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

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: https://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.

Answer: C.
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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)

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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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!

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.
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\(|12x−5|>|7−6x|\)

\((12x-5)^2 > (7-6x)^2\)

solving the quadratics on both the sides we get:
\(9x^2-3x-2 >0\)

now, if the above expression were equal to 0, the product of roots of the equation would have been \(\frac{c}{a} = \frac{-2}{9}\)
since it's not equal to 0 the product would be some value other than \(\frac{-2}{9}\)
Answer C

chetan2u VeritasKarishma Bunuel can you please tell if we can solve this question they way I have done?
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Let’s first solve for when (12x - 5) and (7 - 6x) are both positive.

12x - 5 > 7 - 6x

18x > 12

x > 12/18

x > 2/3

Now let’s solve for when (12x - 5) is negative and (7 - 6x) is positive.

-(12x - 5) > 7 - 6x

-12x + 5 > 7 - 6x

-2 > 6x

-1/3 > x

So we have x < -1/3 or x > 2/3.

If x were -1/3 or 2/3, then the product would be -2/9. However, the inequalities specify that x can be neither -1/3 nor 2/3, so we know the product of two possible values of x cannot be -2/9.

Answer: C
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Hi Bunuel,

i'm a bit confused over here. According to the question, there are 2 CP, 5/12 and 7/6.

so, the three cases are as follows:

1. x <5/12
=> -12x+5>7-6x
=> x<-1/3

2. 5/12<= x <7/6
=> 12x-5>7-6x
=> x>2/3

3. x >=7/6
=>12x-5>6x-7
=> x>-1/3

Could you please help me understand the reason for eliminating the 3rd solution?

Note here that pre-condition for 3rd case is that x >= 7/6. Hence values such as -1/10, 0, 1 are not valid. So solution from third case is that x > 7/6.

Also note that the most convenient way of dealing with such inequalities is squaring both sides. The absolute value sign disappears immediately and your are left with a simple quadratic.
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Yes, that's certainly an interesting way to look at it.
Though think about this: what if I remove -2/9 from the options and put 0 there instead?
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Hi!
As Bunuel mentioned, the graph would be different. These lines are not parallel.
I'll attempt to provide an explanation.

y=|12x-5| is:
y=12x-5 for x>= 5/12; and
y=-12x+5 for x<5/12

y=|7-6x| = |6x-7| is:
y=6x-7 for x>=7/6; and
y=-6x+7 for x<7/6

This can also be written as:
y=6x-7 for x>=14/12;
y=-6x+7 for x<14/12


Note that vertex of 'V' of y=|7-6x| is to the right of the vertex of 'V' of y=|12x-5| (Refer to the attachment)
Also, note that the slope of y=|12x-5| is greater than that of y=|7-6x|, and thus, both lines diverge away towards the right, with no point of intersection.

This means that only 'y=-6x+7' leg of y=|7-6x| interacts with both the legs of y=|12x-5|

Intersecting 'y=-6x+7' with 'y=12x-5' and 'y=-12x+5', we get (2/3,__) and (-1/3,__)

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