Given any two non zero vectors $x,y$ does there exist symmetric matrix $A$ such that $y = Ax$? 
Let $x,y$ be non-zero vectors from $\mathbb{R}^n$. Is it true, that there exists symmetric matrix $A$ such that $y = Ax$.

I was reasoning the following way. Having an equation $y = Ax$ for some symmetric matrix $A$ is equivalent to being able to solve a system of $n + \frac{n^2-n}{2}$ linear equations:
$$\begin{cases} a_{11}x_1 + ... + a_{1n}x_n = y_1 \\ \vdots \\ a_{n1}x_1 + ... + a_{nn}x_n = y_n \\ a_{12} = a_{21} \\ \vdots \\ a_{n,n-1} = a_{n-1,n} \end{cases}$$
First $n$ equations are our constraints for identity $y = Ax$ to be true and the following $\frac{n^2-n}{2}$ for symmetricity of $A$. So, in our system we have $n^2$ variables and $\frac{n^2-n}{2} + n = \frac{n^2+n}{2}$ equations. As $n^2 \geq \frac{n^2+n}{2}$ such system is always solvable. That's why such matrix $A$ always exists.
This solution was marked as wrong on exam, where did I make a mistake?

Note that your system is not homogeneous (since $y \neq 0$). Given a non-homogeneous system of $k$ equations in $n \geq k$ unknowns, it is not always true that this system has a solution. For example, the system
$$ 0 \cdot x + 0 \cdot y = 1 $$
clearly doesn't have a solution. In general, a system of $k$ equations in $n$ unknowns defines a linear map $B \colon \mathbb{F}^n \rightarrow \mathbb{F}^k$. If $n > k$ then this map cannot be one-to-one so you know that the homogeneous system of equations $Bx = 0$ always has a non-trivial solution $x \neq 0$. However, you can't tell automatically that $Bx = y$ has a solution because $y$ might not belong to the image of $B$.
