What is the value of x?

OA is "A".

My query::

|6 - 3x| = x - 2

As a module is always positive, hence we can say that x-2>0
If x-2>0 then it implies x>2. But our answer is showing that x=2

What am I missing here??

I agree A is right. But do not neglect 2nd statement just because it gave 2 answers.

5x+3 = -(2x+ 9) only when 5x+3 <0
so the answer obtained 7x=-12 must satisfy this domain. In this case it is satisfying thus making it insufficient.

Imagine a case when the 2nd solution does not satisfy the domain, in that case it will be sufficient. So do not neglect the statement just because it gave two answers.

Thanks Bunuel!! +1

very nicely explained Bunuel... +1.. thanks

I am not satisfied with the solution.

How do we solve modulus, we take two cases,
1)if (6-3x)>0 then given equation comes out to be 6-3x=x-2 which gives x=2 ,but we are considering the case when 6-3x>0 or x<2, hence this soln x=2 is not possible.

2) When we consider (6-3x)<0 then equation comes out to be -6+3x= x-2 which again gives x=2 , but this time we are considering 6-3x<0 or x>2, hence this soln x=2 is not feasible in this case also.

So according to me answer is E.

Note that OA for this questio is A, not E.

Next, to check whether x=2 is a root of |6 - 3x| = x - 2 just substitute x=2 into it: |6 - 3*2| = 0 = 2 - 2.

Hope it's clear.

Hi Bunuel, cann't it be solved by the following approach?

Statement 1:
6-3x=x-2
X=2
Again,
-(6-3x)=x-2
ie -6+3x=x-2
X= 2

Similar solution from in both scenario. Hence sufficient.

Statement 2:

Similar process and we will get x= 12/7 and 2. Two different solutions. hence Insufficient.

Ans: A

Atal Pandit

Yes, we can do this way too.

Typo: |5x + 3| = 2x + 9 --> x=-12/7 or x=2.

It seems that I have solved some abs. value problems by finding what makes the equation in the abs. value sign = to zero, so for #2, why do we find what makes the equation outside the abs. value sign = to zero? In other words...

|5x + 3| = 2x + 9. ==> When x=-4.5 2x+9 = 2(-4.5)+9 = -9+9 = 0.

Thanks!

We can say that \(x\geq{-4.5}\) by the same logic as we concluded that \(x\geq{2}\) in (1).

How did you get "for X> or = 2, |6x-3| = -6x+3" ?

We have \(|6 - 3x| = x - 2\). The left hand side (|6 - 3x|) is an absolute value. The absolute value is always more than or equal to 0, thus the right hand side (x - 2) must also be \(\geq{0}\) --> \(x-2\geq{0}\) --> \(x\geq{2}\). When \(x\geq{2}\), \(6 - 3x\leq{0}\), thus \(|6 - 3x|=-(6-3x)=-6+3x\), so \(|6 - 3x| = x - 2\) becomes \(-6+3x=x-2\) --> \(x=2\). Sufficient.

Similar questions to practice:
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what-is-x-1-x-2-2-x-3x-126874.html
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Hope it helps.

Hello Bunuel,

I followed the same approach but looking at solution given by you, It is actually not required . I have seen some question types where in we just check for one condition (X>/0 or X<0) and we can conclude the answer (Like you just did in this one)

Are there particular question type where in your approach to this question will work. For ex
| Some expression in x| = x- some no.

Again since LHS >/0, we can say x-some no >/ -----> x>/ some no.

This can be helpful in saving time by avoiding checking expression for 2 possible case of x (<0,>0)

Thanks
Mridul

Yes, in such kind of questions this approach might be helpful.

I just wanted to make sure the way I solved this problem is correct:

What is the value of x?

1. |6-3x|=x-2
2. |5x+3|=2x+9

1. 6-3x is positive when x<2. For example, if x=1 then 6-3x = 6-3(1) = 3. 6-3x is negative when x>2. For example, if x=3 then 6-3(x) = 6-3(3) = -3.

For positive cases: 6-3x=x-2 ==> x=2
For negative cases: -6+3x=x-2 ==> x=2

My issue here is that for the positive case x must be less than 2 and for the negative case x must be greater than 2 but for both cases here, I got x =2. The answer is sufficient, but I thought that for the cases to be sufficient the results had to be in line with what x= (for example, for the positive case here, x=2 but before I found that X must be LESS than 2 for 6-3x to be positive)

Could someone explain? I see why #2 fails (because for the positive case, 2 is greater than -3/5 but for the negative case, -7/12 is NOT greater than -3/5 but I am a bit lost with #1)

Thanks!

You have a point here.

When you study a function, an abs value, you correctly define where is positive or negative
\(x>a\) positve for example
\(x<a\) negative for example
As you see the value \(a\) is not contained in neither interval... How do we solve this?
Just put an \(=\) sign under one of the above inequalities, so \(x\geq{}a\) and \(x<a\) - or \(x>a\) and \(x\leq{}a\).
Where you put the sign, it doesn not matter, the result will not change.

So just rewrite your equation into

For positive cases (\(x\leq{2}\)): 6-3x=x-2 ==> x=2 VALID
For negative cases (\(x>2\)): -6+3x=x-2 ==> x=2

and you have the result! Nothing changes, everything you did is absolutely correct, just put an \(=\) sign in one of the intervals
Hope it's clear

Ok...I see that. But isn't the whole point to test cases where 6-3x is positive and negative? if X=2 then 6-3x is neither positive or negative, it's just zero.