You may forget about $a$. So you have to prove that for every $d$ there exists a $k$ such that if $x>k$, then $f(x)=2x+3>d$. So consider an arbitrary $d$. You need to show that there exists a $k$ such that $x>k$ implies $x>(d-3)/2$. It's sufficient to take $k=(d-3)/2$.