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# If |12x−5|>|7−6x|, which of the following CANNOT be the

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Manager
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If |12x−5|>|7−6x|, which of the following CANNOT be the [#permalink]

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11 Jun 2013, 03:17
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If |12x−5|>|7−6x|, which of the following CANNOT be the product of two possible values of x?

A. -12
B. -7/5
C. -2/9
D. 4/9
E. 17
[Reveal] Spoiler: OA

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Kudos [?]: 1250 [6], given: 17

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Re: If |12x−5|>|7−6x|, which of the following CANNOT be the prod [#permalink]

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11 Jun 2013, 04:19
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emmak wrote:
If |12x−5|>|7−6x|, which of the following CANNOT be the product of two possible values of x?

A. -12
B. -7/5
C. -2/9
D. 4/9
E. 17

Square both sides: $$144x^2-120 x+25>36 x^2-84 x+49$$ --> $$9 x^2-3x-2>0$$ --> factor: $$(x+\frac{1}{3})(x-\frac{2}{3})>0$$ (check here: http://www.purplemath.com/modules/factquad.htm). "$$>$$" sign indicates that the solutions lies to the left of the smaller root and to the right of the greater root (check here: if-x-is-an-integer-what-is-the-value-of-x-1-x-2-4x-94661.html#p731476). Thus $$x<-\frac{1}{3}$$ OR $$x>\frac{2}{3}$$.

$$x=-4<-\frac{1}{3}$$ and $$x=3>\frac{2}{3}$$ are possible values of x --> the product = -12;
$$x=-\frac{7}{5}<-\frac{1}{3}$$ and $$x=1>\frac{2}{3}$$ are possible values of x --> the product = -7/5;
$$x=-\frac{4}{9}<-\frac{1}{3}$$ and $$x=-1<-\frac{1}{3}$$ are possible values of x --> the product = 4/9;
$$x=-1<-\frac{1}{3}$$ and $$x=-17<-\frac{1}{3}$$ are possible values of x --> the product = 17.

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Re: If |12x−5|>|7−6x|, which of the following CANNOT be the prod [#permalink]

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12 Jun 2013, 19:40
Hi.

I thought we cannot square both sides (|12x−5|>|7−6x|) unless we know that they are positive.

Thanks!

Bunuel wrote:
emmak wrote:
If |12x−5|>|7−6x|, which of the following CANNOT be the product of two possible values of x?

A. -12
B. -7/5
C. -2/9
D. 4/9
E. 17

Square both sides: $$144x^2-120 x+25>36 x^2-84 x+49$$ --> $$9 x^2-3x-2>0$$ --> factor: $$(x+\frac{1}{3})(x-\frac{2}{3})>0$$ (check here: http://www.purplemath.com/modules/factquad.htm). "$$>$$" sign indicates that the solutions lies to the left of the smaller root and to the right of the greater root (check here: if-x-is-an-integer-what-is-the-value-of-x-1-x-2-4x-94661.html#p731476). Thus $$x<-\frac{1}{3}$$ OR $$x>\frac{2}{3}$$.

$$x=-4<-\frac{1}{3}$$ and $$x=3>\frac{2}{3}$$ are possible values of x --> the product = -12;
$$x=-\frac{7}{5}<-\frac{1}{3}$$ and $$x=1>\frac{2}{3}$$ are possible values of x --> the product = -7/5;
$$x=-\frac{4}{9}<-\frac{1}{3}$$ and $$x=-1<-\frac{1}{3}$$ are possible values of x --> the product = 4/9;
$$x=-1<-\frac{1}{3}$$ and $$x=17>\frac{2}{3}$$ are possible values of x --> the product = 17.

Kudos [?]: 211 [0], given: 134

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Re: If |12x−5|>|7−6x|, which of the following CANNOT be the prod [#permalink]

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12 Jun 2013, 22:08
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Expert's post
WholeLottaLove wrote:
Hi.

I thought we cannot square both sides (|12x−5|>|7−6x|) unless we know that they are positive.

