Ravshonbek wrote:

alimad wrote:

Is x > 0?

(1) |x + 3| = 4x – 3

(2) |x – 3| = |2x – 3|

D for me.

I will explain if correct.

Yes, I think you are right.. Here is my reasoning.. Please let me know what you think and then we can sync up whose strategy is better ..

I am using the concept of critical point as mentioned in one of the Doc that I have..

St1 : |x + 3| = 4x – 3
the critical point of x here is -3. hence the value of x could be in the range x < -3, -3 < x <3> 3.

a) when x < -3, (x+3) is -ve. hence s1 can be written as

-(x+3) = 4x -3

-x +3 = 4x -3

5x = 0

x = 0 which is not true since x < 3.

b) When -3 < x <3> -3 and <3> 3. Not needed as we already got the value of 2 and this will anyway not hold true since x cannot be equal to 2 and greater than 3 at the same time. hence x = 2 which is greater than 0. hence S1 is suff..

S2 : |x – 3| = |2x – 3|
|x – 3| - |2x – 3| = 0

As above we will check the range of x.

the CP is 3 and 3/2.

Hence 3 scenarios.

a) x > 3, b) x < 3/2 and c) 3/2 < x <3> 3, (x-3) and (2x-3 ) +ve.

(x-3) - (2x-3 ) = 0

=> x = 0 which is not true.

b) when x <3> x =0 Which again is not true..

c) When 3/2 < x < 3, (x-3) -ve and (2x-3) +ve

-(x-3) - (2x -3) = 0

-x + 3 -2x +3 = 0

-3x +6 = 0

x =2. Which is true ..

Hence s2 is also suff.

Hence D.

What say Ravshonbek ??