Abdulla wrote:Is |x| < 1 ?

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

(2) |x - 3| > 0

It is C IMO

Qn can be phrased as "Is -1 < x < 1?"

1) |x+1| = 2|x-1|
Square both sides,

(x+1)^2 = 4 * (x-1)^2
This can be reduced to,
3x^2 - 10x + 3 = 0
x = 3, 1/3

Check both values with qn., For x = 3, you get NO. For x=1/3, you get YES.

Insufficient.

2) |x-3| > 0

Two cases here,
1. x-3 > 0 ==> x > 3
2. -(x-3)>0. which is -x+3 > 0 ==> x < 3

If you combine both, x is NOT equal to 3. x may be any value other than 3.

Insufficient.

Combining,
you have 2 values from 1st stmt., 3 and 1/3. But from 2nd stmt, you know that x is NOT equal to 3.
Then you can conclude that x = 1/3 which answers the question.

Hence C.

Whats the OA?

The second stqtement |x - 3| > 0 does not give any information as it is always true. So in my opinion one can discard it

Taking statement 1 you have 4 cases

1st
x+1 = 2(x-1)
x=3

2nd
x+1 = 2(1-x)
x=1/3

No need to solve further because in the 1st case |x| = |3| = 3 which is >1
and in the 2nd case |x| = |1/3| = 1/3 which is <1

So IMO E

are you okay with that ?

I'm confused with absolute values. If you have an absolute value on each side of the equation, do you only the distribute the negative on one side of the absolute value and not both?

Each time you have an absolute value, in the absence of any more info, it can take 2 different values.

For example |x-3| becomes (x-3) or -(x-3)

|x| yields x or -x. If x is negative, its absolute value is -x. Example if x = -4, |x| = -x = -(-4) = 4

Abs value is always positive.

papgust wrote:

Abdulla wrote:Is |x| < 1 ?

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

(2) |x - 3| > 0

It is C IMO

Qn can be phrased as "Is -1 < x < 1?"

1) |x+1| = 2|x-1|
Square both sides,

(x+1)^2 = 4 * (x-1)^2
This can be reduced to,
3x^2 - 10x + 3 = 0
x = 3, 1/3

Check both values with qn., For x = 3, you get NO. For x=1/3, you get YES.

Insufficient.

2) |x-3| > 0

Two cases here,
1. x-3 > 0 ==> x > 3
2. -(x-3)>0. which is -x+3 > 0 ==> x < 3

If you combine both, x is NOT equal to 3. x may be any value other than 3.

Insufficient.

Combining,
you have 2 values from 1st stmt., 3 and 1/3. But from 2nd stmt, you know that x is NOT equal to 3.
Then you can conclude that x = 1/3 which answers the question.

Hence C.

Whats the OA?

I totally agree with Papgust reasoning ...
The OA is C

Hi papgust,

May I know your reasoning for squaring both modulus? I suppose you meant to remove the modulus parentheses?

Abdulla wrote:Is |x| < 1 ?

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

(2) |x - 3| > 0

(1) |x + 1| = 2|x -1| critical points x=-1 x=1
for x<-1, -(x+1) = -2(x-1) => -x-1 = -2x+2 => x = 3 not feasible as x<-1
for -1<x<1 : (x+1) = -2(x-1) or 3x = 1 or x=1/3 ok
for x>1 : x+1 = 2x-2 or x=3 ok

two possibilities, one >1 another <1

(2) |x-3| > 0 for x>3 so we do not know any thing but x!=3

So C may be picked.

valleeny wrote:Hi papgust,

May I know your reasoning for squaring both modulus? I suppose you meant to remove the modulus parentheses?

True. I squared both sides to remove the modulus. Basically, whenever you see modulus on both sides, it's always a better strategy to square both sides. Because, when you take the modulus as it is, there could be many possible inequalities to handle which will be time-consuming.

papgust wrote:

valleeny wrote:Hi papgust,

May I know your reasoning for squaring both modulus? I suppose you meant to remove the modulus parentheses?

True. I squared both sides to remove the modulus. Basically, whenever you see modulus on both sides, it's always a better strategy to square both sides. Because, when you take the modulus as it is, there could be many possible inequalities to handle which will be time-consuming.

Is there a way to solve this without squaring?

As i mentioned earlier, it will be very time-consuming if you are not squaring both sides. Let's walk through this method if you wish to,

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

As absolute values take +ve and -ve, both LHS and RHS will have 2 scenarios. So, you have totally 4 scenarios [2-LHS, 2-RHS]

(i) -(x+1) = 2(x-1) [LHS -ve, RHS +ve]
-x-1=2x-2
3x=1
x=1/3

(ii) x+1 = 2(x-1) [LHS +ve, RHS +ve]
x+1=2x-2
x=3

(iii) x+1 = 2*-(x-1) [LHS +ve, RHS -ve]
x+1=2*(-x+1)
x+1=-2x+2
3x=1
x=1/3

(iv) -(x+1) = 2*-(x-1) [LHS -ve, RHS -ve]
-x-1=-2x+2
x=3

You get x=1/3, 3 when all scenarios are combined together to form a solution set. As you can see, it is not a good way of solving in real GMAT.

Squaring method is far more easier and quicker when 2 absolutes appear on both sides.

I'm only aware of these 2 methods. Experts can help us if there is any other quicker method.

papgust wrote:As i mentioned earlier, it will be very time-consuming if you are not squaring both sides. Let's walk through this method if you wish to, Experts can help us if there is any other quicker method.

Though I am not an expert but absolute values do not depend on 4 conditions as you have mentioned, neither I have heard of the squaring technique, I am aware about one technique called critical point technique, determined the points at which signs of the absolute changes as within sign changes any value will be same, later you need to check across validity of results, e.g. if x<-1 and values found is 1/2 then discard it.

maihuna wrote:

papgust wrote:As i mentioned earlier, it will be very time-consuming if you are not squaring both sides. Let's walk through this method if you wish to, Experts can help us if there is any other quicker method.

Though I am not an expert but absolute values do not depend on 4 conditions as you have mentioned, neither I have heard of the squaring technique, I am aware about one technique called critical point technique, determined the points at which signs of the absolute changes as within sign changes any value will be same, later you need to check across validity of results, e.g. if x<-1 and values found is 1/2 then discard it.

Hi Maihuna,

Can you explain more about this technique please?

maihuna wrote:

papgust wrote:As i mentioned earlier, it will be very time-consuming if you are not squaring both sides. Let's walk through this method if you wish to, Experts can help us if there is any other quicker method.

Though I am not an expert but absolute values do not depend on 4 conditions as you have mentioned, neither I have heard of the squaring technique, I am aware about one technique called critical point technique, determined the points at which signs of the absolute changes as within sign changes any value will be same, later you need to check across validity of results, e.g. if x<-1 and values found is 1/2 then discard it.

Yeah you are right. 4 conditions isn't needed here. We do not have to consider LHS as -ve.

But i'm sure that squaring method is prevalent for these type of problems. Maybe testluv/stuart could comment!