Let $x_{1}$ and $x_{2}$ be any two elements of $N$.
$\therefore f\left(x_{1}\right)= f\left(x_{2}\right)$
$\Rightarrow 2 x_{1}+3=2 x_{2}+3$
$\Rightarrow x_{1}=x_{2}$
$\therefore f$ is one-one function.
Now, let $y=2 x+3 \Rightarrow x=\frac{y-3}{2}$
It is not onto function, because for $y=3, x=0 \notin N$.
Hence, it is one-one into function.