Please submit your concern

Explanation:

We have,
13x + 1 < 2z                 … (i)
z + 3 = 5y²                    …(ii)
From (i) and (ii), we get,
13x + 1 < 2(5y² – 3)
∴ 13x + 1 < 10y² – 6
∴ x < (10y² – 7)/13
For y = 1, we get x < 3/13
∴ x < y
For  y = 2, we get x < 33/13
∴ x could be greater than or less than y.
∴ None of the given options are necessarily true.
Hence, option (d).