Thanks!

Bunuel wrote:
emmak wrote:
If |12x−5|>|7−6x|, which of the following CANNOT be the product of two possible values of x?

A. -12
B. -7/5
C. -2/9
D. 4/9
E. 17

Square both sides: $$144x^2-120 x+25>36 x^2-84 x+49$$ --> $$9 x^2-3x-2>0$$ --> factor: $$(x+\frac{1}{3})(x-\frac{2}{3})>0$$ (check here: http://www.purplemath.com/modules/factquad.htm). "$$>$$" sign indicates that the solutions lies to the left of the smaller root and to the right of the greater root (check here: if-x-is-an-integer-what-is-the-value-of-x-1-x-2-4x-94661.html#p731476). Thus $$x<-\frac{1}{3}$$ OR $$x>\frac{2}{3}$$.

$$x=-4<-\frac{1}{3}$$ and $$x=3>\frac{2}{3}$$ are possible values of x --> the product = -12;
$$x=-\frac{7}{5}<-\frac{1}{3}$$ and $$x=1>\frac{2}{3}$$ are possible values of x --> the product = -7/5;
$$x=-\frac{4}{9}<-\frac{1}{3}$$ and $$x=-1<-\frac{1}{3}$$ are possible values of x --> the product = 4/9;
$$x=-1<-\frac{1}{3}$$ and $$x=17>\frac{2}{3}$$ are possible values of x --> the product = 17.

We CAN square an inequality if we know that the sides are non-negative, which is the case here.
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Re: If |12x−5|>|7−6x|, which of the following CANNOT be the prod [#permalink]

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13 Jun 2013, 06:57
Thanks for the clarification.

When I factored this problem out, I got 12(9x^2-3x-2)>0 That factored out to:

12(3x+1)(3x-2)>0

So I have two questions:

1.) What happens with the factored out 12?

2.) upon simplifying for the inequalities in the equations, I got:

I.) (3x+1)>0 3x>-1 x>-1/3
II.) (3x-2)>0 3x>2 x>2/3

So my question is, how did you flip the inequality signs to get x<-\frac{1}{3} OR x>\frac{2}{3}. whereas I have x>-1/3 and x>2/3

Thanks!

Kudos [?]: 211 [0], given: 134

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Re: If |12x−5|>|7−6x|, which of the following CANNOT be the prod [#permalink]

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13 Jun 2013, 07:03
WholeLottaLove wrote:
Thanks for the clarification.

When I factored this problem out, I got 12(9x^2-3x-2)>0 That factored out to:

12(3x+1)(3x-2)>0

So I have two questions:

1.) What happens with the factored out 12?

2.) upon simplifying for the inequalities in the equations, I got:

I.) (3x+1)>0 3x>-1 x>-1/3
II.) (3x-2)>0 3x>2 x>2/3

So my question is, how did you flip the inequality signs to get x<-\frac{1}{3} OR x>\frac{2}{3}. whereas I have x>-1/3 and x>2/3

Thanks!

1. 12 is reduced (divide by 12 both sides).
2. Solving inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.
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Re: If |12x−5|>|7−6x|, which of the following CANNOT be the prod [#permalink]

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10 Jul 2013, 09:26
If |12x−5|>|7−6x|, which of the following CANNOT be the product of two possible values of x?

Find two checkpoints:

|12x−5|>|7−6x|

x=5/12, x=7/6

x<5/12, 5/12<x<7/6, x>7/6

x<5/12
|12x−5|>|7−6x|
-(12x-5) > (7-6x)
-12x+5 > 7-6x
-6x > 2
x<-1/3 Valid

5/12<x<7/6
|12x−5|>|7−6x|
(12x-5) > (7-6x)
18x > 12
x > 2/3
2/3<x<7/6

x>7/6
|12x−5|>|7−6x|
(12x-5) > -(7-6x)
12x - 5 > -7 + 6x
6x > -2
x > -1/3
(if the range being tested is >7/6 and x > -1/3 is that valid or invalid?)

