Made minor corrections to my post above.
Jessica please refer to my post
https://www.beatthegmat.com/general-ques ... 20342.html
(1)
I am doing Case 1 and Case 4 (atleast we must do these 2 cases for absolute value problems with x on both sides)
Case 1: Considering the abs value expressions positive on both sides +/+
(1) |x + 1| = 2|x - 1|
x+1 = 2x-2
x=3
Case 4: Considering the abs value expressions -ve on lhs and +ve on rhs
i.e - / +
-(x+1) = 2x-2
-x-1 = 2x-2
-3x = -1
x= 1/3
2 possible values. In these abs value problems u need to substitute thes evalues in the original abs value expression to check if these indeed are solutions. If u
|x + 1| = 2|x - 1|
Put x = 3 and 1/3 seperately and lhs/rhs will be equal.
INSUFFICIENT since we have 2 solutions FOR X where one > 1 and the other is less than 1
(2)
WITH X ON JUST ONE SIDE WE HAVE 2 CASES + AND -
CASE 1:
|x - 3| > 0
X-3>0
X>3
CASE 2
-(X-3)>0
-X+3>0
-X>-3
X<3 (WHEN U DIVIDE/MULTIPLY AN EQUALITY WIHT A NEGATIVE NUMBER THE INEQUALITY REVERSES I.E in this case we are dividing by -1 on both sides)
From x<3 and x>3 we know x can be anytHing but 3 and we cannot still say if |x|<1 INSUFFICIENT
From (1) and (2) combined, we know the only other value for x is 1/3 and hence from both statements we can get a defnite value for X
Let me know if u still have questions!
Good luck!
