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Hi, Lets make this a little complicated. Assume that, in the given problem, its is just mentioned that " x and y are positive integers". Then the approach will be a little different.

LHS : x^2 - 1 = (x+1)(x-1) Since x is positive, we will only consider (x-1)........Eq1 RHS : y^2 -4y +x-1........Eq2

Equating both: (x-1)>y^2 -4y +x-1 We get y(y-4)> 0 Therefore y>0 as the best case and must be true.
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I agree that b must be true. However, i be is true, then A must also be true. If y > 0, then it is also true that Y >= 0.

That's not true. If y=0 and x=y+1=1, then \(x^2-1 \gt y^2-4y+x-1\) does NOT hold true.

This is a problem of semantics. I agree that Y cannot be 0. What I say is that if Y is more than 0, then it must also be true that Y is more than -1 , or -2, or it is equal or more than 0.

If you give me any valid number for Y , answers Y>0 and y>= 0 are always true.
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I agree that b must be true. However, i be is true, then A must also be true. If y > 0, then it is also true that Y >= 0.

That's not true. If y=0 and x=y+1=1, then \(x^2-1 \gt y^2-4y+x-1\) does NOT hold true.

This is a problem of semantics. I agree that Y cannot be 0. What I say is that if Y is more than 0, then it must also be true that Y is more than -1 , or -2, or it is equal or more than 0.

If you give me any valid number for Y , answers Y>0 and y>= 0 are always true.

No, that's not correct.

The question asks: which of the following must be true?

A says \(y\geq{0}\), so it allows y to be 0. But y cannot be 0 because in this case \(x^2-1 \gt y^2-4y+x-1\) does NOT hold true. Thus A is not always true.
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I agree that b must be true. However, if b is true, then A must also be true. If y > 0, then it is also true that Y >= 0.

Quote:

That's not true. If y=0 and x=y+1=1, then \(x^2-1 \gt y^2-4y+x-1\) does NOT hold true.

This is a problem of semantics. I agree that Y cannot be 0. What I say is that if Y is more than 0, then it must also be true that Y is more than -1 , or -2, or it is equal or more than 0.

If you give me any valid number for Y , answers Y>0 and y>= 0 are always true.[/quote]

Quote:

No, that's not correct.

The question asks: which of the following must be true?

A says \(y\geq{0}\), so it allows y to be 0. But y cannot be 0 because in this case \(x^2-1 \gt y^2-4y+x-1\) does NOT hold true. Thus A is not always true.

Of course 0 is not true. You must not test O (zero) because it is not a valid answer. What I said is that for any valid integer of Y , y>0 is true, but also Y>=0 is true. This is what the the question asks for. Let' see it as a sets problem. B is the set of all positive integers, and A is the set of all non-negative integers. This way A has the same elements of B, as well as the 0; therefore, any element that is in B is also in A. B says that y>0, and that is true. A says that Y is equal or more than 0. If Y is more than 0 -excluding 0- it means that Y>= 0 (more or equal, one or the other). We can even say that Y must be more than -10, and it will be true. Y does not have to be 0 in order to be equal or more than 0.

Study the question carefully. The question did not ask for all possible values of Y;

The question is: Is A true for all values of Y? The answer is YES The question is not: : Is Y true for all values of A? The answer is NOT _________________

I agree that b must be true. However, if b is true, then A must also be true. If y > 0, then it is also true that Y >= 0.

Quote:

That's not true. If y=0 and x=y+1=1, then \(x^2-1 \gt y^2-4y+x-1\) does NOT hold true.

This is a problem of semantics. I agree that Y cannot be 0. What I say is that if Y is more than 0, then it must also be true that Y is more than -1 , or -2, or it is equal or more than 0.

If you give me any valid number for Y , answers Y>0 and y>= 0 are always true.

Quote:

No, that's not correct.

The question asks: which of the following must be true?

A says \(y\geq{0}\), so it allows y to be 0. But y cannot be 0 because in this case \(x^2-1 \gt y^2-4y+x-1\) does NOT hold true. Thus A is not always true.

Of course 0 is not true. You must not test O (zero) because it is not a valid answer. What I said is that for any valid integer of Y , y>0 is true, but also Y>=0 is true. This is what the the question asks for. Let' see it as a sets problem. B is the set of all positive integers, and A is the set of all non-negative integers. This way A has the same elements of B, as well as the 0; therefore, any element that is in B is also in A. B says that y>0, and that is true. A says that Y is equal or more than 0. If Y is more than 0 -excluding 0- it means that Y>= 0 (more or equal, one or the other). We can even say that Y must be more than -10, and it will be true. Y does not have to be 0 in order to be equal or more than 0.

Study the question carefully. The question did not ask for all possible values of Y;

The question is: Is A true for all values of Y? The answer is YES The question is not: : Is Y true for all values of A? The answer is NOT[/quote]

Edited option A to avoid ambiguity. Thank you.
_________________

I went about it with a mix of algebra and number plugging: Since X= Y+1

The inequality turns out to be:

Y^2+2Y > Y^2-4y+y 2y>-3y

You might notice right here that Y will increase by 2 on the left and decrease by -3 on the right with any number greater than 0 keeping the statement true. I didn't realize it right away so I just tried a few numbers: 0,1, and finally -1 (made it not true)

Also, you could further reduce the inequality like Bunuel did just to double prove it: 2y > -3y 5y>0 y>0

now since it is a "must be true" question, doesn't 3,4,5 reduce to 2? That is, if y > 8, it MUST be true that y > 0. What am I missing here?

Yes, if y > 8, then y > 0 too. But the correct answer is y > 0 (y must be greater than 0) and this does not mean that y must necessarily be greater than 8.
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