A3. For the decreasing linear function $f(x)=k \cdot x+n$, it holds that $f(x+y)=f(x)+f(y)-3$ and $f(k)=2 f(1)$, for any $x, y \in \mathbb{R}$. What is the value of $f(-1)$?
(A) 4
(B) 0
(C) 2
(D) -2
(E) -4
A3. $f(x+y)=k(x+y)+n=k x+k y+n$ in $f(x)+f(y)-3=k x+n+k y+n-3$. From this we get $n=3$. From $f(k)=2 f(1)$ we get the equation $k \cdot k+3=2 k+6$. We rearrange it to $k^{2}-2 k-3=0$. The solution 3 does not fit. For the solution -1 we get $f(x)=-x+3$. Therefore, $f(-1)=4$. The correct answer is $A$.