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The solution set of the inequality (x-1)/(x+2) >= 2 is the set of all. 16 Oct 2023, 15:21
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The solution set of the inequality \(\frac{x-1}{x+2} ≥ 2\) is the set of all real numbers x such that

(A) \(-5 ≤ x < -2\)
(B) \(-5 ≤ x ≤ -2\)
(C) \(-2 < x ≤ 3\)
(D) \(-2 ≤ x ≤ 3\)
(E) \(x ≤ -5\)
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The solution set of the inequality (x-1)/(x+2) >= 2 is the set of all. 24 Oct 2023, 23:05
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MrWhite
The solution set of the inequality \(\frac{x-1}{x+2} ≥ 2\) is the set of all real numbers x such that

(A) \(-5 ≤ x < -2\)
(B) \(-5 ≤ x ≤ -2\)
(C) \(-2 < x ≤ 3\)
(D) \(-2 ≤ x ≤ 3\)
(E) \(x ≤ -5\)


\(\frac{x-1}{x+2} ≥ 2\)

\(\frac{x-1}{x+2} - 2 ≥ 0\)

\(\frac{x - 1 - 2x - 4}{x+2} ≥ 0\)

\(\frac{-x - 5}{x+2} ≥ 0\)

Multiplying by -1

\(\frac{x + 5}{x+2} \leq 0\)

The critical points are -5, and -2. Using the wavy curve approach we get

------ +ve ----- -5 ------- -ve ------ -2 ---------- +ve -------

We can't have -2 in the denominator as that value will result in zero.

Hence, the correct range is \(-5 ≤ x < -2\)

Option A
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The solution set of the inequality (x-1)/(x+2) >= 2 is the set of all. 31 Dec 2023, 03:24
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tickledpink001

thanks!

Q- why do you have "positive" written on the left side of -5 on the number line?

Hello!

Here is a YouTube video that outlines the process -

General Discussion
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The solution set of the inequality (x-1)/(x+2) >= 2 is the set of all. 16 Oct 2023, 15:29
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Bringing everything to one side \(\frac{x-1}{x+2} - 2 ≥ 0\)

=> \(\frac{-(x+5) }{ x+2} ≥ 0 \)

Multiplying both sides by -1 => \(\frac{x+5 }{ x+2} ≤ 0 \)

Note, x ≠ -2 since denominator can't be 0; Options B & D are out.

Evaluating the inequality in regions around the critical points -5, -2

Option A is the only correct ans.

I was wondering whether this is the most optimal way to approach this problem, open to other solution methods!!
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The solution set of the inequality (x-1)/(x+2) >= 2 is the set of all. 31 Dec 2023, 02:33
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MrWhite
The solution set of the inequality \(\frac{x-1}{x+2} ≥ 2\) is the set of all real numbers x such that

(A) \(-5 ≤ x < -2\)
(B) \(-5 ≤ x ≤ -2\)
(C) \(-2 < x ≤ 3\)
(D) \(-2 ≤ x ≤ 3\)
(E) \(x ≤ -5\)


\(\frac{x-1}{x+2} ≥ 2\)

\(\frac{x-1}{x+2} - 2 ≥ 0\)

\(\frac{x - 1 - 2x - 4}{x+2} ≥ 0\)

\(\frac{-x - 5}{x+2} ≥ 0\)

Multiplying by -1

\(\frac{x + 5}{x+2} \leq 0\)

The critical points are -5, and -2. Using the wavy curve approach we get

------ +ve ----- -5 ------- -ve ------ -2 ---------- +ve -------

We can't have -2 in the denominator as that value will result in zero.

Hence, the correct range is \(-5 ≤ x < -2\)

Option A

thanks!

Q- why do you have "positive" written on the left side of -5 on the number line?
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The solution set of the inequality (x-1)/(x+2) >= 2 is the set of all. 04 Jan 2024, 07:26
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gmatophobia
tickledpink001

thanks!

Q- why do you have "positive" written on the left side of -5 on the number line?

Hello!

Here is a YouTube video that outlines the process -


Thanks! would you know how to solve something like this using this method please:
If (|x| - 2)(x + 5) < 0, then which of the following must be true?

A. x > 2
B. x < 2
C. -2 < x < 2
D. -5 < x < 2
E. x < -5
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The solution set of the inequality (x-1)/(x+2) >= 2 is the set of all. 24 Jun 2024, 00:51
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MrWhite
The solution set of the inequality \(\frac{x-1}{x+2} ≥ 2\) is the set of all real numbers x such that

(A) \(-5 ≤ x < -2\)
(B) \(-5 ≤ x ≤ -2\)
(C) \(-2 < x ≤ 3\)
(D) \(-2 ≤ x ≤ 3\)
(E) \(x ≤ -5\)
­
 \(\frac{x-1}{x+2} ≥ 2\)

 \(\frac{x-1}{x+2} - 2 ≥ 0\)

 \(\frac{-x-5}{x+2} ≥ 0\)

