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# Solve for x: x-7/x+12≥0 A) x∈ [-7, 12] B) x∈ [-7, 11] C) x∈ (-∞, -12)

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Solve for x: x-7/x+12≥0 A) x∈ [-7, 12] B) x∈ [-7, 11] C) x∈ (-∞, -12)  [#permalink]

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Updated on: 14 Oct 2019, 01:45
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Difficulty:

35% (medium)

Question Stats:

61% (01:08) correct 39% (01:22) wrong based on 28 sessions

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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)

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Originally posted by Asad on 13 Oct 2019, 12:26.
Last edited by Asad on 14 Oct 2019, 01:45, edited 1 time in total.
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Re: Solve for x: x-7/x+12≥0 A) x∈ [-7, 12] B) x∈ [-7, 11] C) x∈ (-∞, -12)  [#permalink]

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13 Oct 2019, 14:56
Solve for x: 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)

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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Re: Solve for x: x-7/x+12≥0 A) x∈ [-7, 12] B) x∈ [-7, 11] C) x∈ (-∞, -12)  [#permalink]

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15 Oct 2019, 02:26
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.

Re: Solve for x: x-7/x+12≥0 A) x∈ [-7, 12] B) x∈ [-7, 11] C) x∈ (-∞, -12)   [#permalink] 15 Oct 2019, 02:26
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