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15 Nov 2021, 03:25
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If \(x ≠ 5\) and \(x\) is a non-zero integer, what is the value of \(\frac{|x| + 5}{5 - x}\) ?
(1) \(|y + 5| < 1 - x\)
(2) \(\frac{\sqrt{x^2}}{\sqrt[3]{x^3}}=-1\)
(1) \(|y + 5| < 1 - x\)
(2) \(\frac{\sqrt{x^2}}{\sqrt[3]{x^3}}=-1\)
15 Nov 2021, 03:25
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Official Solution:
If \(x ≠ 5\) and \(x\) is a non-zero integer, what is the value of \(\frac{|x| + 5}{5 - x}\) ?
If \(x > 0\), then \(|x|=x\) and \(\frac{|x| + 5}{5 - x}=\frac{x + 5}{5 - x}\)
If \(x < 0\), then \(|x|=-x\) and \(\frac{-x + 5}{5 - x}=1\)
So, if \(x\) is positive, then the value of \(\frac{|x| + 5}{5 - x}\) depends on exact value of \(x\) but if \(x\) is negative, then he value of \(\frac{|x| + 5}{5 - x}\) is 1.
(1) \(|y + 5| < 1 - x\):
The left hand side (\(|y+5|\)), is the absolute value of a number, so it must be positive or 0. Thus, \(0 < 1 - x\). This gives \(x < 1\). Since given that \(x\) is a non-zero integer, then \(x\) must be negative integer (-1, -2, ...). As we deduced above, if \(x\) is negative, then he value of \(\frac{|x| + 5}{5 - x}\) is 1. Sufficient.
(2) \(\frac{\sqrt{x^2}}{\sqrt[3]{x^3}}=-1\):
\(\frac{|x|}{x}=-1\);
\(|x|=-x\). This means that \(x\) is negative. As we deduced above, if \(x\) is negative, then he value of \(\frac{|x| + 5}{5 - x}\) is 1. Sufficient.
Answer: D
If \(x ≠ 5\) and \(x\) is a non-zero integer, what is the value of \(\frac{|x| + 5}{5 - x}\) ?
If \(x > 0\), then \(|x|=x\) and \(\frac{|x| + 5}{5 - x}=\frac{x + 5}{5 - x}\)
If \(x < 0\), then \(|x|=-x\) and \(\frac{-x + 5}{5 - x}=1\)
So, if \(x\) is positive, then the value of \(\frac{|x| + 5}{5 - x}\) depends on exact value of \(x\) but if \(x\) is negative, then he value of \(\frac{|x| + 5}{5 - x}\) is 1.
(1) \(|y + 5| < 1 - x\):
The left hand side (\(|y+5|\)), is the absolute value of a number, so it must be positive or 0. Thus, \(0 < 1 - x\). This gives \(x < 1\). Since given that \(x\) is a non-zero integer, then \(x\) must be negative integer (-1, -2, ...). As we deduced above, if \(x\) is negative, then he value of \(\frac{|x| + 5}{5 - x}\) is 1. Sufficient.
(2) \(\frac{\sqrt{x^2}}{\sqrt[3]{x^3}}=-1\):
\(\frac{|x|}{x}=-1\);
\(|x|=-x\). This means that \(x\) is negative. As we deduced above, if \(x\) is negative, then he value of \(\frac{|x| + 5}{5 - x}\) is 1. Sufficient.
Answer: D
23 Apr 2024, 07:45
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