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KarishmaB
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For how many integer values of x, is |x – 6| > |3x + 6|? 05 Jun 2020, 05:43
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Kritisood
VeritasKarishma
Bunuel
For how many integer values of x, is |x – 6| > |3x + 6|?

(A) 1

(B) 3

(C) 5

(D) 7

(E) Infinite

Think in terms of the definition of absolute values:

|x - 6| > 3*|x + 2|

Distance from 6 is greater than 3 times the distance from -2

-------- (-2) -------- (0) ------------------------------ (6) -------------

At point 0, the distance from 6 will be equal to 3 times the distance from -2. So on the left of 0, the distance from 6 will be greater.

The distance between -2 and 6 is 8. Distance from 6 will be equal to 3 times the distance from -2 at a point 4 units to the left of -2 i.e. -6. To its left, distance from 6 will be less. So between -6 to 0, distance from 6 is greater than 3 times the distance from -2. We have 5 integer values lying between these two.

Answer (C)

Hi VeritasKarishma i was trying to solve this problem by finding the critical points. the critical points I arrived at are -2 and 6. and I'm not able to find the answer through this. the critical points as mentioned by JeffTargetTestPrep are -6 and 0.

could you help me in where I am going wrong in deducing the critical points?

-- to elaborate how I got the critical points
x-6=0
x=6

and 3x+6=0
x=-2

https://gmatclub.com/forum/is-x-1-1-x-1 ... 82478.html
in this question the statement was |x + 1| = 2|x - 1| hence critical points mentioned are -1 and 1; using the same logic i arrived at the critical points here.

JeffTargetTestPrep VeritasKarishma i am really confused about this. could you please assist?
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The transition points are -2 and 6 as you derived.
But note that the question has a negative between the terms:

|x – 6| - |3x + 6| > 0

If you are trying to use the same method that we do when the terms are added, you will not get the answer. Here, you need to use the method I have discussed above.
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For how many integer values of x, is |x – 6| > |3x + 6|? 02 Jul 2020, 07:19
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I don't understand following problem:

When I raise both sides to 2 and get RHS to LHS I get the following inequality

(x-6)^2-(3x+6)^2>0
(x-6+3x+6)(x-6-3x-6)>0
(4x)(-2X-12)>0

So values for X will be 0 and -6

BUUT

Number line is

....+++.....-6..........---.............0...........+++..........

And as LHS should be BIGGER than 0 I could take infinitive values for X

Where am I wrong?
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For how many integer values of x, is |x – 6| > |3x + 6|? 06 Jul 2020, 01:50
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chetan2u
Bunuel
For how many integer values of x, is |x – 6| > |3x + 6|?

(A) 1

(B) 3

(C) 5

(D) 7

(E) Infinite


Hi sobby,
The actual way would be...

Since both sides are positive, square both sides
\((x-6)^2>(3x+6)^2..........x^2-12x+36>9x^2+36x+36.........8x^2+48x<0\)
8x(x+6)<0.....

For 8x(x+6) to be NEGATIVE, one of 8x and x+6 will be negative and other positive..

If x is positive, 8x will be positive and x+6 will also be positive..
So x should be NEGATIVE..
Then 8x will be negative, and thus x+6 should be positive ..
For that \(x+6>0.....X>-6....\) But x<0.

So range becomes \(0>x>-6\)...
Values are \(-1,-2,-3,-4,-5\)
5 value
C
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That's a nice approach,
I would like to add something too!
Here : x^2+6x<0 --> x^2+6x+9<9
Now take x^2+6x+9 = |x+3|^2 <9 .. after simplifying, we get -3 <x+3<3
= -6<x<0
5 integer values.
Hope this helps!
:D
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For how many integer values of x, is |x – 6| > |3x + 6|? 05 Aug 2022, 06:27
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JeffTargetTestPrep
Bunuel
For how many integer values of x, is \(|x – 6| > |3x + 6|\)?

(A) 1

(B) 3

(C) 5

(D) 7

(E) Infinite

To solve an inequality where the absolute value of an expression is greater than (or less than) the absolute value of another expression, we need to find the values that make the absolute values of the two expressions equal to each other. That is, to solve |x – 6| > |3x + 6|, we first need to solve |x – 6| = |3x + 6|. Recall that if |A| = |B|, then either A = B (Case 1) or A = -B (Case 2).

Case 1: When \(x - 6 = 3x + 6\):

\(x - 6 = 3x + 6\)

\(-2x = 12\)

\(x = -6\)

Case 2: When \(x - 6 = -(3x + 6)\):

\(x - 6 = -(3x + 6)\)

\(x - 6 = -3x - 6\)

\(4x = 0\)

\(x = 0\)

These two values, -6 and 0, are the critical values for determining the solution to the given inequality. Now when we place these two values on the number line, we will see that they partition the number line into 3 intervals:

\(x < -6\)

\(-6 < x < 0\)

\(x > 0\)

For each of these intervals, we will pick a value in that interval (for example, for x < -6, we can pick -7) and test whether it satisfies the given inequality. If it does, then EVERY number in that interval will satisfy the inequality. If it doesn’t, then NO numbers in that interval will satisfy the inequality. Recall that our given inequality is |x – 6| > |3x + 6|. Let’s begin testing the intervals.

1) For \(x < -6\), let’s pick x = -7:

|-7 – 6| > |3(-7) + 6| ?

\(|-13| > |-15|\)?

\(13 > 15\) → No!

Thus, no numbers in the interval \(x < -6\)will satisfy \(|x – 6| > |3x + 6|.\)

2) For \(-6 < x < 0\), let’s pick x = -1:

\(|-1 – 6| > |3(-1) + 6|\)?

\(|-7| > |3|\) ?

\(7 > 3\)→ Yes

Thus, every number in the interval \(-6 < x < 0\) will satisfy \(|x – 6| > |3x + 6|.\)

3) For \(x > 0\), let’s pick x = 1:

\(|1 – 6| > |3(1) + 6|\) ?

\(|-5| > |9|\) ?

\(5 > 9\) → No

Thus, no numbers in the interval \(x > 0\) will satisfy \(|x – 6| > |3x + 6|.\)As we can see, the only interval that satisfies \(|x – 6| > |3x + 6|\) is \(-6 < x < 0\). In other words, the solution to \(|x – 6| > |3x + 6|\) is \(6 < x < 0\). In this interval, there are 5 integers: -5, -4, -3, -2, and -1.

Answer: C
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Hi, Jeff - thank you for your explanation. For Case 2, why do we not also negate A in finding out the value of x?
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For how many integer values of x, is |x – 6| > |3x + 6|? 08 Dec 2024, 06:19
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