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case I) x + 1 >0 and x -1>0 x+1= 2(x-1) --> x=3 case II) x + 1 >0 and x -1<0 x+1= 2(-x+1) --> x=1/3 case III) x + 1 <0 and x -1>0 --> not possible. case IV) x + 1 <0 and x -1<0 -x-1=2(-x+1) -> x=3 multiple solutions -->not suffcient 2) | x - 3 | not equal to 0 itself not suffcieint.

From statment 2: x<>3 from state 1 : x=1/3 or x=3 combined. x=1/3 sufficient.

C
_________________

Your attitude determines your altitude Smiling wins more friends than frowning

case I) x + 1 >0 and x -1>0 x+1= 2(x-1) --> x=3 case II) x + 1 >0 and x -1<0 x+1= 2(-x+1) --> x=1/3 case III) x + 1 <0 and x -1>0 --> not possible. case IV) x + 1 <0 and x -1<0 -x-1=2(-x+1) -> x=3 multiple solutions -->not suffcient

2) | x - 3 | not equal to 0 itself not suffcieint.

From statment 2: x<>3 from state 1 : x=1/3 or x=3 combined. x=1/3 sufficient.

C

Thanks Suresh. Thats awesome.
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----------------------------------------------------------- 'It's not the ride, it's the rider'

There's another way to deal with absolute (in)equalities.

(i) | x + 1 | = 2| x -1 |

square on both sides x²+1+2x = 4 ( x² + 1 -2x) 3x²-10x+3=0 (3x-1)(x-3) = 0 so x can be 1/3 or 3 so (i) is insufficient

(ii) |x-3| is not 0, in other words x is not 3 . This is insufficient because x can be one of billions and billions of numbers except 3. so (ii) is not sufficient.

combining (i) and (ii) (i) says x can be 1/3 or 3 (ii) says x cannot be 3

so i and ii combined say x=1/3. So answer is C
_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

There's another way to deal with absolute (in)equalities.

(i) | x + 1 | = 2| x -1 |

square on both sides x²+1+2x = 4 ( x² + 1 -2x) 3x²-10x+3=0 (3x-1)(x-3) = 0 so x can be 1/3 or 3 so (i) is insufficient

(ii) |x-3| is not 0, in other words x is not 3 . This is insufficient because x can be one of billions and billions of numbers except 3. so (ii) is not sufficient.

combining (i) and (ii) (i) says x can be 1/3 or 3 (ii) says x cannot be 3

so i and ii combined say x=1/3. So answer is C

Good method !!! +1 for u this saves time and actually clears the fundamental
_________________

Does this always work? Squaring both sides? absolute values, equalities, and inequalities?

amitdgr wrote:

IgnitedMind wrote:

Is | x | < 1 ?

1) | x + 1 | = 2| x -1 |

2) | x - 3 | not equal to 0

There's another way to deal with absolute (in)equalities.

(i) | x + 1 | = 2| x -1 |

square on both sides x²+1+2x = 4 ( x² + 1 -2x) 3x²-10x+3=0 (3x-1)(x-3) = 0 so x can be 1/3 or 3 so (i) is insufficient

(ii) |x-3| is not 0, in other words x is not 3 . This is insufficient because x can be one of billions and billions of numbers except 3. so (ii) is not sufficient.

combining (i) and (ii) (i) says x can be 1/3 or 3 (ii) says x cannot be 3

Just to add my late two cents perhaps sombody may find this method easier.

Statement 1: |x+1| = 2*|x-1|

The easiest way to see the various scenarios is to track the critical points of the two absolute value expressions on a number line. The critical point of the expression /x + 1/ is -1 and the critical point of the expression /x - 1/ is 1 How many distinct regions are there on the number line with the points -1 and/or 1 as boundaries? The answer is three: (-1<x), ( x<1) and (-1 < x < 1).

Now let’s test this scenario:

SCENARIO 1: when x < -1 (x+1) is –ve in this reason. So is (x-1) so the equation becomes: -x-1 = 2*(-x+1) => -x-1 = -2x+2 => x=3

SCENARIO 2: when -1 < x < 1, (x + 1) is positive in this region and (x - 1) is negative, so the equation becomes: (x+1) = 2*(-x+1) => x+1 = -2x+2 => 3x = 1 => X = 1/3

SCENARIO 3: when x > 1, (x + 1) is positive in this region and (x - 1) is also positive, so the equation becomes: x+1 = 2*(x-1) => x+1 = 2x-2 => x = 3

So we have 2 possible values of x. i.e. 1/3 and 3 Statement1 insufficient

Statement 2 is clearly insufficient

Combine: st.1 tells us that x is either 1/3 or 3 st. 2 tells us that x is not equal to 3 Then x must be 1/3 Sufficient IMO C

Does this always work? Squaring both sides? absolute values, equalities, and inequalities?

When you have mods on both sides of an equality or inequality then this method is best, as squaring removes so many options that you would otherwise have to try. Plugging numbers or Tracking critical points works. But when you have so many scenarios, as worked out in above post, squaring is better in my opinion. Time yourself and check.

I am no math Guru, I just happened to learn this method somewhere and this works for me
_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Tracking point works, I agree, but it takes me way too long.

For the squaring, do you need absolute values on both side? Or this is this the time that you choose to apply it?

amitdgr wrote:

bigfernhead wrote:

Does this always work? Squaring both sides? absolute values, equalities, and inequalities?

When you have mods on both sides of an equality or inequality then this method is best, as squaring removes so many options that you would otherwise have to try. Plugging numbers or Tracking critical points works. But when you have so many scenarios, as worked out in above post, squaring is better in my opinion. Time yourself and check.

I am no math Guru, I just happened to learn this method somewhere and this works for me