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08 Oct 2021, 03:15
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C
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Which of the options must be true if x satisfies the two inequalities given below?
\(\frac{5x−1}{x+3}<1\) and \(|2x − 5| ≤ 5\)?
A. x < 0
B. x > 0
C. x < 2
D. x > 2
E. None of the above
\(\frac{5x−1}{x+3}<1\) and \(|2x − 5| ≤ 5\)?
A. x < 0
B. x > 0
C. x < 2
D. x > 2
E. None of the above
23 Oct 2021, 04:31
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I think it's easiest to start with the second inequality. If |2x - 5| < 5, then, dividing by 2 throughout, |x - 2.5| < 2.5. Since |a - b| is just the distance between a and b on the number line, our inequality says in words "the distance between x and 2.5 is at most 2.5", or in other words, x is between 0 and 5 (inclusive).
Now since we know x is at least 0, we can be sure x+3 is at least 3, and so we can be sure x+3 is positive. So we can now safely multiply both sides of the first inequality by x+3 without worrying about whether we might need to reverse the inequality. Doing that, we find 5x - 1 < x + 3, so 4x < 4, and x < 1. Note here that we assumed in advance that x+3 > 0, or that x > -3, so we haven't even looked at what happens when x < -3 (it turns out in that case there are no solutions, so the complete solution set to the first inequality is -3 < x < 1).
So when we combine the two inequalities, the complete solution set for x is: 0 < x < 1. The question does not, however, ask for the complete solution set for x. Instead it simply asks what must be true. And certainly if x is between 0 and 1, it must be true that x < 2, so C is correct. Answer C would also have been correct if it said "x < 1000" or "-5 < x < 5", or "x is a number"; we're only looking for an answer that must be true when we know 0 < x < 1 is true, and A, B and D can all be false (in the case of A, only because x = 0 is possible), so they're wrong answers, and E is wrong because one answer, C, must indeed be true.
I'd add that if you begin solving by looking at the first inequality rather than at the second, you need to separate it into two cases if you want to multiply by x+3 on both sides, because x+3 might be negative, and if we multiply both sides of an inequality by something negative, we must reverse the inequality. So to analyze the first inequality correctly, we can, as our first case, assume x+3 > 0, or that x > -3, which is what I did above, and from there we find x < 1. Combining the assumption "x > -3" with the solution "x < 1", we learn that -3 < x < 1 is one set of solutions to the inequality. We then must consider the second case, where x + 3 < 0, or where x < -3. You might be able to see by inspection that the inequality won't work in this case (the value of the fraction on the left side will be bigger than 1 for sure), or you can do algebra: we can multiply on both sides by x+3, reversing the inequality when we do because we're multiplying by something negative, to get 5x - 1 > x + 3. Solving, we find x > 1. Since we assumed x < -3, and learned, under that assumption, that x > 1, we've arrived at a contradictory situation, and there can be no solutions in this case, so there are only solutions to the first inequality when x + 3 is positive, and so when -3 < x < 1.
Now since we know x is at least 0, we can be sure x+3 is at least 3, and so we can be sure x+3 is positive. So we can now safely multiply both sides of the first inequality by x+3 without worrying about whether we might need to reverse the inequality. Doing that, we find 5x - 1 < x + 3, so 4x < 4, and x < 1. Note here that we assumed in advance that x+3 > 0, or that x > -3, so we haven't even looked at what happens when x < -3 (it turns out in that case there are no solutions, so the complete solution set to the first inequality is -3 < x < 1).
So when we combine the two inequalities, the complete solution set for x is: 0 < x < 1. The question does not, however, ask for the complete solution set for x. Instead it simply asks what must be true. And certainly if x is between 0 and 1, it must be true that x < 2, so C is correct. Answer C would also have been correct if it said "x < 1000" or "-5 < x < 5", or "x is a number"; we're only looking for an answer that must be true when we know 0 < x < 1 is true, and A, B and D can all be false (in the case of A, only because x = 0 is possible), so they're wrong answers, and E is wrong because one answer, C, must indeed be true.
I'd add that if you begin solving by looking at the first inequality rather than at the second, you need to separate it into two cases if you want to multiply by x+3 on both sides, because x+3 might be negative, and if we multiply both sides of an inequality by something negative, we must reverse the inequality. So to analyze the first inequality correctly, we can, as our first case, assume x+3 > 0, or that x > -3, which is what I did above, and from there we find x < 1. Combining the assumption "x > -3" with the solution "x < 1", we learn that -3 < x < 1 is one set of solutions to the inequality. We then must consider the second case, where x + 3 < 0, or where x < -3. You might be able to see by inspection that the inequality won't work in this case (the value of the fraction on the left side will be bigger than 1 for sure), or you can do algebra: we can multiply on both sides by x+3, reversing the inequality when we do because we're multiplying by something negative, to get 5x - 1 > x + 3. Solving, we find x > 1. Since we assumed x < -3, and learned, under that assumption, that x > 1, we've arrived at a contradictory situation, and there can be no solutions in this case, so there are only solutions to the first inequality when x + 3 is positive, and so when -3 < x < 1.
General Discussion
08 Oct 2021, 22:22
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Bunuel
\(\frac{5x−1}{x+3}<1\)
Or, \(5x - 1 < x + 3\)
Or, \(4x < 4\)
So, \(x < 1\)
\(|2x − 5| ≤ 5\) can have 2 possibilities
Possibbility - I
\(2x - 5 ≤ 5\)
Or, \(2x ≤ 10\)
Or, \(x ≤ 5\)
Possibbility - II
\(-2x + 5 ≤ 5\)
Or, \(-2x ≤ 0\)
Or, \(0 ≤ 2x\)
Or, \(0 ≤ x\)
Combining the above we can safely claim that answer must be (C) \(x < 2\)
