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What is the value of x?

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What is the value of x? [#permalink]

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What is the value of x?

(1) |6 - 3x| = x - 2

(2) |5x + 3| = 2x + 9
[Reveal] Spoiler: OA

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Re: Absolute Value [#permalink]

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New post 30 May 2010, 03:24
(1) |6-3x| = x-2 <=> 6-3x = x-2 or 6-3x = -(x-2)

solving these equations you will have x=2 or x = 2 then (1) is sufficient. X = 2

(2) 5x+3 = 2x+9 or 5x+3 = -(2x+9)

3x = 6 or 7x = -12 we will have two values for x. (2) is not sufficient

Would pick (A)

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Re: Absolute Value [#permalink]

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New post 30 May 2010, 03:39
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??
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Re: Absolute Value [#permalink]

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(1) |6-3x| = x-2 <=> 6-3x = x-2 or 6-3x = -(x-2)

solving these equations you will have x=2 or x = 2 then (1) is sufficient. X = 2

(2) 5x+3 = 2x+9 or 5x+3 = -(2x+9)

3x = 6 or 7x = -12 we will have two values for x. (2) is not sufficient

Would pick (A)


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.
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Re: Absolute Value [#permalink]

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Hussain15 wrote:
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??


Red part is not correct: absolute value is never negative (absolute value \(\geq{0}\)).

Step-by-step solution:

(1) \(|6 - 3x| = x - 2\) --> as LHS (left hand side) is absolute value, which is never negative (absolute value \(\geq{0}\)), RHS also must be \(\geq{0}\) --> \(x-2\geq{0}\) --> \(x\geq{2}\). For \(x\geq{2}\), \(|6 - 3x|=-6+3x\), so \(|6 - 3x| = x - 2\) becomes \(-6+3x=x-2\) --> \(x=2\). Sufficient.

(2) \(|5x + 3| = 2x + 9\). (note that by the same logic as above \(x\geq-\frac{9}{2}=-4.5\)) One check point \(x=-\frac{3}{5}\) (check point(s) the value of x when absolute value equals to zero).

\(x\leq{-\frac{3}{5}}\) --> \(-5x-3=2x+9\) --> \(x=-\frac{12}{7}\) - valid solution as \(-\frac{9}{2}<-\frac{12}{7}<-\frac{3}{5}\);

\(x>{-\frac{3}{5}}\) --> \(5x+3=2x+9\) --> \(x=2\) - valid solution as \(-\frac{9}{2}<-\frac{3}{5}<2\);


Two valid solutions. Not sufficient.

Answer: A.

Hope it helps.
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Re: Absolute Value [#permalink]

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Re: Absolute Value [#permalink]

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New post 31 May 2010, 07:26
very nicely explained Bunuel... +1.. thanks

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Re: Absolute Value [#permalink]

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New post 27 May 2013, 05:20
Bunuel wrote:
Hussain15 wrote:
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??


Red part is not correct: absolute value is never negative (absolute value \(\geq{0}\)).

Step-by-step solution:

(1) \(|6 - 3x| = x - 2\) --> as LHS (left hand side) is absolute value, which is never negative (absolute value \(\geq{0}\)), RHS also must be \(\geq{0}\) --> \(x-2\geq{0}\) --> \(x\geq{2}\). For \(x\geq{2}\), \(|6 - 3x|=-6+3x\), so \(|6 - 3x| = x - 2\) becomes \(-6+3x=x-2\) --> \(x=2\). Sufficient.


(2) \(|5x + 3| = 2x + 9\). (note that by the same logic as above \(x\geq-\frac{9}{2}=-4.5\)) One check point \(x=-\frac{3}{5}\) (check point(s) the value of x when absolute value equals to zero).

\(x\leq{-\frac{3}{5}}\) --> \(-5x-3=2x+9\) --> \(x=-\frac{12}{7}\) - valid solution as \(-\frac{9}{2}<-\frac{12}{7}<-\frac{3}{5}\);

\(x>{-\frac{3}{5}}\) --> \(5x+3=2x+9\) --> \(x=2\) - valid solution as \(-\frac{9}{2}<-\frac{3}{5}<2\);


Two valid solutions. Not sufficient.

Answer: A.

Hope it helps.



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.

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Re: Absolute Value [#permalink]

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New post 27 May 2013, 05:26
gchawla123 wrote:
Bunuel wrote:
Hussain15 wrote:
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??


Red part is not correct: absolute value is never negative (absolute value \(\geq{0}\)).

Step-by-step solution:

(1) \(|6 - 3x| = x - 2\) --> as LHS (left hand side) is absolute value, which is never negative (absolute value \(\geq{0}\)), RHS also must be \(\geq{0}\) --> \(x-2\geq{0}\) --> \(x\geq{2}\). For \(x\geq{2}\), \(|6 - 3x|=-6+3x\), so \(|6 - 3x| = x - 2\) becomes \(-6+3x=x-2\) --> \(x=2\). Sufficient.


(2) \(|5x + 3| = 2x + 9\). (note that by the same logic as above \(x\geq-\frac{9}{2}=-4.5\)) One check point \(x=-\frac{3}{5}\) (check point(s) the value of x when absolute value equals to zero).

\(x\leq{-\frac{3}{5}}\) --> \(-5x-3=2x+9\) --> \(x=-\frac{12}{7}\) - valid solution as \(-\frac{9}{2}<-\frac{12}{7}<-\frac{3}{5}\);

\(x>{-\frac{3}{5}}\) --> \(5x+3=2x+9\) --> \(x=2\) - valid solution as \(-\frac{9}{2}<-\frac{3}{5}<2\);


Two valid solutions. Not sufficient.

Answer: A.

Hope it helps.



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.
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Re: What is the value of x? [#permalink]

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New post 29 May 2013, 21:05
Hussain15 wrote:
What is the value of x?

(1) |6 - 3x| = x - 2

(2) |5x + 3| = 2x + 9


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
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Re: What is the value of x? [#permalink]

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New post 30 May 2013, 01:33
atalpanditgmat wrote:
Hussain15 wrote:
What is the value of x?

(1) |6 - 3x| = x - 2

(2) |5x + 3| = 2x + 9


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.
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Re: What is the value of x? [#permalink]

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New post 04 Jun 2013, 17:15
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!

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New post 04 Jun 2013, 17:33
WholeLottaLove wrote:
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).
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How did you get "for X> or = 2, |6x-3| = -6x+3" ?

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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.

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Re: What is the value of x? [#permalink]

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New post 05 Jun 2013, 01:40
Bunuel wrote:
atalpanditgmat wrote:
Hussain15 wrote:
What is the value of x?

(1) |6 - 3x| = x - 2

(2) |5x + 3| = 2x + 9


Hi Bunuel, can'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.



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
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New post 05 Jun 2013, 01:43
mridulparashar1 wrote:
Bunuel wrote:
atalpanditgmat wrote:
Yes, we can do this way too.

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



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.
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Re: What is the value of x? [#permalink]

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New post 05 Jun 2013, 11:54
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!

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Re: What is the value of x? [#permalink]

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New post 05 Jun 2013, 12:02
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WholeLottaLove wrote:
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
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Re: What is the value of x? [#permalink]

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New post 05 Jun 2013, 12:18
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.

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Re: What is the value of x?   [#permalink] 05 Jun 2013, 12:18

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