I've got $\space$ $V$ $K$ - vectorial space, and $H$ which is a subspace of $V$. We say that $H$ is a hyperplane when $dimH=n-1$. If we've got $\space a_1,a_2,...,a_n\in{\mathbb{R}}$ which are not all of them null.
We have to prove that $H=${$(x_1,...,x_n)\in{\mathbb{R^n}}|a_1x_1+...+a_nx_n=0$} is an hyperplane of $\mathbb{R^n}$.
What I understand is that what we have to prove is that the dimension of $H$ is $n-1$. As I know that $H$ is a subspace of $V$ and the dimension of $V$ is n, we know that $dimH\le{n}$. And know, I suppose that I would have to deduce something else owing to the fact that $\space a_1,a_2,...,a_n\in{\mathbb{R}}$ are not all of them null. But I don't know which information to take from there...
If you don't know what the rank-nullity theorem is or don't want to use kernels or linear maps.
Let $ H = \{(x_1,x_2,...,x_n) \in V : a_1x_1 + a_2x_2 + ...a_nx_n = 0.\}$ where at least one of the $a_i$'s is non zero WLOG consider $a_1 \neq 0$ then the equation defining $H$ can be rewritten as $$ x_1 = \frac{-1}{a_1}\sum_{k=2}^na_ix_i.$$
Therefore $\forall (x_1,...,x_n) \in H$:
\begin{align*} (x_1,x_2,...,x_n) &= \left(\frac{-1}{a_1}\sum_{k=2}^na_ix_i ,\; x_2\,...,x_n \right) \\ &= x_2(-a_2/a_1,1,0,...,0) + x_3(-a_3/a_1,0,1,0,...,0) + ... + x_n(-a_n/a_1,0,...,0,1).\\ \end{align*}
This means $H$ is generated by the vectors $\frac{-a_i}{a_1}e_1 + e_i$ for $i = 2,3,...n.$ Since these vectors are also a linearly independent they form a basis of $H$ with $n-1$ elements. Therefore $\dim H = n-1.$