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

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

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11 Jun 2013, 04:17
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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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Re: If |12x−5|>|7−6x|, which of the following CANNOT be the prod [#permalink]

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11 Jun 2013, 05:19
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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, 20:40
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.

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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, 23:08
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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, 07:57
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!
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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, 08:03
Expert's post
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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13 Jun 2013, 08:54
1
KUDOS
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.

Hi Bunuel,
Please correct me if i am wrong here, but dont you mean
$$x=-1<-\frac{1}{3}$$ and $$x=-17<\frac{-1}{3}$$ are possible values of x --> the product = 17.
For the last point??

Since -1 * 17 would be -17??
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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, 09:10
Expert's post
kpali 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 $$x=17>\frac{2}{3}$$ are possible values of x --> the product = 17.

Hi Bunuel,
Please correct me if i am wrong here, but dont you mean
$$x=-1<-\frac{1}{3}$$ and $$x=-17<\frac{-1}{3}$$ are possible values of x --> the product = 17.
For the last point??

Since -1 * 17 would be -17??

Sure. Typo edited. Thank you.
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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, 10:26
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, 15:16
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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)

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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, 22:08
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 9630 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, 22:42
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 10006 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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24 Jul 2013, 00:20
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, 01:52
Expert's post
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, 01:59
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 prod [#permalink]

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27 Mar 2014, 12:33
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.

Dear Bunuel

Pl enlighten us in what all cases we can square the modulus

Thanks
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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, 22:15
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Expert's post
2
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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, 09:17
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$$|12x-5|>|7-6x|$$
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 prod [#permalink]

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23 Apr 2014, 10:02
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.

Hi Bunnel,

I have a doubt. how you are getting 7/5, 4/9, 1/3 etc ion above post.

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

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23 Apr 2014, 10:10
thanks Mahendru. the other explanations are all so weird! No offence*
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Re: If |12x−5|>|7−6x|, which of the following CANNOT be the   [#permalink] 23 Apr 2014, 10:10

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