[tex]3x + 3 > 48[/tex]
[tex]3x > 48 - 3 \\ 3x > 45 \\ x > \frac{45}{3} \Longrightarrow x > \frac{ \cancel{45}}{ \cancel{3}} \Longrightarrow x > \frac{15}{1} \\ x > 15[/tex]
Because x must be greater than 15. We will substitute x = 16 in the Inequality.
[tex]3(16) + 3 > 48 \\ 48 + 3 > 48 \\ 51 > 48 \: \: \checkmark[/tex]
The Inequality is true when x > 15.
[tex] \large \boxed{x > 15}[/tex]