 \(\frac{x+5}{x+2}  \leq 0\)

Using the wavy line method discussed in this video:
https://youtu.be/PWsUOe77__E

we know that x will lie between -5 and -2 but it cannot be -2 because denominator cannot be 0. Hence,

\(-5 \leq x < -2\)

Answer (A)­
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The solution set of the inequality (x-1)/(x+2) >= 2 is the set of all. 04 Jul 2024, 21:24
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wanted to ask, so whenever we have an variable, we cannot multiply by that variable in a inequality because that variable could be negative or positive, so we wouldn't know what to do with the sign. so to do ---> x≥2x+5---> x≤5 would be incorrect, since we are multiplying by x+2 which we don't know whether it is positive or negative?­ Is my thinking correct?
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The solution set of the inequality (x-1)/(x+2) >= 2 is the set of all. 05 Jul 2024, 00:14
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mamoonus
wanted to ask, so whenever we have an variable, we cannot multiply by that variable in a inequality because that variable could be negative or positive, so we wouldn't know what to do with the sign. so to do ---> x≥2x+5---> x≤5 would be incorrect, since we are multiplying by x+2 which we don't know whether it is positive or negative?­ Is my thinking correct?
­
You are partially correct. In inequalities, you cannot multiply or divide by a variable if you don't know its sign, as it could change the direction of the inequality. However, you can add or subtract the variable without changing the direction. So, for x ≥ 2x + 5, you can subtract x from both sides to get 0 ≥ x + 5.

­

9. Inequalities



For more check Ultimate GMAT Quantitative Megathread



Hope it helps.­
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The solution set of the inequality (x-1)/(x+2) >= 2 is the set of all. 09 Aug 2024, 09:32
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\(x \neq{-2}\)

\(\frac{­x - 1}{x + 2} \geq{2}\)

=> \(\frac{x - 1}{x +2} - \frac{2x + 4}{x + 2} \geq{0}\)

=> \(\frac{- x - 5}{x + 2} \geq{0}\)

=> \(\frac{x + 5}{x + 2} \leq{0}\)

______ -5 _______ -2 ________
...(+).............(-)..............(+)

=> \(-5 \leq{x} \leq{-2}\) (coz \(x \neq{-2}\))­
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The solution set of the inequality (x-1)/(x+2) >= 2 is the set of all. 23 Oct 2024, 02:59
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While i understand the wavy line method, i need more clarity on why it's correct in this case since if we put x=0, the equation doesn't hold true

MrWhite
Bringing everything to one side \(\frac{x-1}{x+2} - 2 ≥ 0\)

=> \(\frac{-(x+5) }{ x+2} ≥ 0 \)

Multiplying both sides by -1 => \(\frac{x+5 }{ x+2} ≤ 0 \)

Note, x ≠ -2 since denominator can't be 0; Options B & D are out.

Evaluating the inequality in regions around the critical points -5, -2

Option A is the only correct ans.

I was wondering whether this is the most optimal way to approach this problem, open to other solution methods!!
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The solution set of the inequality (x-1)/(x+2) >= 2 is the set of all. 23 Oct 2024, 06:04
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AryakiSekera
While i understand the wavy line method, i need more clarity on why it's correct in this case since if we put x=0, the equation doesn't hold true


x = 0 should not satisfy the inequality because the correct answer is (A) \(-5 ≤ x < -2\), and this range does not include 0.
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The solution set of the inequality (x-1)/(x+2) >= 2 is the set of all. 02 Feb 2025, 07:12
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Slightly different approach:

1) If x + 2 > 0, or x > -2, we can multiply the denominator to the other side without flipping the sign:
x-1 ≥ 2*(x+2) --> x ≤ -5 which cannot be since we assumed x > -2. No solution.

2) If x + 2 < 0, or x < -2, we can multiple the denominator the the other side at the cost of flipping the sign:
x - 1 ≤ 2*(x+2) --> -5 ≤ x, which together with our initial requirement becomes -5 ≤ x < -2
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The solution set of the inequality (x-1)/(x+2) >= 2 is the set of all. 02 Feb 2025, 07:35
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MrWhite
The solution set of the inequality \(\frac{x-1}{x+2} ≥ 2\) is the set of all real numbers x such that

(A) \(-5 ≤ x < -2\)
(B) \(-5 ≤ x ≤ -2\)
(C) \(-2 < x ≤ 3\)
(D) \(-2 ≤ x ≤ 3\)
(E) \(x ≤ -5\)
Let's look at it logically.

For any positive value of x greater than 1, the value of the given expression will be positive but since numerator is less than the denominator hence the result can NEVER be greater than 2

hence,

Option C and Option D are eliminated. (which have positive values in solution set)


Now for the negative values, lets just try a few values
@x=-1, result is 2 which satisfies
x=-2 can't substitute as the function will be undefined. (Eliminated option B)

x=-6 doesn't satisfy so option E eliminated


Left with just option A

answer: Option A
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