SAT Prep: Test 4 Section 2 Part 6 Video

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SAT Prep: Test 4 Section 2 Part 6

We are on problem number 19, if A, B, C and F are four non-zero numbers that all of the following proportions are equivalent except. So this is interesting, this is kind of when we have to work through all of them so choice A. A/F = V/C, choice B says, F/C = B/A so the real simple way of doing this, we could take the first. Let’s cross-multiply all of them essentially, so we can cross-multiply. If we multiply both sides by essentially F, C and that is what cross-multiplying is. So we say A(C), A/C = B/F, and you could get there and multiply both sides by F and both multiplying both sides by C. If we cross-multiply in this side, we have A/F =B/C, so choice C is C/A = F/B what is this tell us? This tells us that B/C = A/F so this two are equal, these are the same thing. And D, we have A/C = B/F this is if you cross multiply it you get A/F is equal to B/C. This are equal, I think you see a pattern and I suspected E is going to look very similar to these three and this, the different one. But just to do it I will show you, choice E has A/F over B/C = 1 essentially. So that is A/F = B/C which is the same thing as all D, so A is the one choice that is different. A/C is equal to B/F is very different than A/F = B/C right? We are multiplying different things, these are all the same. Hopefully you see that and that is you know multiplication is, communicative and all that. So it doesn’t matter what order you F/A and A/F are the same thing. But here on multiplying, it was in A/C=B/F, let’s do problem number 20. And you could actually in the last problem. Let me try it with the four different non-zero numbers and then you will actually – it should work out if you pick good numbers. If you don’t want to deal with the abstract letters. Problem number 20, for all numbers X and Y let the operations square be defined as X²Y= XY – Y. If A and B are positive integers which of the following can be equal to zero? So AB positive integers, so which can be equal to zero? So they are saying choice number one is A²B. And that we’ve learned from here is the same thing as A(B) so this pattern matching where we see the X from A and where we see the B or Y you put a B. So that is AB – B and I can come up with the situation where this comes out to be zero? Well sure, if A is one, what happens when A is one? If A is one and this translate to B – B = 0. And why did I think what happens when A is one, well this could also be written as A – 1(B) if you just kind of factor about the B’s. So it says well if A is one and this term becomes zero. So one, this definitely can be equal to zero, it is not always zero but it can be especially if A is one. That is our choice two, they wrote A+B²B and what is that equal too? Well everytime you saw a X with basically with A+B everytime you see a Y replace it with B. So that is equal to A+B(B) times the second term, minus the second term minus B. So we can’t do what we did the previous time because now the term A+B can’t equal one. There is no circumstances under A+B=1 how? Because they are both positive, they are both integers. So in order for this to equal one, one would have to be one and one would have to be zero. And neither of them can be negative, so this can’t equal one, so that is not a situation in which where this is equal to zero. So let’s see if we can solve it, -- and of course B can’t be equal to zero, so let’s say A+B. We could factor it out, we can say A+B – 1 (B) = 0, we know B can’t equal to zero under any circumstances. We know B can’t equal zero because we are saying it’s a positive integer. And so this term would have to be equal to zero and all I did. Just you could multiply this out, you see where I got, I just factored out just the same. I divide each of these terms by B and factor it out and then you are left with A+B – 1(B). So if B can’t equal zero, can A+B – 1 = 0? Well A+B-1=0 then A+B would have to equal one which I said it just can’t happen. So under no circumstances can choice two B=0, and all those choice three say? Choice three is A² A+B and if we map it that is A(A+B – A+B) and lot here. We once again we can factor out the A+B so we can write this as A-1(A+B). And we want to set that equals zero – see if this possible. Well for this to B=0 either this term would have to be zero or this term or both of them really. Can A+B=0? No, because A is positive, B is positive, there are both integers, I thing it’s negative they have to be greater than zero so there is no way that this is going to be equals zero. Well can A-1=0? Well sure, A could be one, if A is one that this expression becomes zero. So this one also works, so our choice are the ones that could be zero are one and three. And those are choices that is choice E, I will see you on the next section.

