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Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3 | This is not an equation. If you're dealing with an inequality, and if you multiply or divide both sides of the equation by a negative number, you have to swap the inequality. So in this case, the less than becomes greater than since we're dividing by a negative number. And so negative 4 divided by negative 4, those can... |
Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3 | And so negative 4 divided by negative 4, those cancel out. We have p is greater than 16 divided by negative 4, which is negative 4. And we can plot this solution set right over here. And then we can try out some values to help us feel good about the idea of it working. So let's say this is negative 5, negative 4, negat... |
Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3 | And then we can try out some values to help us feel good about the idea of it working. So let's say this is negative 5, negative 4, negative 3, negative 2, 1, negative 1, I should say, 0. Let me write that a little bit neater. Negative 1, 0. And then we can keep going to the right. And so our solution is p is not great... |
Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3 | Negative 1, 0. And then we can keep going to the right. And so our solution is p is not greater than or equal, so we have to exclude negative 4. p is greater than negative 4, so all the values above that. So negative 3.99999999 will work. Negative 4 will not work. And let's just try some values out to feel good that th... |
Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3 | So negative 3.99999999 will work. Negative 4 will not work. And let's just try some values out to feel good that this is really the solution set. So first let's try out when p is equal to negative 3. This should work. The way I've drawn it, this is in our solution set. p equals negative 3 is greater than negative 4. |
Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3 | So first let's try out when p is equal to negative 3. This should work. The way I've drawn it, this is in our solution set. p equals negative 3 is greater than negative 4. So let's try that out. We have negative 3 times negative 3. The first negative 3 is this one, and then we're saying p is negative 3. |
Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3 | p equals negative 3 is greater than negative 4. So let's try that out. We have negative 3 times negative 3. The first negative 3 is this one, and then we're saying p is negative 3. Minus 7 should be less than, instead of a p, we're going to put a negative 3, should be less than negative 3 plus 9. Negative 3 times negat... |
Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3 | The first negative 3 is this one, and then we're saying p is negative 3. Minus 7 should be less than, instead of a p, we're going to put a negative 3, should be less than negative 3 plus 9. Negative 3 times negative 3 is 9. Minus 7 should be less than negative 3 plus 9 is 6. 9 minus 7 is 2. 2 should be less than 6, of ... |
Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3 | Minus 7 should be less than negative 3 plus 9 is 6. 9 minus 7 is 2. 2 should be less than 6, of which, of course, it is. Now let's try a value that definitely should not work. So let's try negative 5. Negative 5 is not in our solution set, so it should not work. So we have negative 3 times negative 5 minus 7. |
Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3 | Now let's try a value that definitely should not work. So let's try negative 5. Negative 5 is not in our solution set, so it should not work. So we have negative 3 times negative 5 minus 7. Let's see whether it is less than negative 5 plus 9. Negative 3 times negative 5 is 15. Minus 7, it really should not be less than... |
Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3 | So we have negative 3 times negative 5 minus 7. Let's see whether it is less than negative 5 plus 9. Negative 3 times negative 5 is 15. Minus 7, it really should not be less than negative 5 plus 9. So we're just seeing if p equals negative 5 works. 15 minus 7 is 8. And so we get 8 is less than 4, which is definitely no... |
Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3 | Minus 7, it really should not be less than negative 5 plus 9. So we're just seeing if p equals negative 5 works. 15 minus 7 is 8. And so we get 8 is less than 4, which is definitely not the case. So p equals negative 5 doesn't work, and it shouldn't work because that's not in our solution set. And now if we want to fee... |
Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3 | And so we get 8 is less than 4, which is definitely not the case. So p equals negative 5 doesn't work, and it shouldn't work because that's not in our solution set. And now if we want to feel really good about it, we can actually try this boundary point. Negative 4 should not work, but it should satisfy the related equ... |
Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3 | Negative 4 should not work, but it should satisfy the related equation. When I talk about the related equation, negative 4 should satisfy negative 3 minus 7 is equal to p plus 9. It'll satisfy this, but it won't satisfy this because when we get the same value on both sides, the same value is not less than the same valu... |
Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3 | So let's try it out. Let's see whether negative 4 at least satisfies the related equation. So if we get negative 3 times negative 4 minus 7, this should be equal to negative 4 plus 9. So this is 12 minus 7 should be equal to negative 4 plus 9 should be equal to 5. And this, of course, is true. 5 is equal to 5. So it sa... |
Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3 | So this is 12 minus 7 should be equal to negative 4 plus 9 should be equal to 5. And this, of course, is true. 5 is equal to 5. So it satisfies the related equation, but it should not satisfy this. If you put negative 4 for p here, and I encourage you to do so, actually we could do it over here instead of an equal sign... |
Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3 | So it satisfies the related equation, but it should not satisfy this. If you put negative 4 for p here, and I encourage you to do so, actually we could do it over here instead of an equal sign. If you put it into the original inequality, let me delete all of that, it really just becomes this. The original inequality is... |
Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3 | The original inequality is this right over here. If you put negative 4, you have less than, less than, and then you get 5 is less than 5, which is not the case. And that's good because we did not include that in the solution set. We put an open circle. If negative 4 was included, we would fill that in. But the only rea... |
Finding the slope of a line from its graph Algebra I Khan Academy.mp3 | And just as a bit of a review, slope is just telling us how steep a line is. And the best way to view it, slope is equal to change in y, change in y over change in x. Over change in x. And for a line, this will always be constant. And sometimes you might see it written like this. You might see this triangle. That's a c... |
Finding the slope of a line from its graph Algebra I Khan Academy.mp3 | And for a line, this will always be constant. And sometimes you might see it written like this. You might see this triangle. That's a capital delta. That means change in y over change in x. That's just a fancy way of saying change in y over change in x. So let's see what this change in y is for any change in x. |
Finding the slope of a line from its graph Algebra I Khan Academy.mp3 | That's a capital delta. That means change in y over change in x. That's just a fancy way of saying change in y over change in x. So let's see what this change in y is for any change in x. So let's start at some point that seems pretty reasonable to read from this table right here, from this graph. So let's see, we're s... |
Finding the slope of a line from its graph Algebra I Khan Academy.mp3 | So let's see what this change in y is for any change in x. So let's start at some point that seems pretty reasonable to read from this table right here, from this graph. So let's see, we're starting here. Let me do it in a more vibrant color. So let's say we start at that point right there. And we want to go to another... |
Finding the slope of a line from its graph Algebra I Khan Academy.mp3 | Let me do it in a more vibrant color. So let's say we start at that point right there. And we want to go to another point that's pretty straightforward to read. So we can move to that point right there. We can literally pick any two points on this line. I'm just picking ones that are at nice integer coordinates so it's... |
