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We're on problem-- where are we? Problem 232. Only a few left. If 1/x minus 1 over x plus 1 is equal to 1 over x plus 4, then x could be which of the following? And you might have a temptation to just invert both sides, but you can't do that just yet. Because if you took the inverse of this, it's not like you just get x minus x plus 1. You actually have to subtract, or you could say add these two fractions together. So let's do that. So if we simplify this left-hand side, a common denominator would be x times x plus 1. And then a numerator would be, let's see-- 1/x is the same thing as x plus 1 over this thing, because that cancels out and you get 1/x, minus-- 1 over x plus 1, that's the same thing as x/x times x plus 1. The x's would cancel out, you get 1 over x plus 1. So this left-hand side, the x's cancel out and you get 1 over-- well, we could just make that x squared plus x, right? So now this simplifies to-- the left-hand side is 1 over x squared plus x, and the right-hand side is 1 over x plus 4. And now we can invert both sides or just set the denominators as equal to each other. But let's just invert both. It's the same thing, actually. So if you take the inverse of both sides and you get-- and I'll just switch colors just randomly-- x squared plus x is equal to x plus 4. Let's subtract x from both sides. You get x squared is equal to 4. So x is equal to plus or minus 2. And they say then x could be-- well, choice C is minus 2, so that's one of the values that x could be. Next question, 233. All right, they have 1/2 to the minus 3 power times 1/4 to the minus 2 power times 1/16 to the minus 1. The first thing I do when I see negative exponents, just so they don't confuse me, is that I just invert all of the numbers. So this is the same thing. This is the exact same thing as equal to-- this is the same thing as 2 to the third power, right? When you take something to a negative exponent, that's the same thing as the inverse of that thing to the positive exponent, so this is 2 to the third power. This is times 4 squared times 16. And so 2 to the third is 8 times 16 times 16, right? That's 8 times 256. And then-- oh, wait. I'm going down the wrong path, because they have it in completely different format. So actually, maybe I should look at the choices before I continue these problems. So let me rewrite it. So they all have it as -- you know, 1/2 to some power, 1/8 to some power, so let me rewrite this. I went down the wrong road. So let's write all of these as powers of 1/2. So 1/2 to the minus 1/3, fine. That's still 1/2 to the minus 1/3. 1/4, what is that? 1/4 is the same thing as 1/2 squared. We're raising that to the minus 2. And then 1/16, that's the same thing as 1/2 to the third power is 1/8, so it's to the fourth power. We're taking that to the negative 1. And then this is-- we'll just use what we know about exponent rules, and we've proven this in other Khan Academy videos. 1/2 squared to the minus 2. You can just multiply these exponents, so that's equal to 1/2 to the minus 4. And then 1/2 to the 4 to the minus 1, you multiply them and that's equal to minus 4. And now we have the same base, and we're multiplying them, so we can just add the exponents. This is equal to 1/2 to the minus 3 plus minus 4, minus 7, minus 4, is equal to minus 11. So it's 1/2 to the negative 11th power, which is choice B. That was a good lesson there. Look at your choices before you go down some path because I was actually going to evaluate this thing, which would have taken a little bit of extra time. Problem 234. In a certain game, a large container is filled with red, yellow, green-- so let's see, there's red, yellow, green, and blue beads, worth respectively-- so respectively means the same order; we said the colors are in the same order that they're worth -- so worth respectively 7, 5, 3, and 2 points each. OK, a number of beads are then removed from the container. If the product of the point of values of the removed beads is 147,000, how many red beads were removed? So this is interesting. So there's something that hopefully will jump out at you about these numbers. They're all prime numbers, right? So if you look at any number, you can tell how many of a certain prime number there is in that number by just doing a prime factorization. So let's do that for 147,000. And you'd be surprised how quickly a number can get smaller. So let's see. How many times-- let's see, does 3 go 147? I think we've done this before. Sure, 3 times -- let me just do it up here. 3 goes into 147 4, 12, 27, 49 times. So you get a 3 and you get a 49, right? 