For how many integer values of x, is |x – 6| |3x + 6|?

Question

For how many integer values of x, is |x – 6| > |3x + 6|?

(A) 1

(B) 3

(C) 5

(D) 7

(E) Infinite

chetan2u · Accepted Answer

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

JeffTargetTestPrep · Answer

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

sobby · Answer

by substituting values i can see 5 as answer ...
No positive value for x can satisfy the equation...------------1
x=0 can equalize the equation-----------2
-1 to -5 will satisfy the equation...and -6 will equalize the condition-----3
so the answer is c - --5 ..

is there any other way to solve i'e without doing actual substitution ...
Thanks..

sobby · Answer

2nd approach. ......

we can do it by squaring both side and using number line to solve it ...
we will get the roots as 0 and -6 and using this we can deduce only number between 0 and -6 will satisfy the actual solution..

Bunuel  : Is this approach correct

I have one doubt ---while squaring in absolute ....what necessary condition we need to consider....

broall · Answer

In this question, in my opinion, there are 2 possible solutions. Solution 1. $|x – 6| > |3x + 6| \implies |x – 6| - |3x + 6|>0$ If $x<-2$ we have $(6-x)-(-3x-6)=6-x+3x+6=2x+12 > 0 \implies x >-6$. Hence $-6 0 \implies x<0$. Hence $-2 \leq x < 0$ If $x \geq 6$ we have $(x-6)-(3x+6)=-2x-12>0 \implies x<-6$. There is no satisfied value of $x$ in this case. Combine all cases we have $-6 < x < 0$. Thus, $x$ could receive 5 integer value $\{-5;-4;-3;-2;-1\}$ The answer is C. Solution 2. $|x – 6| > |3x + 6| \implies (x-6)^2 > (3x+6)^2$ $\implies x^2-12x+36 > 9x^2+36x+36 \implies 8x^2+48x<0$ $\implies 8x(x+6)<0 \implies -6

Bunuel · Answer

We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality). So, if you have absolute values on both sides of the inequality then you can safely square.

Check for more the following topics:
 Inequalities Made Easy!

Solving Quadratic Inequalities - Graphic Approach 
 Inequality tips

DS Inequalities Problems  
 PS Inequalities Problems

700+ Inequalities problems

Hope it helps.

RMD007 · Answer

KarishmaB · Answer

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)

KarishmaB · Answer

Responding to a pm: First consider the equation: |x - 6| = 3*|x + 2| That is, the point where distance from 6 is equal to three times the distance from -2. There is such a point between 6 and -2. That point is x = 0. Here the distance from 6 is 6 which is equal to three times 2, the distance of 0 from -2. How do we get x = 0? Distance from 6 should be thrice the distance from -2 Distance from 6 : Distance from -2 = 3:1 Split the distance between 6 and -2 into the ratio 3:1. Since there are 8 units between 6 and -2, you get that x is 6 units away from 6 and hence at 0. For the inequality, we want the distance from 6 to be greater than the distance from -2 so we move to the left (closer to -2). The distance from 6 will keep increasing till we reach -2 when the distance from 6 is 8. Note that now also distance from 6 is greater than 3 times the distance from -2. Now as we go further to the left, the distance from 6 keeps increasing but so does the distance from -2. There will be some point again where distance from 6 will be equal to 3 times the distance from -2. At this point, the distance of 8 units between 6 and -2 will make up twice the distance of x from -2 (x)......................(-2).................(0)...................................................(6)................... < Distance from -2> <-------------------Distance from 6 -----------------------> So x must be 4 units to the left of -2. So at x = -6, again, distance from 6 will be three times the distance from -2. Hence, for our inequality, x can take values between -6 and 0. This gives us 5 integer values.

RMD007 · Answer

Thank you so much  VeritasPrepKarishma  for the explanation. +1

A good thoughtful explanation. It took a little time for me to understand but I guess I need to practice more to internalize it.

Just one question here, for the below inequality -
|x-6| > |x+2|
can we say that the solution for this is x<2?

KarishmaB · Answer

Yes, that is correct.

|x-6| > |x+2|

The distance of x from 6 is more than the distance of x from -2.

The two distances will be equal at the mid point that is at x = 2. As you move the left, the distance of x from 6 will keep increasing. To the left of -2 too, the distance from 6 will always be more than the distance from -2.

Hence answer will be x < 2.

manojhanagandi · Answer

In this question, in my opinion, there are 2 possible solutions.

Solution 1.

|x–6|>|3x+6|⟹|x–6|−|3x+6|>0|x–6|>|3x+6|⟹|x–6|−|3x+6|>0

If x<−2x<−2 we have (6−x)−(−3x−6)=6−x+3x+6=2x+12>0⟹x>−6(6−x)−(−3x−6)=6−x+3x+6=2x+12>0⟹x>−6. Hence −6<x<−2−6<x<−2

If −2≤x<6−2≤x<6 we have (6−x)−(3x+6)=−4x>0⟹x<0(6−x)−(3x+6)=−4x>0⟹x<0. Hence −2≤x<0−2≤x<0

If x≥6x≥6 we have (x−6)−(3x+6)=−2x−12>0⟹x<−6(x−6)−(3x+6)=−2x−12>0⟹x<−6. There is no satisfied value of xx in this case.