Finding the slope of a line from its graph Algebra I Khan Academy.mp3 | So we can move to that point right there. We can literally pick any two points on this line. I'm just picking ones that are at nice integer coordinates so it's easy to read. So what is the change in y and what is the change in x? So first let's look at the change in x. So if we go from there to there, what is the chang... |
Finding the slope of a line from its graph Algebra I Khan Academy.mp3 | So what is the change in y and what is the change in x? So first let's look at the change in x. So if we go from there to there, what is the change in x? My change in x is equal to what? Well, I could just count it out. I went one step, two steps, three steps. My change in x is 3. |
Finding the slope of a line from its graph Algebra I Khan Academy.mp3 | My change in x is equal to what? Well, I could just count it out. I went one step, two steps, three steps. My change in x is 3. And you could even see it from the x values. If I go from negative 3 to 0, I went up by 3. So my change in x is 3. |
Finding the slope of a line from its graph Algebra I Khan Academy.mp3 | My change in x is 3. And you could even see it from the x values. If I go from negative 3 to 0, I went up by 3. So my change in x is 3. So let me write this. Change in x delta x is equal to 3. And what's my change in y? |
Finding the slope of a line from its graph Algebra I Khan Academy.mp3 | So my change in x is 3. So let me write this. Change in x delta x is equal to 3. And what's my change in y? Well, my change in y, I'm going from negative 3 up to negative 1. Or you could just say 1, 2. So my change in y is equal to positive 2. |
Finding the slope of a line from its graph Algebra I Khan Academy.mp3 | And what's my change in y? Well, my change in y, I'm going from negative 3 up to negative 1. Or you could just say 1, 2. So my change in y is equal to positive 2. So let me write that down. Change in y is equal to 2. So what is my change in y for a change in x? |
Finding the slope of a line from its graph Algebra I Khan Academy.mp3 | So my change in y is equal to positive 2. So let me write that down. Change in y is equal to 2. So what is my change in y for a change in x? Well, when my change in x was 3, my change in y is 2. So this is my slope. And one thing I want to do, I want to show you that I could have really picked any two points here. |
Finding the slope of a line from its graph Algebra I Khan Academy.mp3 | So what is my change in y for a change in x? Well, when my change in x was 3, my change in y is 2. So this is my slope. And one thing I want to do, I want to show you that I could have really picked any two points here. Let's say I didn't pick those two points. Let me pick some other points. And I'll even go in a diffe... |
Finding the slope of a line from its graph Algebra I Khan Academy.mp3 | And one thing I want to do, I want to show you that I could have really picked any two points here. Let's say I didn't pick those two points. Let me pick some other points. And I'll even go in a different direction. And I want to show you that you're going to get the same answer. Let's say I viewed this as my starting ... |
Finding the slope of a line from its graph Algebra I Khan Academy.mp3 | And I'll even go in a different direction. And I want to show you that you're going to get the same answer. Let's say I viewed this as my starting point. And I want to go all the way over there. So what is my, well, let's think about the change in y first. So the change in y, I am going down by how many units? 1, 2, 3,... |
Finding the slope of a line from its graph Algebra I Khan Academy.mp3 | And I want to go all the way over there. So what is my, well, let's think about the change in y first. So the change in y, I am going down by how many units? 1, 2, 3, 4 units. So my change in y in this example is negative 4. I went from 1 to negative 3. That's negative 4. |
Finding the slope of a line from its graph Algebra I Khan Academy.mp3 | 1, 2, 3, 4 units. So my change in y in this example is negative 4. I went from 1 to negative 3. That's negative 4. That's my change in y. Change in y is equal to negative 4. Now what is my change in x? |
Finding the slope of a line from its graph Algebra I Khan Academy.mp3 | That's negative 4. That's my change in y. Change in y is equal to negative 4. Now what is my change in x? Well, I'm going from this point, or from this x value, all the way back like this. So I'm going to the left. So it's going to be a negative change in x. |