I think I am approaching finding x the right way, but I am not sure how I can figure out what CANNOT be the product of two possible values of x. Can anyone help? Thanks!

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Re: If |12x−5|>|7−6x|, which of the following CANNOT be the prod [#permalink]

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15 Jul 2013, 14:16
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Another way to solve...

If |12x−5|>|7−6x|, which of the following CANNOT be the product of two possible values of x?

|12x−5|>|7−6x|
(12x-5)>(7-6x)
12x-5>7-6x
18x>12
x>2/3 Valid as a number greater than 2/3 will make |12x−5|>|7−6x| true

|12x−5|>|7−6x|
12x-5>-(7-6x)
12x-5>-7+6x
6x>-2
x>-1/3 Invalid as a number greater than -1/3 may or may not make |12x−5|>|7−6x| true

|12x−5|>|7−6x|
-(12x-5)>-(7-6x)
-12x+5>-7+6x
-18x>-12
x<2/3 Invalid as a number less than 2/3 may or may not make |12x−5|>|7−6x| true

|12x−5|>|7−6x|
-(12x-5)>(7-6x)
-12x+5>7-6x
-6x>2
x<-1/3 Valid as every number less than -1/3 will make |12x−5|>|7−6x| true

The two invalid values of x are -1/3 and 2/3. (-1/3)*(2/3) = (-2/9)

(C)

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Re: If |12x−5|>|7−6x|, which of the following CANNOT be the prod [#permalink]

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23 Jul 2013, 21:08
Hi Folks,

I have a query on this and see the attachment for the same.Please ignore the untidy drawing of mine, couldn't help due to time constraint and poor word knowledge.

I have drawn graphs for |12x+5| and |7-6x| and compared where the first modulus is greater than the second modulus. However, against the above solutions , I got only x> 2/3 solution which doesn't satisfies above solution.

The line in red is the graph for |7-6x| and the line in black is for |12x-5|. The pink line shows intersection point of the lines and has (2/3) as x - coordinate.So by the graph we can see that |12x-5| > |7-6x| only after x= 2/3.

Please tell where I am going wrong !!!

Regards,
TGC !
Attachments

query.JPG [ 13.53 KiB | Viewed 15195 times ]

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Re: If |12x−5|>|7−6x|, which of the following CANNOT be the prod [#permalink]

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23 Jul 2013, 21:42
1
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Expert's post
targetgmatchotu wrote:
Hi Folks,

I have a query on this and see the attachment for the same.Please ignore the untidy drawing of mine, couldn't help due to time constraint and poor word knowledge.

I have drawn graphs for |12x+5| and |7-6x| and compared where the first modulus is greater than the second modulus. However, against the above solutions , I got only x> 2/3 solution which doesn't satisfies above solution.

The line in red is the graph for |7-6x| and the line in black is for |12x-5|. The pink line shows intersection point of the lines and has (2/3) as x - coordinate.So by the graph we can see that |12x-5| > |7-6x| only after x= 2/3.

Please tell where I am going wrong !!!

Regards,
TGC !

The graphs drawn are not correct. The proper drawing is below:
Attachment:

MSP2441f260916666dabe40000498a5d50ib182823.gif [ 9.23 KiB | Viewed 16021 times ]
Hope it helps.
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Re: If |12x−5|>|7−6x|, which of the following CANNOT be the prod [#permalink]

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23 Jul 2013, 23:20
Bunuel wrote:
targetgmatchotu wrote:
Hi Folks,

I have a query on this and see the attachment for the same.Please ignore the untidy drawing of mine, couldn't help due to time constraint and poor word knowledge.

I have drawn graphs for |12x+5| and |7-6x| and compared where the first modulus is greater than the second modulus. However, against the above solutions , I got only x> 2/3 solution which doesn't satisfies above solution.

The line in red is the graph for |7-6x| and the line in black is for |12x-5|. The pink line shows intersection point of the lines and has (2/3) as x - coordinate.So by the graph we can see that |12x-5| > |7-6x| only after x= 2/3.

