This is equivalent to finding out if -1 < x < 1.
1. Let's split this up into cases:
a. x < -1, then |x + 1|= -x - 1 and |x - 1| = 1 - x. Solve the equation:
-x - 1 = 2 - 2x
x = 3. This however is not consistent with the initial assumption, of x < -1.

b. x is between -1 and 1. You get that |x + 1| = x + 1 and |x - 1| = 1 - x. Solve for x:
x + 1 = 2 - 2x
3x = 1, with x = 1/3. This is consistent with -1 < x < 1. So x = 1/3 is a solution of the equation.

c. x > 1 making our equation:
x + 1 = 2x - 2
x = 3 - consistent with x > 1. So x = 3 is also a solution of the equation.
So 1 has two solutions: 1/3 and 3, which means that 1 is insufficient.

2. is insufficient, since this only tells us that x is not equal to 3.

Take both stmts together and you eliminate one of the solutions of stmt 1, meaning x = 3. Then you are left with x = 1/3, and you can tell that |x| <1.


So C is indeed correct.

DanaJ wrote:This is equivalent to finding out if -1 < x < 1.
1. Let's split this up into cases:
a. x < -1, then |x + 1|= -x - 1 and |x - 1| = 1 - x. Solve the equation:
-x - 1 = 2 - 2x
x = 3. This however is not consistent with the initial assumption, of x < -1.

b. x is between -1 and 1. You get that |x + 1| = x + 1 and |x - 1| = 1 - x. Solve for x:
x + 1 = 2 - 2x
3x = 1, with x = 1/3. This is consistent with -1 < x < 1. So x = 1/3 is a solution of the equation.

c. x > 1 making our equation:
x + 1 = 2x - 2
x = 3 - consistent with x > 1. So x = 3 is also a solution of the equation.
So 1 has two solutions: 1/3 and 3, which means that 1 is insufficient.

2. is insufficient, since this only tells us that x is not equal to 3.

Take both stmts together and you eliminate one of the solutions of stmt 1, meaning x = 3. Then you are left with x = 1/3, and you can tell that |x| <1.


So C is indeed correct.

Perfect....maria you can't ask for more than this...