The solution set of the inequality (x-1)/(x+2) = 2 is the set of all.

Question

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

gmatophobia · Accepted Answer

\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

gmatophobia · Answer

Hello!

Here is a YouTube video that outlines the process -

https://www.youtube.com/watch?v=t-ZYEXvSX04

MrWhite · Answer

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!!

tickledpink001 · Answer

thanks!

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

tickledpink001 · Answer

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

KarishmaB · Answer

\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)

mamoonus · Answer

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?

Bunuel · Answer

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 
     Theory   
 Inequalities Made Easy! 
 Inequalities: Tips and hints 
 Arithmetic with Inequalities 
 Wavy Line Method Application - Complex Algebraic Inequalities 
 Solving Quadratic Inequalities: Graphic Approach 
 Inequalities - Quadratic Inequalities 
 Graphic approach to problems with inequalities 
 Inequalities trick 
 INEQUATIONS(Inequalities) Part 1 
 INEQUATIONS(Inequalities) Part 2 
     Questions   
 Inequality and absolute value questions from my collection 
 DS Questions 
 PS Questions

For more check  Ultimate GMAT Quantitative Megathread

Hope it helps.

Gemmie · Answer

x 
eq{-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 
eq{-2} )

AryakiSekera · Answer

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

Bunuel · Answer

x = 0 should not satisfy the inequality because the correct answer is (A)  -5 ≤ x < -2 , and this range does not include 0.

JuvenalUrbino · Answer

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

GMATinsight · Answer

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

ravjaz · Answer

But wait, why can't we bring the x+2 over to the other side and multiply into 2, why do we have to move the 2 into the left side instead?

Krunaal · Answer

Your doubt is answered here: https://gmatclub.com/forum/the-solution ... l#p3418359

rajdeep212003 · Answer

A bit clumsy but hope everyone will get it as think a bit different than other ans

speedemon · Answer

I see many people bringing the 2 to LHS, however, my first instinct was to multiply with x + 2 on both side to eliminate the denominator, am I using an incorrect approach? Did end up with the correct answer though.

Bunuel · Answer

You can’t just multiply both sides by (x + 2) directly because you don’t know whether (x + 2) is positive or negative. If it’s positive, the inequality direction stays the same; if it’s negative, it flips.

Also, the solution set for x - 1 ≥ 2(x + 2) is x ≤ -5, not -5 ≤ x < -2, which is the correct set for  \frac{x-1}{x+2} ≥ 2 .

speedemon · Answer

Got it, thanks bunuel!

The solution set of the inequality (x-1)/(x+2) >= 2 is the set of all.

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