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We are on problem number 19, if A, B, C and F are four non-zero numbers that all of the following proportions are equivalent except. So this is interesting, this is kind of when we have to work through all of them so choice A. A/F = V/C, choice B says, F/C = B/A so the real simple way of doing this, we could take the first.

Let’s cross-multiply all of them essentially, so we can cross-multiply. If we multiply both sides by essentially F, C and that is what cross-multiplying is. So we say A(C), A/C = B/F, and you could get there and multiply both sides by F and both multiplying both sides by C. If we cross-multiply in this side, we have A/F =B/C, so choice C is C/A = F/B what is this tell us? This tells us that B/C = A/F so this two are equal, these are the same thing.

And D, we have A/C = B/F this is if you cross multiply it you get A/F is equal to B/C. This are equal, I think you see a pattern and I suspected E is going to look very similar to these three and this, the different one. But just to do it I will show you, choice E has A/F over B/C = 1 essentially. So that is A/F = B/C which is the same thing as all D, so A is the one choice that is different.

A/C is equal to B/F is very different than A/F = B/C right? We are multiplying different things, these are all the same. Hopefully you see that and that is you know multiplication is, communicative and all that. So it doesn’t matter what order you F/A and A/F are the same thing. But here on multiplying, it was in A/C=B/F, let’s do problem number 20.

And you could actually in the last problem. Let me try it with the four different non-zero numbers and then you will actually – it should work out if you pick good numbers. If you don’t want to deal with the abstract letters. Problem number 20, for all numbers X and Y let the operations square be defined as X²Y= XY – Y. If A and B are positive integers which of the following can be equal to zero?

So AB positive integers, so which can be equal to zero? So they are saying choice number one is A²B. And that we’ve learned from here is the same thing as A(B) so this pattern matching where we see the X from A and where we see the B or Y you put a B. So that is AB – B and I can come up with the situation where this comes out to be zero? Well sure, if A is one, what happens when A is one? If A is one and this translate to B – B = 0.

And why did I think what happens when A is one, well this could also be written as A – 1(B) if you just kind of factor about the B’s. So it says well if A is one and this term becomes zero. So one, this definitely can be equal to zero, it is not always zero but it can be especially if A is one. That is our choice two, they wrote A+B²B and what is that equal too? Well everytime you saw a X with basically with A+B everytime you see a Y replace it with B. So that is equal to A+B(B) times the second term, minus the second term minus B.

So we can’t do what we did the previous time because now the term A+B can’t equal one. There is no circumstances under A+B=1 how? Because they are both positive, they are both integers. So in order for this to equal one, one would have to be one and one would have to be zero. And neither of them can be negative, so this can’t equal one, so that is not a situation in which where this is equal to zero. So let’s see if we can solve it, -- and of course B can’t be equal to zero, so let’s say A+B. We could factor it out, we can say A+B – 1 (B) = 0, we know B can’t equal to zero under any circumstances.

We know B can’t equal zero because we are saying it’s a positive integer. And so this term would have to be equal to zero and all I did. Just you could multiply this out, you see where I got, I just factored out just the same. I divide each of these terms by B and factor it out and then you are left with A+B – 1(B). So if B can’t equal zero, can A+B – 1 = 0? Well A+B-1=0 then A+B would have to equal one which I said it just can’t happen.

So under no circumstances can choice two B=0, and all those choice three say? Choice three is A² A+B and if we map it that is A(A+B – A+B) and lot here. We once again we can factor out the A+B so we can write this as A-1(A+B). And we want to set that equals zero – see if this possible.

Well for this to B=0 either this term would have to be zero or this term or both of them really. Can A+B=0? No, because A is positive, B is positive, there are both integers, I thing it’s negative they have to be greater than zero so there is no way that this is going to be equals zero. Well can A-1=0? Well sure, A could be one, if A is one that this expression becomes zero. So this one also works, so our choice are the ones that could be zero are one and three.

And those are choices that is choice E, I will see you on the next section.

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