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So, for this case we have \(x < 0\) and \(-1 < y < 0\).

Check the options:

A. \(x ≤ 1\). This is not always true. For example, x can be 1.5 (from case 1)

B. \(0 ≤ y\). This is not always true. For example, y can be -0.5 (from case 2)

C. \(x ≤ 3y + 2\). This is not always true. For example, x = -0.1 and y = - 0.9 (from case 2)

D. \(|x − 2| ≤ 2 − x\). This implies that \(x - 2 ≤ 0\) or \(x ≤ 2\). Now, \(x ≤ 2\) covers all values of x possible (\(0 < x < 2\) from case 1 as well as \(x < 0\) from case 2). So, in any case \(x ≤ 2\) must be true.

E. |y| ≤ |y + 1|. This is not always true. For example, consider y = - 0.9 (from case 2).

This inequality question is best solved by thinking in terms of number properties. A hint to that effect comes from the fact that each inequality discusses whether a product is positive or negative. Let's turn to the positive/negative properties now.

In order to get \(xy<0\), we must either have

\(x>0\) and \(y>0\)

or

\(x<0\) and \(y<0\)

In order to get \((x−2)(y+1)<0\), we must either have

\(x−2>0\) and \((y+1)<0\)

\(x>2\) and \(y<−1\)

or

\(x−2<0\) and \((y+1)>0\)

\(x<2\) and \(y>−1\)

However, note that the case in which \(x>2\) and \(y<−1\) contradicts the other given inequality, since in this case x and y would have different signs and xy would have to be negative (violating the other given inequality).

Conversely, the case in which \(x<2\) and \(y>−1\) is not problematic. If \(0<x<2\) and \(y>0\) then both inequalities will be satisfied. Or if \(x<0\) and \(−1<y<0\) then again both inequalities will work.

We turn now to the answer choices.

Answer A is close, but, since x could be between 1 and 2 with y positive, it is not necessarily true that \(x≤1\). (E.g. \(x=1.1\), \(y=10\))

Answer B is close as well, but, since y could be between −1 and 0 with x negative, it is not necessarily true that \(0≤y\). (E.g. \(x=−5\), \(y=−0.1\))

Answer C is close yet again, but we cannot quite obtain this inequality algebraically (we can get \(x<2y+2\) by solving the system using elimination), and in fact this answer choice can be false if y is negative and close to −1. For instance, \(x=−0.1\), \(y=−0.9\) satisfies both given inequalities but does not satisfy answer choice C since \(3y+2=−0.7<−0.1\).

Answer D is correct. Since we've established that \(x<2\), we also know that \(x−2<0\), and therefore \(|x−2|=−(x−2)=2−x\). It is therefore true that \(|x−2|≤2−x\).

Answer E is again close but wrong. In the case of y close to −1, adding 1 will actually move y closer to 0. For example, if \(x=−0.1\) and \(y=−0.9\), then both given inequalities are satisfied but \(|y|=0.9\) is greater than \(|y+1|=|−0.9+1|=|0.1|=0.1\).
_________________

So, for this case we have \(x < 0\) and \(-1 < y < 0\).

Check the options:

A. \(x ≤ 1\). This is not always true. For example, x can be 1.5 (from case 1)

B. \(0 ≤ y\). This is not always true. For example, y can be -0.5 (from case 2)

C. \(x ≤ 3y + 2\). This is not always true. For example, x = -0.1 and y = - 0.9 (from case 2)

D. \(|x − 2| ≤ 2 − x\). This implies that \(x - 2 ≤ 0\) or \(x ≤ 2\). Now, \(x ≤ 2\) covers all values of x possible (\(0 < x < 2\) from case 1 as well as \(x < 0\) from case 2). So, in any case \(x ≤ 2\) must be true.

E. |y| ≤ |y + 1|. This is not always true. For example, consider y = - 0.9 (from case 2).

Answer: D.

Hello Bunuel, But option D says x can be equal to 2 also. If we put the value of x=2 in the premise given the x-2 will become 0 and (x-2)(y+1)<0 will not hold true. How do I approach such questions? Can you provide more such questions for practice?

So, for this case we have \(x < 0\) and \(-1 < y < 0\).

Check the options:

A. \(x ≤ 1\). This is not always true. For example, x can be 1.5 (from case 1)

B. \(0 ≤ y\). This is not always true. For example, y can be -0.5 (from case 2)

C. \(x ≤ 3y + 2\). This is not always true. For example, x = -0.1 and y = - 0.9 (from case 2)

D. \(|x − 2| ≤ 2 − x\). This implies that \(x - 2 ≤ 0\) or \(x ≤ 2\). Now, \(x ≤ 2\) covers all values of x possible (\(0 < x < 2\) from case 1 as well as \(x < 0\) from case 2). So, in any case \(x ≤ 2\) must be true.

E. |y| ≤ |y + 1|. This is not always true. For example, consider y = - 0.9 (from case 2).

Answer: D.

