(A) for me

Stat 1
|x + 3| = 4x – 3

o If x >= -3 then
|x + 3| = 4x – 3

<=> x+3 = 4*x - 3

<=> x = 2 >>> Ok, as x >= -3

o If x < -3 then
|x + 3| = 4x – 3

<=> -(x+3) = 4*x - 3

<=> x = 0 >>> Out as x < -3

Thus, x=2 is the only solution and it's positive.

SUFF.

Stat 2
|x – 3| = |2x – 3|

We have to look at 3 intervals : x < 3/2, 3/2 =< x < 3 and x >=3

o If x < 3/2 then
|x – 3| = |2x – 3|

<=> -(x-3) = -(2*x-3)

<=> x-3 = 2*x - 3

<=> x = 0 >>>> OK as x < 3/2

o If 3/2 =< x < 3 then
|x – 3| = |2x – 3|

<=> -(x-3) = 2*x-3

<=> 3*x = 6

<=> x = 2 >>>> OK as 3/2 =< x < 3

o If x >= 3 then
|x – 3| = |2x – 3|

<=> x-3 = 2*x - 3

<=> x = 0 >>>> Out as x >= 3

Finally, we have x = 0 or x = 2

INSUFF.