Please tell where I am going wrong !!!

Regards,
TGC !

The graphs drawn are not correct. The proper drawing is below:
Attachment:
MSP2441f260916666dabe40000498a5d50ib182823.gif
Hope it helps.

12x -5 , the graph touches y axis where x=0 => y = -5 taking modulus => y=5

7-6x, x=0 => y =7 .

Further, touches x axis where y=0 hence x=5/12 and x=7/6 (7/6 > 5/12)

Rgds,
TGC !

So by any chance the graph cross??????
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Re: If |12x−5|>|7−6x|, which of the following CANNOT be the prod [#permalink]

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24 Jul 2013, 00:52
targetgmatchotu wrote:
Bunuel wrote:
targetgmatchotu wrote:
Hi Folks,

I have a query on this and see the attachment for the same.Please ignore the untidy drawing of mine, couldn't help due to time constraint and poor word knowledge.

I have drawn graphs for |12x+5| and |7-6x| and compared where the first modulus is greater than the second modulus. However, against the above solutions , I got only x> 2/3 solution which doesn't satisfies above solution.

The line in red is the graph for |7-6x| and the line in black is for |12x-5|. The pink line shows intersection point of the lines and has (2/3) as x - coordinate.So by the graph we can see that |12x-5| > |7-6x| only after x= 2/3.

Please tell where I am going wrong !!!

Regards,
TGC !

The graphs drawn are not correct. The proper drawing is below:
Attachment:
MSP2441f260916666dabe40000498a5d50ib182823.gif
Hope it helps.

12x -5 , the graph touches y axis where x=0 => y = -5 taking modulus => y=5

7-6x, x=0 => y =7 .

Further, touches x axis where y=0 hence x=5/12 and x=7/6 (7/6 > 5/12)

Rgds,
TGC !

So by any chance the graph cross??????

I don't understand your question... As I said the graph of |12x−5|>|7−6x| is:

x<-1/3 and x>2/3.
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Re: If |12x−5|>|7−6x|, which of the following CANNOT be the prod [#permalink]

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24 Jul 2013, 00:59
Bunuel wrote:
I don't understand your question... As I said the graph of |12x−5|>|7−6x| is:

x<-1/3 and x>2/3.

Got it,

there was a mistake in drawing the graph !!

Thanks,
TGC !
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Re: If |12x−5|>|7−6x|, which of the following CANNOT be the [#permalink]

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27 Mar 2014, 21:15
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emmak wrote:
If |12x−5|>|7−6x|, which of the following CANNOT be the product of two possible values of x?

A. -12
B. -7/5
C. -2/9
D. 4/9
E. 17

You can also use the number line method of mods to solve this question:

|12x−5|>|7−6x|

12|x - 5/12| > 6|x - 7/6|

2|x - 5/12| > |x - 14/12|

Twice of the distance from 5/12 should be more than distance from 14/12.

___________0_______5/12_________________14/12__________

We find the points where the two distances are equal.
The distance between 5/12 and 14/12 is 9/12 which gets divided into 1:2 i.e. the point where the distances will be equal will be 3/12 away from 5/12 i.e. at 8/12 = 2/3. At any point to the right of 2/3, twice the distance from 5/12 will be more than the distance from 14/12.

Another point will be 9/12 to the left of 5/12 i.e. at -4/12 = -1/3. At any left to the left of -1/3, twice the distance from 5/12 will be more than the distance from 14/12.

x < -1/3 OR x > 2/3

Then proceed as given above.
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Re: If |12x−5|>|7−6x|, which of the following CANNOT be the [#permalink]

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02 Apr 2014, 08:17
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Another Method
$$|12x-5|>|7-6x|$$
Only 2 cases can arise; the other 2 cases are the same as these ones
Ist Case
$$|12x-5|>|7-6x|$$
$$2x-5>7-6x$$
$$18x>12$$
$$x>2/3$$

IInd Case
$$|12x-5|>|7-6x|$$
$$12x-5<6x-7$$
$$x<-2/6$$
$$x<-1/3$$

Now we know that$$x>2/3$$ and $$x<-1/3$$
So from the above we can deduce that the answer has to be negative, thus we can cross out D and E.
From the next 3 options which are all negative, Option A and B both can be formed, but option C is between $$-1/3$$ and $$2/3$$ which is not in the range of x. Thus C is your answer.

