We have $\mathit{f}:\mathit{Z}\to \mathit{Z}$ defined by $\mathit{f}\left(\mathit{x}\right)=3\mathit{x}-5$
Let us check if the function is injective
Assume, $\mathit{f}\left({\mathit{x}}_1\right)=\mathit{f}\left({\mathit{x}}_2\right)\forall \mathrm{}{\mathit{x}}_1,{\mathit{x}}_2\in \mathit{Z}$
$\Rightarrow 3{\mathit{x}}_1-5=3{\mathit{x}}_2-5$
$\Rightarrow 3{\mathit{x}}_1=3{\mathit{x}}_2$
$\Rightarrow {\mathit{x}}_1={\mathit{x}}_2$
Thus, f is injective
Now, let us check if the function is surjective.
For f to be surjective, for every $\mathit{y}\in \mathit{Z},$ there must exist on $\mathit{x}\in \mathit{Z}$ such that $\mathit{f}\left(\mathit{x}\right)=\mathit{y}$
y = 3x - 5
$\Rightarrow \mathit{x}=\frac{\mathit{y}+5}{3}$
If y = 1, then $\mathit{x}=\frac{1+5}{3}=2,$ which is an integer.
If y = 2, then $\mathit{x}=\frac{2+5}{3}=\frac{7}{2},$ which is not an integer.
Since, x is not always an integer for every integer y, f is not surjective.
f is not bijective because it is not surjective.
The reason is correct, as a bijective function must be injective and surjective.