Finding the slope of a line from its graph Algebra I Khan Academy.mp3 | Now what is my change in x? Well, I'm going from this point, or from this x value, all the way back like this. So I'm going to the left. So it's going to be a negative change in x. And I went 1, 2, 3, 4, 5, 6 units back. So my change in x is equal to negative 6. Change in x is equal to negative 6. |
Finding the slope of a line from its graph Algebra I Khan Academy.mp3 | So it's going to be a negative change in x. And I went 1, 2, 3, 4, 5, 6 units back. So my change in x is equal to negative 6. Change in x is equal to negative 6. And you can even see I started at an x is equal to 3. And I went all the way to x is equal to a negative 3. That's a change of negative 6. |
Finding the slope of a line from its graph Algebra I Khan Academy.mp3 | Change in x is equal to negative 6. And you can even see I started at an x is equal to 3. And I went all the way to x is equal to a negative 3. That's a change of negative 6. I went 6 to the left, or a change of negative 6. So what is my change in y over change in x? My change in y over change in x is equal to negative... |
Finding the slope of a line from its graph Algebra I Khan Academy.mp3 | That's a change of negative 6. I went 6 to the left, or a change of negative 6. So what is my change in y over change in x? My change in y over change in x is equal to negative 4 over negative 6. The negatives cancel out. And what's 4 over 6? Well, that's just 2 over 3. |
Finding the slope of a line from its graph Algebra I Khan Academy.mp3 | My change in y over change in x is equal to negative 4 over negative 6. The negatives cancel out. And what's 4 over 6? Well, that's just 2 over 3. So it's the same value. You just have to be consistent. If this is my start point, I went down 4. |
Finding the slope of a line from its graph Algebra I Khan Academy.mp3 | Well, that's just 2 over 3. So it's the same value. You just have to be consistent. If this is my start point, I went down 4. And then I went back 6. Negative 4 over negative 6. If I viewed this as my starting point, I could say that I went up 4. |
Finding the slope of a line from its graph Algebra I Khan Academy.mp3 | If this is my start point, I went down 4. And then I went back 6. Negative 4 over negative 6. If I viewed this as my starting point, I could say that I went up 4. So it would be a change in y would be 4. And then my change in x would be 6. And either way, once again, change in y over change in x is going to be 4 over 6... |
Basic linear equation word problem 7th grade Khan Academy.mp3 | She makes $27 on every computer she sells, and her monthly expenses are $10,000. What is the minimum number of computers she needs to sell in a month to make a profit? So I'll let you think about that for a second. Well, let's think about what we have to figure out. We have to figure out the minimum number of computers... |
Basic linear equation word problem 7th grade Khan Academy.mp3 | Well, let's think about what we have to figure out. We have to figure out the minimum number of computers she needs to sell. So let's set that to a variable or set a variable to represent that. So let's let x equal the number of computers she sells. Number of computers sold. Now let's think about how much net profit sh... |
Basic linear equation word problem 7th grade Khan Academy.mp3 | So let's let x equal the number of computers she sells. Number of computers sold. Now let's think about how much net profit she will make in a month. And that's what we're thinking about. How many computers, the minimum number she needs to sell in order to make a net profit. So I'll write her profit. Her profit is goin... |
Basic linear equation word problem 7th grade Khan Academy.mp3 | And that's what we're thinking about. How many computers, the minimum number she needs to sell in order to make a net profit. So I'll write her profit. Her profit is going to be how much money she brings in from selling the computers. And she makes $27 on every computer she sells. So her profit is going to be $27 times... |
Basic linear equation word problem 7th grade Khan Academy.mp3 | Her profit is going to be how much money she brings in from selling the computers. And she makes $27 on every computer she sells. So her profit is going to be $27 times the number of computers she sells. She gets $27 per computer times the number she sells. But we're not done yet. She still has expenses of $10,000 per ... |
Basic linear equation word problem 7th grade Khan Academy.mp3 | She gets $27 per computer times the number she sells. But we're not done yet. She still has expenses of $10,000 per month. So we're going to have to subtract out the $10,000. What we care about is making a profit. We want this number right over here to be greater than 0. So let's just think about what number of compute... |
