negative 1. Or negative 1y is the same thing as negative y. So those simplify to this right over here. Now let's look at the xy terms. If I have 4 of this, 4 xy's, and I were to take away 4 xy's, how many xy's am I left with? Well, I'm left with no xy's. Or you could say, add the coefficients, 4 plus negative 4 gives you 0 xy's. |
Combining like terms, but more complicated Introduction to algebra Algebra I Khan Academy.mp3 | Now let's look at the xy terms. If I have 4 of this, 4 xy's, and I were to take away 4 xy's, how many xy's am I left with? Well, I'm left with no xy's. Or you could say, add the coefficients, 4 plus negative 4 gives you 0 xy's. Either way, these two cancel out. If I have 4 of something and I take away those 4 of that something, I'm left with none of them. So I'm left with no xy's. |
Combining like terms, but more complicated Introduction to algebra Algebra I Khan Academy.mp3 | Or you could say, add the coefficients, 4 plus negative 4 gives you 0 xy's. Either way, these two cancel out. If I have 4 of something and I take away those 4 of that something, I'm left with none of them. So I'm left with no xy's. And then I have right over here, I could have written 0 xy, but that seems unnecessary. Then right over here, I have my x squared terms. Negative 2 plus 3 is 1. |
Combining like terms, but more complicated Introduction to algebra Algebra I Khan Academy.mp3 | So I'm left with no xy's. And then I have right over here, I could have written 0 xy, but that seems unnecessary. Then right over here, I have my x squared terms. Negative 2 plus 3 is 1. Or another way of saying, if I have 3 x squared's and I were to take away 2 of those x squared's, I'm left with 1 x squared. So this right over here simplifies to 1 x squared. Or I could literally just write x squared. |
Combining like terms, but more complicated Introduction to algebra Algebra I Khan Academy.mp3 | Negative 2 plus 3 is 1. Or another way of saying, if I have 3 x squared's and I were to take away 2 of those x squared's, I'm left with 1 x squared. So this right over here simplifies to 1 x squared. Or I could literally just write x squared. 1 x squared is the same thing as x squared. So plus x squared. And then these, there's nothing really left to simplify. |
Combining like terms, but more complicated Introduction to algebra Algebra I Khan Academy.mp3 | Or I could literally just write x squared. 1 x squared is the same thing as x squared. So plus x squared. And then these, there's nothing really left to simplify. So plus 2x plus y squared. And we're done. And obviously, you might have gotten an answer in some other order, but the order in which I write these terms don't matter. |
Slope-intercept equation from two solutions example Algebra I Khan Academy.mp3 | What is the equation of the line? So let's just try to visualize this. So that is my x-axis. And you don't have to draw it to do this problem, but it always helps to visualize. That is my y-axis. And the first point is (-1, 6). So (-1, 1, 2, 3, 4, 5, 6). |
Slope-intercept equation from two solutions example Algebra I Khan Academy.mp3 | And you don't have to draw it to do this problem, but it always helps to visualize. That is my y-axis. And the first point is (-1, 6). So (-1, 1, 2, 3, 4, 5, 6). So it's this point right over there. It's (-1, 6). And the other point is (-5, 4). |
Slope-intercept equation from two solutions example Algebra I Khan Academy.mp3 | So (-1, 1, 2, 3, 4, 5, 6). So it's this point right over there. It's (-1, 6). And the other point is (-5, 4). So 1, 2, 3, 4, 5. And then we go down 4. So 1, 2, 3, 4. |
Slope-intercept equation from two solutions example Algebra I Khan Academy.mp3 | And the other point is (-5, 4). So 1, 2, 3, 4, 5. And then we go down 4. So 1, 2, 3, 4. So it's right over there. And so the line that connects them will look something like this. The line will draw a rough approximation. |
Slope-intercept equation from two solutions example Algebra I Khan Academy.mp3 | So 1, 2, 3, 4. So it's right over there. And so the line that connects them will look something like this. The line will draw a rough approximation. I could draw a straighter line than that. I'll draw a little dotted line maybe. Easier to do dotted lines. |
Slope-intercept equation from two solutions example Algebra I Khan Academy.mp3 | The line will draw a rough approximation. I could draw a straighter line than that. I'll draw a little dotted line maybe. Easier to do dotted lines. So the line will look something like that. So let's find its equation. So a good place to start is we could find its slope. |
Slope-intercept equation from two solutions example Algebra I Khan Academy.mp3 | Easier to do dotted lines. So the line will look something like that. So let's find its equation. So a good place to start is we could find its slope. Remember, we can find equation y is equal to mx plus b. This is a slope-intercept form where m is a slope and b is a y-intercept. We can first try to solve for m. We could find the slope of this line. |
Slope-intercept equation from two solutions example Algebra I Khan Academy.mp3 | So a good place to start is we could find its slope. Remember, we can find equation y is equal to mx plus b. This is a slope-intercept form where m is a slope and b is a y-intercept. We can first try to solve for m. We could find the slope of this line. So m, or the slope, is the change in y over the change in x. Or we could view it as the y values of our end point minus the y value of our starting point over the x values of our end point minus the x values of our starting point. Let me make that clear. |
Slope-intercept equation from two solutions example Algebra I Khan Academy.mp3 | We can first try to solve for m. We could find the slope of this line. So m, or the slope, is the change in y over the change in x. Or we could view it as the y values of our end point minus the y value of our starting point over the x values of our end point minus the x values of our starting point. Let me make that clear. So this is equal to change in y over change in x, which is the same thing as rise over run, which is the same thing as the y value of your ending point minus the y value of your starting point. This is the same exact thing as change in y. And that over the x value of your ending point minus the x value of your starting point. |
Slope-intercept equation from two solutions example Algebra I Khan Academy.mp3 | Let me make that clear. So this is equal to change in y over change in x, which is the same thing as rise over run, which is the same thing as the y value of your ending point minus the y value of your starting point. This is the same exact thing as change in y. And that over the x value of your ending point minus the x value of your starting point. This is the exact same thing as change in x. And you just have to pick one of these as a starting point and one as the ending point. So let's just make this over here our starting point and make that our ending point. |
Slope-intercept equation from two solutions example Algebra I Khan Academy.mp3 | And that over the x value of your ending point minus the x value of your starting point. This is the exact same thing as change in x. And you just have to pick one of these as a starting point and one as the ending point. So let's just make this over here our starting point and make that our ending point. So what is our change in y? So our change in y, we started at y is equal to 6. We started at y is equal to 6. |