Suppose we have a system $S$ of homogeneous linear equations in $n$ variables, and an extra homogeneous equation $E$ in $n$ variables. If $E$ a linear combination of the equations in $S$, then it's a logical consequence of them, in particular it gives no further information in trying to work out the values of the variables. Is the converse true? That is, if $E$ is linearly independent of the equations of $S$, does $E$ automatically give extra information on the values of the variables?
More precisely, if $E$ is not a linear combination of the equations in $S$, does that imply that the solution space of $E$ does not contain the solution space of $S$ (so that the implication "any solution of $S$ is a solution of $E$" fails)?
Yes, it is true. To be more precise: If $V$ is a $n$-dimensional vector space over a field $F$, if $E$ is a vector space of linear of linear maps form $V$ into $F$ and if $\alpha$ is linear form from $V$ into $F$ which does not belong to $E$, then there's some element of $\ker\alpha$ which is does not belong to $\bigcap_{\beta\in E}\ker\beta$.
In order to prove it, let $\beta_1,\ldots,\beta_k$ be a basis if $E$. Then$$\bigcap_{\beta\in E}\ker\beta=\bigcap_{j=1}^k\ker\beta_j$$and therefore$$\dim\bigcap_{\beta\in E}\ker\beta=\dim\bigcap_{j=1}^k\ker\beta_j=n-k.$$If you add $\beta$ to the set $\{\beta_1,\ldots,\beta_k\}$, the dimension of the intersection of the kernels will be $n-k-1$, which is strictly smaller than $n-k$. Therefore, not all elements of $\ker\beta$ belong to $\bigcap_{j=1}^k\ker\beta_j$.