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Re: If |12x−5|>|7−6x|, which of the following CANNOT be the [#permalink]

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21 Jun 2014, 02:30
VeritasPrepKarishma wrote:
emmak wrote:
If |12x−5|>|7−6x|, which of the following CANNOT be the product of two possible values of x?

A. -12
B. -7/5
C. -2/9
D. 4/9
E. 17

You can also use the number line method of mods to solve this question:

|12x−5|>|7−6x|

12|x - 5/12| > 6|x - 7/6|

2|x - 5/12| > |x - 14/12|

Twice of the distance from 5/12 should be more than distance from 14/12.

___________0_______5/12_________________14/12__________

We find the points where the two distances are equal.
The distance between 5/12 and 14/12 is 9/12 which gets divided into 1:2 i.e. the point where the distances will be equal will be 3/12 away from 5/12 i.e. at 8/12 = 2/3. At any point to the right of 2/3, twice the distance from 5/12 will be more than the distance from 14/12.

Another point will be 9/12 to the left of 5/12 i.e. at -4/12 = -1/3. At any left to the left of -1/3, twice the distance from 5/12 will be more than the distance from 14/12.

x < -1/3 OR x > 2/3

Then proceed as given above.

I apply this distance inference. But in a different way,

The nearest the number to 5/12 must be the answer, as it yields the shortest distance to 5/12. Then answer is C.

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Re: If |12x−5|>|7−6x|, which of the following CANNOT be the [#permalink]

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08 Dec 2016, 21:21
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happyface101 wrote:
Bunuel wrote:
emmak wrote:
If |12x−5|>|7−6x|, which of the following CANNOT be the product of two possible values of x?

A. -12
B. -7/5
C. -2/9
D. 4/9
E. 17

Square both sides: $$144x^2-120 x+25>36 x^2-84 x+49$$ --> $$9 x^2-3x-2>0$$ --> factor: $$(x+\frac{1}{3})(x-\frac{2}{3})>0$$ (check here: http://www.purplemath.com/modules/factquad.htm). "$$>$$" sign indicates that the solutions lies to the left of the smaller root and to the right of the greater root (check here: if-x-is-an-integer-what-is-the-value-of-x-1-x-2-4x-94661.html#p731476). Thus $$x<-\frac{1}{3}$$ OR $$x>\frac{2}{3}$$.

$$x=-4<-\frac{1}{3}$$ and $$x=3>\frac{2}{3}$$ are possible values of x --> the product = -12;
$$x=-\frac{7}{5}<-\frac{1}{3}$$ and $$x=1>\frac{2}{3}$$ are possible values of x --> the product = -7/5;
$$x=-\frac{4}{9}<-\frac{1}{3}$$ and $$x=-1<-\frac{1}{3}$$ are possible values of x --> the product = 4/9;
$$x=-1<-\frac{1}{3}$$ and $$x=-17<-\frac{1}{3}$$ are possible values of x --> the product = 17.

Bunuel - this is really helpful, thank you! Can you let us know if a question like this is possible on the GMAT? How is it possible to interpret the question and do all this math / testing in 2mins or 2.5mins max?

To solve the question faster, we don't have to look at all the choices.
Once we know that x>2/3 and x<−1/3, we can say that x can neither be 2/3 nor -1/3 (x!=2/3 && x!=-1/3)

so the product of two possible values of x can never be 2/3*(-1/3)=-2/9, Hence the correct answer is C.