Basic linear equation word problem 7th grade Khan Academy.mp3 | So we're going to have to subtract out the $10,000. What we care about is making a profit. We want this number right over here to be greater than 0. So let's just think about what number of computers would get us to 0 if maybe she needs to sell a little bit more than that. So let's see what gets her to break even. So b... |
Basic linear equation word problem 7th grade Khan Academy.mp3 | So let's just think about what number of computers would get us to 0 if maybe she needs to sell a little bit more than that. So let's see what gets her to break even. So break even, that's 0 profit, neither positive or negative, is equal to 27 times, and I'll do it all in one color now, 27x minus 10,000. Well, we've se... |
Basic linear equation word problem 7th grade Khan Academy.mp3 | Well, we've seen equations like this before. We can add 10,000 to both sides. So let's do both sides. Add 10,000 to both sides so it's no longer on the right-hand side. And we are left with 10,000 is equal to 27x. And then to solve for x, we just have to divide both sides by 27. Let's do that. |
Basic linear equation word problem 7th grade Khan Academy.mp3 | Add 10,000 to both sides so it's no longer on the right-hand side. And we are left with 10,000 is equal to 27x. And then to solve for x, we just have to divide both sides by 27. Let's do that. Divide both sides by 27. On our right-hand side, we have x. So let me just write this down. |
Basic linear equation word problem 7th grade Khan Academy.mp3 | Let's do that. Divide both sides by 27. On our right-hand side, we have x. So let me just write this down. So we have x on our right-hand side is going to be equal to 10,000 over 27. I've switched the right and the left-hand sides here. Now, what is this going to be? |
Basic linear equation word problem 7th grade Khan Academy.mp3 | So let me just write this down. So we have x on our right-hand side is going to be equal to 10,000 over 27. I've switched the right and the left-hand sides here. Now, what is this going to be? Well, we can do a little bit of long division to handle that. So 27 goes into 10,000. So 27 doesn't go into 1, doesn't go into ... |
Basic linear equation word problem 7th grade Khan Academy.mp3 | Now, what is this going to be? Well, we can do a little bit of long division to handle that. So 27 goes into 10,000. So 27 doesn't go into 1, doesn't go into 10. It goes into 100 three times. 3 times 27 is what, 81. 100 minus 81 is 19. |
Basic linear equation word problem 7th grade Khan Academy.mp3 | So 27 doesn't go into 1, doesn't go into 10. It goes into 100 three times. 3 times 27 is what, 81. 100 minus 81 is 19. Then we can bring down a 0. 27 goes into 190. It looks like it'll go into it about 6 times. |
Basic linear equation word problem 7th grade Khan Academy.mp3 | 100 minus 81 is 19. Then we can bring down a 0. 27 goes into 190. It looks like it'll go into it about 6 times. Let's see if that's right. 6 times 7 is 42. 6 times 2 is 12 plus 4 is 16. |
Basic linear equation word problem 7th grade Khan Academy.mp3 | It looks like it'll go into it about 6 times. Let's see if that's right. 6 times 7 is 42. 6 times 2 is 12 plus 4 is 16. Let's see, 90 minus 62 is actually 28. Oh, yes. Let me turn this back. |
Basic linear equation word problem 7th grade Khan Academy.mp3 | 6 times 2 is 12 plus 4 is 16. Let's see, 90 minus 62 is actually 28. Oh, yes. Let me turn this back. So it goes 7 times. 7 times 7 is 49. 7 times 2 is 14 plus 4 is 18. |
Basic linear equation word problem 7th grade Khan Academy.mp3 | Let me turn this back. So it goes 7 times. 7 times 7 is 49. 7 times 2 is 14 plus 4 is 18. There you go. Look at that. So 190 minus 189, we get 1. |
Basic linear equation word problem 7th grade Khan Academy.mp3 | 7 times 2 is 14 plus 4 is 18. There you go. Look at that. So 190 minus 189, we get 1. Let's bring down another 0. We have a 0 right there. 27 goes into 10 how many times? |
Basic linear equation word problem 7th grade Khan Academy.mp3 | So 190 minus 189, we get 1. Let's bring down another 0. We have a 0 right there. 27 goes into 10 how many times? Well, it doesn't go into 10 at all. So we put a 0 right there. 0 times 27 is 0. |
