1. Since |a| is always greater than or equal to 0, this means that, since |2x - 6| = x, x will always be positive.
Now we have two options:
a. |2x - 6| = 2x - 6, which means that 2x - 6 is positive or equal to zero. This means that 2x - 6 >=0 or that 2x >= 6. In the end you get that x >= 3. Now let's solve the equation and see what we get:
2x - 6 = x
x = 6, which is consistent with the fact that x >=3.
b. |2x - 6| = 6 - 2x, when x < 3, but x > 0. Solve this equation and we get that:
6 - 2x = x
6 = 3x or x = 2, which, again, is consistent with 0 < x < 3.
So we have two solutions to the inital equation, which means that 1 is insufficient.

2. sqrt(1 + 2x^2) = x + 1 means that, first of all, x + 1 is positiveor equal to 0, since the sqrt of a number can only be positive or equal to 0. So x >= -1. Now let's raise everything to the second power:
1 + 2x^2 = x^2 + 2x + 1, which means that x^2 - 2x = 0, or that x(x - 2) = 0. This means that x is either 0 or 2, and, since both values are consistent with x >= -1. Again, 2 is insufficient.

Taken together, the two equations have only one common solution, x = 2. This means that answer is C.

Ans is C

1)
|2x-6| = x
this implies,
(2x-6) = x (or)
(2x-6) = -x

solving first we get x = 6, while solving the second we get x = 2
so insuff

2) sqrt(1 + 2x^2) = x + 1
sq on both sides,
(1+2x^2) = x^2 + 1 + 2x
x^2 = 2x
x(x-2) = 0

x= 0, 2
insuff

conbining both we have x=2 in common - suff.