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Re: If |12x−5|>|7−6x|, which of the following CANNOT be the [#permalink]

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10 Dec 2016, 06:58
VeritasPrepKarishma wrote:
emmak wrote:
If |12x−5|>|7−6x|, which of the following CANNOT be the product of two possible values of x?

A. -12
B. -7/5
C. -2/9
D. 4/9
E. 17

You can also use the number line method of mods to solve this question:

|12x−5|>|7−6x|

12|x - 5/12| > 6|x - 7/6|

2|x - 5/12| > |x - 14/12|

Twice of the distance from 5/12 should be more than distance from 14/12.

___________0_______5/12_________________14/12__________

We find the points where the two distances are equal.
The distance between 5/12 and 14/12 is 9/12 which gets divided into 1:2 i.e. the point where the distances will be equal will be 3/12 away from 5/12 i.e. at 8/12 = 2/3. At any point to the right of 2/3, twice the distance from 5/12 will be more than the distance from 14/12.

Another point will be 9/12 to the left of 5/12 i.e. at -4/12 = -1/3. At any left to the left of -1/3, twice the distance from 5/12 will be more than the distance from 14/12.

x < -1/3 OR x > 2/3

Then proceed as given above.

Hi VeritasPrepKarishma,

I could not understand how did you solve it. I could not comprehend the line

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Re: If |12x−5|>|7−6x|, which of the following CANNOT be the [#permalink]

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14 Dec 2017, 22:31
emmak wrote:
If |12x−5|>|7−6x|, which of the following CANNOT be the product of two possible values of x?

A. -12
B. -7/5
C. -2/9
D. 4/9
E. 17

We have x<−1/3 OR x>2/3.
Therefore, x cannot be -1/3 and 2/3.

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If |12x−5|>|7−6x|, which of the following CANNOT be the [#permalink]

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31 Dec 2017, 13:02
emmak wrote:
If |12x−5|>|7−6x|, which of the following CANNOT be the product of two possible values of x?

A. -12
B. -7/5
C. -2/9
D. 4/9
E. 17

VERITAS PREP OFFICIAL SOLUTION:

Solution: C. Algebraically, the inequalities give you four different possibilities:

12x – 5 > 7 – 6x (which works out cleanly to 18x > 12, meaning x > 2/3)
12x – 5 < -(7 – 6x)
-(12x – 5) < 7 – 6x
-(12x -5) > -(7 – 6x)

With the last three all involving negatives and inequalities, it can be helpful to simply find the inequality point and then test values on either side to determine whether it's greater than or less than:

12x – 5 < -(7 – 6x) works out to:

12x - 5 < 6x - 7

6x < -2

x < −1/3, but try a value like -1 (less than −1/3) and like 0 (greater than −1/3) to ensure the - vs. + of a tricky inequality with absolute values. -1 fits with the given information and 0 does not, so it should be clear that x < −1/3.

-(12x – 5) < 7 – 6x works out to 5 - 12x < 7 - 6x, which gives you -2 < 6x, and x > −1/3. This is why testing negative/positive is so important...the four "original" inequalities allow for really all sets of possible values other than −1/3 and 2/3, so some quick trial and error can help you determine which side of the inequalities are valid and which are not. Again, it should be clear from a quick plug-in of 0 and -1 that x < −1/3.

-(12x -5) > -(7 – 6x) works out to 5 - 12x > 6x - 7, which leads you to 12 > 18x, and x < 2/3. This confirms the "break point" of 2/3, so again a quick plug in of easy numbers on either side (0 and 1) will help you determine that x must be greater than 2/3.

So you know that x is either greater than 2/3 or less than −1/3. Certain answer choices, then, are easy to pick off: -12 could be -1*12. 17 could be 1*17. 4/9 could be -1*(-4/9) and -7/5 could be -1*7/5. But -2/9 cannot be done.
_________________

Kudos [?]: 139624 [0], given: 12794

If |12x−5|>|7−6x|, which of the following CANNOT be the   [#permalink] 31 Dec 2017, 13:02

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