Prove that $\forall t \in \mathbb R$ the set $f^{-1}(\{t\})$ is a hyperplane of $X$ Exercise :

Let $X$ be a vector space and $f:X \to \mathbb R$ be a linear functional. Show that for all $t \in \mathbb R$, the set $f^{-1}(\{t\})$ is a hyperplane of $X$.

Attempt :
I have proved a slightly different example, showing that if $W$ is a hyperplane of $X$ then there exists a linear functional such that $W=f^{-1}(\{t\})$. This was carried out by using the trick and setting $f(x) = f(\lambda x + y)= \lambda$, since we just needed to show that there exists some linear functional that would fullfill the given condition for some $t \in \mathbb R$
The case in this exercise though, is different, since we need to generally prove that for any linear functional and all $t \in \mathbb R$ the hyperplane condition holds.
Essentialy what I need to prove is that $f^{-1}(\{t\})$ is a subspace of $X$ of $\text{co}\dim=1$ which then means that it is a hyperplane. Or, to prove that every element in this image can be written as $x = \lambda x_0 + y$ with $x_0 \notin Y$.
Question - Request : I can't see how to proceed proving the fact above though, as the only $\text{co}\dim$ statement that I recall is the kernel one. I would really appreciate any tips, hints or elaboration to help me work over this exercise and understand it.

Assume that $f$ is nonzero. Then $f$ must be surjective, so $f^{-1}(t)$ is non empty. Pick any $v\in f^{-1}(t)$. Then we have $f^{-1}(t)=v + \ker f$: Clearly the right hand side is contained in the left hand side. Conversely, for any $w\in f^{-1}(t)$ we have $v-w \in \ker f$ by linearity of $f$, hence the left hand side is contained in the right hand side. Therefore $f^{-1}(t)$ is an affine translation of the kernel and you can apply the dimension theorem you mentioned.