49,000, I should say. You could do a 7 and a 7,000. And then you can get another 7, and you can get a 1,000. And then you could do 2 a bunch of times. So you get a 2, you get a -- well, the main thing, just so we don't waste time is they want to know-- what color do they want to know? They want to know how many red beads were removed, right? So red corresponds to 7. So all the beads, you multiply them together and you get 147,000. So we've essentially factored out all the 7's that we can fit into 147,000. 7 does not go into 1,000 at all. 7 is not a factor. So we've already solved our problem. There are exactly two 7 prime factors in 147,000. You could keep going with this. You know, you get a 2. 2 times 500. You get another 2. 2 times 250. There's a bunch of 2's in here. And then you'd probably get-- I don't know what else you're going to get. You'd probably get a couple of 5's in here, and then you'd be done. But these are all the 7's. There are two 7's in 147,000. Two 7 prime factors, so there would be 2 red beads. And that is choice D. Problem 235. Switch colors. 235. If 2 over 1 plus 2 over y is equal to 1, then what does y equal? Let's take the inverse of both. Let's multiply both sides of the equation times 1 plus 2/y. And so we get 2 is equal to 1 plus 2/y. And let's multiply both sides of the equation by y. So you get 2y. And actually, you could probably already eyeball this. Actually there's an easier way. Let's subtract 1 from both sides. You get 1 is equal to 2/y. Multiply both sides by y and you get y is equal to 2. And that is choice D. And you could've actually eyeballed it, because, say, 1 plus 1 is equal to 2, so this has to be 1. So how am I going to make that a 1? That equals 2, but sometimes it's easier just to mechanically go through the algebra, because sometimes, you can't eyeball it, and it's good to just keep moving forward. Problem 236. If a, b, and c are consecutive positive integers, and a is less than b, which is less than c, which of the following must be true? Consecutive positive, so, you know, we could say that b would be equal to a plus 1, and c would be equal to a plus 2 or b plus 1. And, of course, a is equal to a. I don't know if that helps, I just felt like writing that down. So in statement 1, they say c minus a is equal to 2. Sure. They're consecutive positive integers, so c is going to be 2 more than a. So this is right. Statement 2, they tell us abc is even. Well, instead of going into some algebraic thing where I represent even and odd integers and all of that, let's just try it with two cases, one where a is odd and one where a is even. So let's say a is even. So the numbers would be 2-- sorry, would be 2 times 3 times 4, right? So then we would get 8. Well, you know what? This has to be even because one of these numbers are going to be even, right? And so if you're divisible, and by definition, if one of these numbers are even, it's divisible by 2. So the whole thing, when you multiply it together, it's got to be divisible by 2, right? Either a and c are even, in which case this whole thing has to be even, or b is even, in which case this whole thing is even, right? If you multiply a string of numbers and at least one of them is even, then that whole number is going to be even, because that number that's even is a factor of that whole string of numbers. But anyway, you could try it with 2, 3, 4, or you could try it with 3, 4, 5. Either way, you're going to get an even number. So this is also right. Statement 3. a plus b plus c over 3 is an integer. Well, let's think about it. I don't know. Let's say if we were to take a plus b plus c. I don't know, even if we were to take-- let's see, 3 plus 4 plus-- no, that works. 2 plus 3-- no, that works. Let's write it as-- so this is a plus a plus 1, right? B is a plus 1, plus a plus 2 over 3, right? And so this is equal to 3a plus 3 over 3. b is just a plus 1, c is a plus 2. You add them all together. Sure enough, you know, you could divide the top and the bottom by 3 and you get a plus 1. So yes, statement 3 is also correct. This is always going to evaluate to a plus 1, which is an integer. So all three are true, so that is choice E. And I'm all out of time. See you in the next video.