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25 Apr 2023, 07:33
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If \(x\) and \(y\) are consecutive integers, \(x > y\), and \(x^2 - 1 > y^2 - 4y + x - 1\), which of the following must be true?
A. \(x < 0\)
B. \(x ≠ 2\)
C. \(y > -1\)
D. \(y ≠ 1\)
E. \(x + y > 3\)
A. \(x < 0\)
B. \(x ≠ 2\)
C. \(y > -1\)
D. \(y ≠ 1\)
E. \(x + y > 3\)
25 Apr 2023, 07:33
Kudos
Bookmarks
Official Solution:
If \(x\) and \(y\) are consecutive integers, \(x > y\), and \(x^2 - 1 > y^2 - 4y + x - 1\), which of the following must be true?
A. \(x < 0\)
B. \(x ≠ 2\)
C. \(y > -1\)
D. \(y ≠ 1\)
E. \(x + y > 3\)
Given that \(x\) and \(y\) are consecutive integers, and \(x > y\), we can say that \(x = y + 1\). Substituting this into the given inequality, we get:
\((y + 1)^2 - 1 > y^2 - 4y + (y + 1) - 1\);
\(y^2 + 2y + 1 - 1 > y^2 - 4y + y\);
\(2y > - 3y\);
\(5y > 0\);
\(y > 0\). \(y\) can be 1, 2, 3, 4, and so on.
Since \(y = x - 1\), we can rewrite this inequality in terms of \(x\): \(x - 1 > 0\), which gives \(x > 1\). \(x\) can be 2, 3, 4, 5, and so on.
Since \(y > 0\), then \(y\) must also be greater than -1, making option C correct.
Even though we have already determined that option C is correct, let's elaborate on the other options for the sake of understanding why they are not necessarily true:
A. \(x < 0\). We found that \(x > 1\), so this option is not true. Discard.
B. \(x ≠ 2\). Since we found that \(x > 1\), this option could be true, but it's not necessarily true. Discard.
D. \(y ≠ 1\). Since we found that \(y > 0\), this option could be true, but it's not necessarily true. Discard.
E. \(x + y > 3\). Since we found that \(y > 0\) and \(x > 1\), the smallest possible value for \(x + y\) is \(1 + 2 = 3\). Therefore, this option is not necessarily true. Discard.
Answer: C
If \(x\) and \(y\) are consecutive integers, \(x > y\), and \(x^2 - 1 > y^2 - 4y + x - 1\), which of the following must be true?
A. \(x < 0\)
B. \(x ≠ 2\)
C. \(y > -1\)
D. \(y ≠ 1\)
E. \(x + y > 3\)
Given that \(x\) and \(y\) are consecutive integers, and \(x > y\), we can say that \(x = y + 1\). Substituting this into the given inequality, we get:
\((y + 1)^2 - 1 > y^2 - 4y + (y + 1) - 1\);
\(y^2 + 2y + 1 - 1 > y^2 - 4y + y\);
\(2y > - 3y\);
\(5y > 0\);
\(y > 0\). \(y\) can be 1, 2, 3, 4, and so on.
Since \(y = x - 1\), we can rewrite this inequality in terms of \(x\): \(x - 1 > 0\), which gives \(x > 1\). \(x\) can be 2, 3, 4, 5, and so on.
Since \(y > 0\), then \(y\) must also be greater than -1, making option C correct.
Even though we have already determined that option C is correct, let's elaborate on the other options for the sake of understanding why they are not necessarily true:
A. \(x < 0\). We found that \(x > 1\), so this option is not true. Discard.
B. \(x ≠ 2\). Since we found that \(x > 1\), this option could be true, but it's not necessarily true. Discard.
D. \(y ≠ 1\). Since we found that \(y > 0\), this option could be true, but it's not necessarily true. Discard.
E. \(x + y > 3\). Since we found that \(y > 0\) and \(x > 1\), the smallest possible value for \(x + y\) is \(1 + 2 = 3\). Therefore, this option is not necessarily true. Discard.
Answer: C
05 May 2025, 07:10
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Bunuel
I did the same till arriving at y>0. I eliminated all options by this. In your answer when testing the options I plugged in y=0 and x=1 to test option c and the output is 0>0 which is not possible.