Hello Bunuel, But option D says x can be equal to 2 also. If we put the value of x=2 in the premise given the x-2 will become 0 and (x-2)(y+1)<0 will not hold true. How do I approach such questions? Can you provide more such questions for practice?

Regards

It is the other way around. We know that \(x < 0\) or \(0 < x < 2\). These are the ranges for all possible values of x. Now, for any possible x (again from \(x < 0\) or from \(0 < x < 2\)) \(|x − 2| ≤ 2 − x\) will be true.

Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be [#permalink]

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07 May 2017, 09:17

Bunuel wrote:

It is the other way around. We know that \(x < 0\) or \(0 < x < 2\). These are the ranges for all possible values of x. Now, for any possible x (again from \(x < 0\) or from \(0 < x < 2\)) \(|x − 2| ≤ 2 − x\) will be true.

Hope it's clear.

But the inequality is strict for x. X cannot be 2 else the premise will go for a toss. Regards
_________________

It is the other way around. We know that \(x < 0\) or \(0 < x < 2\). These are the ranges for all possible values of x. Now, for any possible x (again from \(x < 0\) or from \(0 < x < 2\)) \(|x − 2| ≤ 2 − x\) will be true.

Hope it's clear.

But the inequality is strict for x. X cannot be 2 else the premise will go for a toss. Regards

I'll try to explain it once more. Again, x CANNOT be 2 because we know that \(x < 0\) or \(0 < x < 2\). For any value of x possible \(|x − 2| ≤ 2 − x\) will be true.
_________________

Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be [#permalink]

Show Tags

07 May 2017, 09:23

Bunuel wrote:

gmatexam439 wrote:

Bunuel wrote:

It is the other way around. We know that \(x < 0\) or \(0 < x < 2\). These are the ranges for all possible values of x. Now, for any possible x (again from \(x < 0\) or from \(0 < x < 2\)) \(|x − 2| ≤ 2 − x\) will be true.

Hope it's clear.

But the inequality is strict for x. X cannot be 2 else the premise will go for a toss. Regards

I'll try to explain it once more. Again, x CANNOT be 2 because we know that \(x < 0\) or \(0 < x < 2\). For any value of x possible \(|x − 2| ≤ 2 − x\) will be true.

Ok ok, I understand. Sorry for commotion. Can you please share more such questions or point out a few links for practice of such types of question. Thanks a lot Bunuel.
_________________

So, for this case we have \(x < 0\) and \(-1 < y < 0\).

Check the options:

A. \(x ≤ 1\). This is not always true. For example, x can be 1.5 (from case 1)

B. \(0 ≤ y\). This is not always true. For example, y can be -0.5 (from case 2)

C. \(x ≤ 3y + 2\). This is not always true. For example, x = -0.1 and y = - 0.9 (from case 2)

D. \(|x − 2| ≤ 2 − x\). This implies that \(x - 2 ≤ 0\) or \(x ≤ 2\). Now, \(x ≤ 2\) covers all values of x possible (\(0 < x < 2\) from case 1 as well as \(x < 0\) from case 2). So, in any case \(x ≤ 2\) must be true.

E. |y| ≤ |y + 1|. This is not always true. For example, consider y = - 0.9 (from case 2).

So, for this case we have \(x < 0\) and \(-1 < y < 0\).

Check the options:

A. \(x ≤ 1\). This is not always true. For example, x can be 1.5 (from case 1)

B. \(0 ≤ y\). This is not always true. For example, y can be -0.5 (from case 2)

C. \(x ≤ 3y + 2\). This is not always true. For example, x = -0.1 and y = - 0.9 (from case 2)

D. \(|x − 2| ≤ 2 − x\). This implies that \(x - 2 ≤ 0\) or \(x ≤ 2\). Now, \(x ≤ 2\) covers all values of x possible (\(0 < x < 2\) from case 1 as well as \(x < 0\) from case 2). So, in any case \(x ≤ 2\) must be true.

E. |y| ≤ |y + 1|. This is not always true. For example, consider y = - 0.9 (from case 2).

Answer: D.

How did u get 0<x<2 when x-2<0

Because x<2, x can take any value below 2

pls explain

Thanks in advance

Case 1 considers the case when: [/b] if \(x > 0\) and \(y > 0\). So, x < 2 and x > 0 --> 0 < x < 2.
_________________

Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be [#permalink]

Show Tags

07 Oct 2017, 03:51

xy>0 means x&y have same sign.

(x-2)(y+1)<0 means (x-2) & (y+1) have opposite sign.

Case 1: If x&y are both positive- In such case, it's (x-2) that led to the sign change by shifting two units to left on the number line and entering negative territory. This means that x is any value between 0 and 2 when y>0.

Case 2: If x&y are both negative- In such case, it's (y+1) that led to the sign change by shifting one units to right on the number line and entering positive territory. This means that y is any value between -1 and 0 when x<0.

Combining Case 1&2, we can deduce that irrespective of the sign, x cannot be greater than +2 and y cannot be smaller than -1.

Now, the only Answer choice that satisfies one of the above two conditions is D.