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Question about critical method in modulus inequaliy

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by Mo2men » Tue Jan 10, 2017 2:22 am
For how many integer values of x, is |x - 6| > |3x + 6|?

(A) 1

(B) 3

(C) 5

(D) 7

(E) Infinite

[spoiler]When I approached this question I tried to apply critical method but the method failed to recognize all points I did the follows:

|x - 6| = |3x + 6|........then x=-6 but 0 is also a solution.

What is the mistake above?

When I squared both sides ans simplified it becomes:

X ( X+6) <0 henece x= 0 or -6 the I could solve it through number line to find that they are 5 integers

Answer: C[/spoiler]
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by melguy » Tue Jan 10, 2017 3:05 am
|x - 6| > |3x + 6| means

x - 6 > 3x + 6
and
x - 6 < -(3x + 6)

Rearranging and solving both equations

x - 6 > 3x + 6
x - 3x > 6 + 6
-2x > 12
x > -6

x - 6 < -(3x + 6)
x - 6 < -3x -6
x + 3x < 6 - 6
4 x < 0
x < 0

So -6 < x < 0 (i.e 5 integers)

Answer C
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by GMATGuruNY » Tue Jan 10, 2017 4:36 am
Mo2men wrote:For how many integer values of x, is |x - 6| > |3x + 6|?

(A) 1

(B) 3

(C) 5

(D) 7

(E) Infinite
The CRITICAL POINTS are the values of x that make the two sides of the inequality EQUAL.

Case 1: Signs unchanged
x-6 = 3x+6
-12 = 2x
x = -6.

Case 2: Signs changed on ONE SIDE
-x+6 = 3x+6
0 = 4x
x = 0.

The critical points are -6 and 0.
To determine where |x-6| > |3x+6|, test one value to the LEFT OF -6, one value BETWEEN -6 and 0, and one value to the RIGHT OF 0.

If x = -10, then |x-6| < |3x+6|.
If x = -1, then |x-6| > |3x+6|.
If x=6, then |x-6| < |3x+6|.

Only the blue case -- a value between -6 and 0 -- satisfies |x-6| > |3x+6|, implying that x must be BETWEEN -6 AND 0.
Integer values between -6 and 0:
-5, -4, -3, -2 and -1.

The correct answer is C.
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by [email protected] » Tue Jan 10, 2017 2:51 pm
Hi Mo2men,

This prompt is rather similar to the one that you posted here:

https://www.beatthegmat.com/simple-way-t ... 93999.html

It can be also be solved in the same general way - with a bit of 'brute force.' To start, consider the two absolute values...

|X - 6| vs. |3X + 6|

IF... X is any POSITIVE value, then the second absolute value will clearly be larger than the first. We're asked to find INTEGER values for X that will make the FIRST larger than the second, so positive values are clearly NOT an option. As such, we should start with 0 and work DOWN...

IF... X = 0, then the values are 6 and 6, which does NOT match what we're looking for
IF... X = -1, then the values are 7 and 3, which IS a match
IF... X = -2, then the values are 8 and 0, which IS a match
IF... X = -3, then the values are 9 and 3, which IS a match
IF... X = -4, then the values are 10 and 6, which IS a match
IF... X = -5, then the values are 11 and 9, which IS a match
IF... X = -6, then the values are 12 and 12, which does NOT match

From the increases in the two values, we can see that the second term will continue to increase at a much faster rate, so the second term will always be greater when X is less than -6. As such, there are just 5 integer values that fit the given inequality.

Final Answer: C

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