Basic linear equation word problem 7th grade Khan Academy.mp3 | 27 goes into 10 how many times? Well, it doesn't go into 10 at all. So we put a 0 right there. 0 times 27 is 0. Then we subtract. And then we get 10 again. And now we're in the decimal range, or we're going to start getting decimal values. |
Basic linear equation word problem 7th grade Khan Academy.mp3 | 0 times 27 is 0. Then we subtract. And then we get 10 again. And now we're in the decimal range, or we're going to start getting decimal values. But we could bring down another 0. We could get 27 goes into 100 three times. So our x value is going to be approximately 370.3. |
Basic linear equation word problem 7th grade Khan Academy.mp3 | And now we're in the decimal range, or we're going to start getting decimal values. But we could bring down another 0. We could get 27 goes into 100 three times. So our x value is going to be approximately 370.3. And then we're going to keep going on and on and on and on. But this is enough information for us to answer... |
Basic linear equation word problem 7th grade Khan Academy.mp3 | So our x value is going to be approximately 370.3. And then we're going to keep going on and on and on and on. But this is enough information for us to answer our question. What is the minimum number of computers she needs to sell in a month to make a profit? Well, she can't sell a decimal number of computers, a third ... |
Adding fractions with different signs Fractions Pre-Algebra Khan Academy.mp3 | Let's just do the first part first. It's pretty straightforward. We have two positive numbers. Let me draw a number line. And I'll try to focus in. So we're going to start at 3 1 8ths. So let's make this 0. |
Adding fractions with different signs Fractions Pre-Algebra Khan Academy.mp3 | Let me draw a number line. And I'll try to focus in. So we're going to start at 3 1 8ths. So let's make this 0. So you have 1, 2, 3, and then you have 4. 3 1 8ths is going to be right about there. So let me just draw its absolute value. |
Adding fractions with different signs Fractions Pre-Algebra Khan Academy.mp3 | So let's make this 0. So you have 1, 2, 3, and then you have 4. 3 1 8ths is going to be right about there. So let me just draw its absolute value. So this 3 1 8ths is going to be 3 1 8ths to the right of 0. So it's going to be exactly that distance from 0 to the right. So this right here, the length of this arrow, you ... |
Adding fractions with different signs Fractions Pre-Algebra Khan Academy.mp3 | So let me just draw its absolute value. So this 3 1 8ths is going to be 3 1 8ths to the right of 0. So it's going to be exactly that distance from 0 to the right. So this right here, the length of this arrow, you could view it as 3 1 8ths. Now, whenever I like to deal with fractions, especially when they have different... |
Adding fractions with different signs Fractions Pre-Algebra Khan Academy.mp3 | So this right here, the length of this arrow, you could view it as 3 1 8ths. Now, whenever I like to deal with fractions, especially when they have different denominators and all of that, I like to deal with them as improper fractions. It makes the addition and the subtraction and actually the multiplication and the di... |
Adding fractions with different signs Fractions Pre-Algebra Khan Academy.mp3 | So 3 1 8ths is the same thing as 8 times 3 is 24, plus 1 is 25 over 8. So this is 25 over 8, which is the same thing as 3 1 8ths. Another way to think about it, 3 is 24 over 8. And you add 1 8th to that, so you get 25 over 8. So this is our starting point. Now, to that, we're going to add 3 4ths. So we're going to move... |
Adding fractions with different signs Fractions Pre-Algebra Khan Academy.mp3 | And you add 1 8th to that, so you get 25 over 8. So this is our starting point. Now, to that, we're going to add 3 4ths. So we're going to move another 3 4ths. We are going to move another 3 4ths. It's hard drawing these arrows. We're going to move another 3 4ths to the right. |
Adding fractions with different signs Fractions Pre-Algebra Khan Academy.mp3 | So we're going to move another 3 4ths. We are going to move another 3 4ths. It's hard drawing these arrows. We're going to move another 3 4ths to the right. So this right here, the length of this that we're moving to the right is 3 4ths, so plus 3 4ths. Now, where does this put us? Well, both of these are positive inte... |
Adding fractions with different signs Fractions Pre-Algebra Khan Academy.mp3 | We're going to move another 3 4ths to the right. So this right here, the length of this that we're moving to the right is 3 4ths, so plus 3 4ths. Now, where does this put us? Well, both of these are positive integers, so we can just add them. We just have to find a like denominator. So we have 25 over 8. We have 25 ove... |
