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TheUltimateWinner
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![Solve for x: x-7/x+12≥0 A) x∈ [-7, 12] B) x∈ [-7, 11] C) x∈ (-∞, -12)](/forum/styles/gmatclub_light/imageset/icon_post_target_unread.svg "Solve for x: x-7/x+12≥0 A) x∈ [-7, 12] B) x∈ [-7, 11] C) x∈ (-∞, -12)"){kind=icon}
Updated on: 14 Oct 2019, 01:45
Originally posted by TheUltimateWinner on 13 Oct 2019, 12:26.
Last edited by TheUltimateWinner on 14 Oct 2019, 01:45, edited 1 time in total.
Last edited by TheUltimateWinner on 14 Oct 2019, 01:45, edited 1 time in total.
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
74% (01:23) correct
26%
(01:23)
wrong
based on 94
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Solve for x: \(\frac{x-7}{x+12}\)≥0
A) x∈ [-7, 12]
B) x∈ [-7, 11]
C) x∈ (-∞, -12) ∪ [7, ∞)
D) x∈ (-∞, -7) ∪ (12, ∞)
E) x∈ (-12,7)
A) x∈ [-7, 12]
B) x∈ [-7, 11]
C) x∈ (-∞, -12) ∪ [7, ∞)
D) x∈ (-∞, -7) ∪ (12, ∞)
E) x∈ (-12,7)
13 Oct 2019, 14:56
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Asad
First, note that the 'set notation' used in the answers is very unGMAT-y. Limit answer choices to the standard <, >, = and so on.
To the question:
Under the assumption that what you intended to write was (x-7)/(x+12) then it is sufficient for the numerator to equal or for both numerator and denominator to have equal signs.
The first case gives x = 7
The second case gives x > 7 (both positive) or x < -12 (both negative).
Putting them together gives answer choice (C)
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15 Oct 2019, 02:26
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This inequality can be turned into a quadratic inequality.
Before that, there's a tricky bit to these type of questions. The denominator.
\(\frac{x-7}{x+12} ≥ 0\)
Test takers could potentially make mistakes because it is so easy to go wrong when you're in a rush. But this can be overcome with a little bit of observation.
The RHS is \(0\).
So we ought to understand that the RHS should comply with this information. If the denominator of \(\frac{x-7}{x+12}\) were to be zero, the LHS would be undefined. Thus, an important inference - \(x\)can never be equal to \(-12\)
keeping this information in mind, one can proceed to simplify this expression.
multiply both sides of the inequality with \((x+12)^2\).
\(=> (x-7)(x+12) ≥ 0\)
You'll arrive at the roots 7 and -12. Look at the inequality again and determine the sign. The ranges for x will be of the form \(x ≥ 7\) and \(x < -12\).
In terms of the format that the answer choices take
\(x ∈ (-∞, -12)\) and \([7, ∞)\)
Notice the circular bracket in\(x ∈ (-∞, -12)\) indicating that x is not equal to 12.
The answer is C.
Before that, there's a tricky bit to these type of questions. The denominator.
\(\frac{x-7}{x+12} ≥ 0\)
Test takers could potentially make mistakes because it is so easy to go wrong when you're in a rush. But this can be overcome with a little bit of observation.
The RHS is \(0\).
So we ought to understand that the RHS should comply with this information. If the denominator of \(\frac{x-7}{x+12}\) were to be zero, the LHS would be undefined. Thus, an important inference - \(x\)can never be equal to \(-12\)
keeping this information in mind, one can proceed to simplify this expression.
multiply both sides of the inequality with \((x+12)^2\).
\(=> (x-7)(x+12) ≥ 0\)
You'll arrive at the roots 7 and -12. Look at the inequality again and determine the sign. The ranges for x will be of the form \(x ≥ 7\) and \(x < -12\).
In terms of the format that the answer choices take
\(x ∈ (-∞, -12)\) and \([7, ∞)\)
Notice the circular bracket in\(x ∈ (-∞, -12)\) indicating that x is not equal to 12.
The answer is C.