Combine all cases we have −6<x<0−6<x<0.

Thus, xx could receive 5 integer value {−5;−4;−3;−2;−1}{−5;−4;−3;−2;−1}

The answer is C.

Solution 2.

|x–6|>|3x+6|⟹(x−6)2>(3x+6)2|x–6|>|3x+6|⟹(x−6)2>(3x+6)2

⟹x2−12x+36>9x2+36x+36⟹8x2+48x<0⟹x2−12x+36>9x2+36x+36⟹8x2+48x<0

⟹8x(x+6)<0⟹−6<x<0⟹8x(x+6)<0⟹−6<x<0.

hi nguyendinhtuong .
could u please explain why x<-6 has no satisfied values in solution 1.

thank you..

EMPOWERgmatRichC · Answer

Hi All,

One of the great 'design aspects' of most GMAT questions is that they can be approached in a variety of ways. As such, just because you got a question correct doesn't necessarily mean that you solved it in the fastest/easiest way possible (and if you're doing a lot of complex math to get to the solution, then you are likely NOT using the fastest approach). Sometimes the simplest approach to certain questions is just 'brute force' arithmetic - plow through the basic math necessary to prove the solution.

Here, we're asked for the number of INTEGER solutions to |X - 6| > |3X + 6|.

From the answer choices, it seems clear that there aren't that many possibilities (and it's unlikely that there's an 'unlimited' number of solutions), so we just have to find the 1, 3, 5 or 7 options that 'fit' this inequality.

To start, let's TEST a couple of simple values for X: 0 and 1....

IF...
X = 0, then we end up with...
|-6| > |6| ..... 6 > 6.... which is NOT correct - so X=0 is NOT a solution

IF...
X = 1, then we end up with...
|-5| > |9| ..... 5 > 9.... which is NOT correct - so X=1 is NOT a solution

Increasing the value of X will just make the "right side' absolute value a lot bigger, so there's no point in raising X. Thus, let's decrease it and see what happens....

IF...
X = -1, then we end up with...
|-7| > |3| ..... 7 > 3.... which IS correct - so X=-1 IS a solution

At this point, how long would it take you to TEST -2, -3, -4, -5 and -6? What happens when you try -6 though? And what about -7? Notice how you would stop working for the same reason why you wouldn't both trying 2, 3, 4, etc... Now, how many total solutions do you have?

Final Answer:  C

GMAT assassins aren't born, they're made,
Rich

BillyZ · Answer

Bunuel Why my solution is incorrect? (1) Take positive |x – 6|=x - 6 |3x + 6|=3x + 6 x-6>3x+6 x<-6 (2) Take negative |x-6| = -x+6 |3x+6| = -3x-6 -x+6>-3x-6 x>-6 Does this approach correct? https://gmatclub.com/forum/for-how-many ... l#p1804531

stonecold · Answer

Hi ziyuen It is not that easy. This is an extremely difficult problem. You highlighted part is incorrect. Here is what did => Firstly the definition of |x| --> |x|= x for x>0 = -x for x<0 =0 for x=0 Now lets look at the two modulus equations --> |x-6| = x-6 for x>6 =-(x-6)=-x+6 for x<6 =0 for x=6 Similarly |3x+6| => 3x+6 for x>-2 => -3x-6 for x<-2 => 0 for x=-2 Hence the three boundaries are => (-∞,-2] , and chetan2u 's blog here --> https://gmatclub.com/forum/absolute-modu ... l#p1622372 That would surely help you out to clear the Modulus basics. Regards Stone Cold

Ravindra.here · Answer

MY way simple and sober and within a minute you get the answer

equate x – 6 = 3x + 6 
we get x=-6
Now if you will decrease x let's say then |x – 6| > |3x + 6| wil not hold  ((LHS WILL BE LESS THAN RHS

However if you DECREASE X then |x – 6| > |3x + 6| holds 
this continues till -1 and then again fails 
therefore  -5.-4,-3,-2,-1  check for 0 
Ans =5

danielarasan · Answer

|x-6| > |3x + 6|?

(X - 6)^2 > (3x + 6)^2
x^2 - 12x + 36 > 9x^2 + 36x + 36
8x^2 + 48x < 0
8x ( x + 6) < 0

Now, for it to be negative, either 1) 8x is positive and (x+6) is negative or 2) 8x is negative and (x+6) is positive. Both cannot be positive and both cannot be negative

1) For 8x to be positive and x+6 to be negative is impossible. As you cannot have a positive number x that when multiplied by 8 is positive and when added to 6 is negative, then no results here.
2) For 8x is negative and (x+6) is positive: Any negative number multiplied by 8 is negative, but not any negative number added to 6 is positive.
The only negative integers that when added to 6 become positive are -1, -2, -3, -4 and -5

Therefore, answer C, 5 integers

SchruteDwight · Answer

By far the easiest approach to these super rare, super hard inequality problems is number plugging

0: NO
1: NO
2: NO
.
.
.
100000:NO

-1: YES
-2: YES
-3: YES
-4: YES
-5: YES
-6: NO
.
.
.
-100000: NO

done (t=45sec)

Kritisood · Answer

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?

For how many integer values of x, is |x – 6| > |3x + 6|?

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For how many integer values of x, is |x – 6| > |3x + 6|?

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