Adding fractions with different signs Fractions Pre-Algebra Khan Academy.mp3 | Well, both of these are positive integers, so we can just add them. We just have to find a like denominator. So we have 25 over 8. We have 25 over 8 plus 3 4ths. That's the same thing as we need to find a common denominator here. The common denominator, or the least common multiple of 4 and 8 is 8. So it's going to be ... |
Adding fractions with different signs Fractions Pre-Algebra Khan Academy.mp3 | We have 25 over 8 plus 3 4ths. That's the same thing as we need to find a common denominator here. The common denominator, or the least common multiple of 4 and 8 is 8. So it's going to be something over 8. To get from 4 to 8, we multiply by 2. So we have to multiply 3 by 2 as well. So you get 6. |
Adding fractions with different signs Fractions Pre-Algebra Khan Academy.mp3 | So it's going to be something over 8. To get from 4 to 8, we multiply by 2. So we have to multiply 3 by 2 as well. So you get 6. So 3 4ths is the same thing as 6 8ths. If we have 25 8ths and we're adding 6 8ths to that, that gives us 25 plus 6 is 31 over 8. So this number right over here is 31 over 8. |
Adding fractions with different signs Fractions Pre-Algebra Khan Academy.mp3 | So you get 6. So 3 4ths is the same thing as 6 8ths. If we have 25 8ths and we're adding 6 8ths to that, that gives us 25 plus 6 is 31 over 8. So this number right over here is 31 over 8. And it makes sense, because 32 over 8 would be 4. So it should be a little bit less than 4. So this number right over here is 31 ove... |
Adding fractions with different signs Fractions Pre-Algebra Khan Academy.mp3 | So this number right over here is 31 over 8. And it makes sense, because 32 over 8 would be 4. So it should be a little bit less than 4. So this number right over here is 31 over 8. Or the length of this arrow, the absolute value of that number is 31 over 8. A little bit less than 4. If you wanted to write that as a mi... |
Adding fractions with different signs Fractions Pre-Algebra Khan Academy.mp3 | So this number right over here is 31 over 8. Or the length of this arrow, the absolute value of that number is 31 over 8. A little bit less than 4. If you wanted to write that as a mixed number, it would be 3 and 7 8ths. So that's that right over here. This is 31 over 8. That's that part right over there. |
Adding fractions with different signs Fractions Pre-Algebra Khan Academy.mp3 | If you wanted to write that as a mixed number, it would be 3 and 7 8ths. So that's that right over here. This is 31 over 8. That's that part right over there. Now to that, we want to add a negative 2 and 1 6. So we're going to add a negative number. So think about what negative 2 and 1 6 is going to be like. |
Adding fractions with different signs Fractions Pre-Algebra Khan Academy.mp3 | That's that part right over there. Now to that, we want to add a negative 2 and 1 6. So we're going to add a negative number. So think about what negative 2 and 1 6 is going to be like. Let me do this in a new color. Do it in pink. Negative 2 and 1 6. |
Adding fractions with different signs Fractions Pre-Algebra Khan Academy.mp3 | So think about what negative 2 and 1 6 is going to be like. Let me do this in a new color. Do it in pink. Negative 2 and 1 6. So we're going to subtract, or I guess we're going to add a negative 1. We're going to add a negative 2 and then a negative 1 6. Let me draw. |
Adding fractions with different signs Fractions Pre-Algebra Khan Academy.mp3 | Negative 2 and 1 6. So we're going to subtract, or I guess we're going to add a negative 1. We're going to add a negative 2 and then a negative 1 6. Let me draw. So negative 2 and 1 6, we could literally draw like this. Negative 2 and 1 6, we can draw with an arrow that looks something like that. So this is negative 2 ... |
Adding fractions with different signs Fractions Pre-Algebra Khan Academy.mp3 | Let me draw. So negative 2 and 1 6, we could literally draw like this. Negative 2 and 1 6, we can draw with an arrow that looks something like that. So this is negative 2 and 1 6. Now there's a couple of ways to think about it. We could just say, hey look, when you add this arrow, this thing that's moving to the left, ... |
