Inequalities in modulus are tricky.

If a > b, where both a and b are positive quantities then
a^2 > b^2

so
|2x + 3| > |7x - 2| (and both quantities are positive)
(squaring both the sides)
= |2x + 3|^2 > |7x - 2|^2
|2x + 3| is either (2x + 3) or -(2x + 3) but the square of both quantities is (2x + 3)^2
So |2x + 3|^2 = (2x + 3)^2
and
|7x - 2|^2 = (7x - 2)^2

Now we have
(2x + 3)^2 > (7x - 2)^2
(2x + 3)^2 - (7x - 2)^2 > 0
(2x + 3 +7x -2)(2x + 3 - 7x + 2) > 0
(9x + 1)(5 - 5x) > 0
(multiplying both sides with - 1 and changing the sign of the inequality)
(9x + 1)(5x - 5) < 0
((x + 1/9)(x - 1) < 0
Using Critical Points Method (as explained here: https://www.beatthegmat.com/critical-poi ... 10450.html)
solution:
[spoiler]-1/9 < x < 1[/spoiler]

[spoiler](C)[/spoiler] is the answer

Here is one approach of solving the problem.

Since,|2x + 3| > |7x - 2|,
we know that (|2x + 3|)^2 >(|7x - 2|)^2

4x^2 + 12x + 9 > 49x^2 + 4 - 28x
0 > 45x^2 - 40x - 5 (Subtracting 4x^2 + 12x + 9 from both sides)
0 > 5*(9x^2 -8x - 1)
0 > 5*(9x+1)*(x-1)

For the value of 5*(9x+1)*(x-1) to be less than zero (< 0), x must be a number between -1/9 and 1. i.e. (-1/9) < x < 1
How? - Here is a very useful post by aneesh that explains the critical points method https://www.beatthegmat.com/critical-poi ... 10450.html

Hope that helps.