Adding fractions with different signs Fractions Pre-Algebra Khan Academy.mp3 | So this is negative 2 and 1 6. Now there's a couple of ways to think about it. We could just say, hey look, when you add this arrow, this thing that's moving to the left, we could put it over here and you would get straight to negative 2 and 1 6. But we're adding this negative 2 and 1 6. It's the same thing as subtract... |
Adding fractions with different signs Fractions Pre-Algebra Khan Academy.mp3 | But we're adding this negative 2 and 1 6. It's the same thing as subtracting a positive 2 and 1 6. We're moving 2 and 1 6 to the left. And we're going to end up with a number whose absolute value is going to look something like that. And it's actually going to be to the right. So it's not going to only be its absolute ... |
Adding fractions with different signs Fractions Pre-Algebra Khan Academy.mp3 | And we're going to end up with a number whose absolute value is going to look something like that. And it's actually going to be to the right. So it's not going to only be its absolute value. It's going to be what's absolute value is going to be the number, since it's going to be a positive number. So let's just think ... |
Adding fractions with different signs Fractions Pre-Algebra Khan Academy.mp3 | It's going to be what's absolute value is going to be the number, since it's going to be a positive number. So let's just think about what it is. This value right here, which is going to be the answer to our problem, is just going to be the difference of 31 over 8 and 2 and 1 6. And it's the positive difference, becaus... |
Adding fractions with different signs Fractions Pre-Algebra Khan Academy.mp3 | And it's the positive difference, because we're dealing with a positive number. So we just take 31 over 8. And from that, we will subtract 2 and 1 6. So let's do this. So this orange value is going to be 31 over 8 minus 2 and 1 6. So 2 and 1 6 is the same thing as 6 times 2 is 12 plus 1 is 13 minus 13 over 6. And this ... |
Adding fractions with different signs Fractions Pre-Algebra Khan Academy.mp3 | So let's do this. So this orange value is going to be 31 over 8 minus 2 and 1 6. So 2 and 1 6 is the same thing as 6 times 2 is 12 plus 1 is 13 minus 13 over 6. And this is equal to, once again, we need to get a common denominator over here. And it looks like 24 will be the common denominator. Let me make it very clear... |
Adding fractions with different signs Fractions Pre-Algebra Khan Academy.mp3 | And this is equal to, once again, we need to get a common denominator over here. And it looks like 24 will be the common denominator. Let me make it very clear. This is the 31 over 8. And this is the 2 and 1 6. This right here is the 2 and 1 6. So 31 over 8 over 24, you have to multiply by 3 to get to the 24 over here. |
Adding fractions with different signs Fractions Pre-Algebra Khan Academy.mp3 | This is the 31 over 8. And this is the 2 and 1 6. This right here is the 2 and 1 6. So 31 over 8 over 24, you have to multiply by 3 to get to the 24 over here. So we multiply by 3 on the 31. That gives us 93. And then to go from 6 to 24, you have to multiply by 4. |
Adding fractions with different signs Fractions Pre-Algebra Khan Academy.mp3 | So 31 over 8 over 24, you have to multiply by 3 to get to the 24 over here. So we multiply by 3 on the 31. That gives us 93. And then to go from 6 to 24, you have to multiply by 4. Let me do that in another color. You have to multiply it by 4. So we have to multiply by 4 up here as well. |
Adding fractions with different signs Fractions Pre-Algebra Khan Academy.mp3 | And then to go from 6 to 24, you have to multiply by 4. Let me do that in another color. You have to multiply it by 4. So we have to multiply by 4 up here as well. So 4 times 13. Let's see. 4 times 10 is 40. |
Adding fractions with different signs Fractions Pre-Algebra Khan Academy.mp3 | So we have to multiply by 4 up here as well. So 4 times 13. Let's see. 4 times 10 is 40. 4 times 3 is 12. So that's 52. So this is going to be equal to 93 minus 52 over 24. |
Adding fractions with different signs Fractions Pre-Algebra Khan Academy.mp3 | 4 times 10 is 40. 4 times 3 is 12. So that's 52. So this is going to be equal to 93 minus 52 over 24. And that is, so 93 minus 52. 3 minus 2 is 1. 9 minus 5 is 4. |
Adding fractions with different signs Fractions Pre-Algebra Khan Academy.mp3 | So this is going to be equal to 93 minus 52 over 24. And that is, so 93 minus 52. 3 minus 2 is 1. 9 minus 5 is 4. So it is 41 over 24. Positive. And you can see that here just by looking at the number line. |
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