$4x+7y-\pi z=0\\2x-y+z=0$, basis for solution space 
Find the dimension over $\mathbb{R}$ of the space of solutions of the following systems of equations. Also find a basis for this space of solutions.
$4x+7y-\pi z=0\\2x-y+z=0$

Using the formula $\text{row rank}+\dim \text{space of solutions}=n$, in which n is the number of variables.
I figured out that $\dim \text{space of solutions}=1$
However I cannot find the basis for the solution.
Questions:
What is the basis of the solution space? Am I doing something wrong?
Thanks in advance!
 A: You have to solve the system :
$$\left\{ \begin{matrix}4x+7y-\pi z = 0 \\ 2x-y+z = 0\end{matrix}\right. \Leftrightarrow
 \left\{\begin{matrix} 4x +7(2x+z)-\pi z=0 \\ y=2x+z\end{matrix}\right. \Leftrightarrow 
\left\{\begin{matrix}18x+(7-\pi)z = 0 \\ y=2x+z  \end{matrix}\right.\Leftrightarrow
\left\{\begin{matrix}x={\pi-7\over 18}z \\ y=2({\pi-7\over18}z)+z = {\pi+2\over9}z\end{matrix}\right. $$
Now that you solved it you can deduce a basis for the solution space for example by choosing $z=1$:
$$\vec{u}=\begin{pmatrix}{\pi -7\over18}\\ {\pi+2\over9} \\ 1\end{pmatrix}$$
As a side note, and for you to well understand what a solution basis is, you have to understand that the solutions of a system of equations live in a space, and you have to find that space. 
Here, by solving the system, we see that both $x$ and $y$ are defined by $z$ so we know for a particular value of $z$ there is only one point $\begin{pmatrix}x\\y\\z\end{pmatrix}$ in $\mathbb{R^3}$ that satisfies the system. So we know the solution space is a line (which corrolates to your $\dim=1$). 
Now since this is a particular type of system with constants $=0$ (I don't know the right term in english), we know the point $\begin{pmatrix}0\\0\\0\end{pmatrix}$ is a solution so all we need is a different point that satisfies the system and this point will also be the coordinates of a vector which is a basis for the solution space. That's why we substitute $z=1$ to find this vector. Hope this helped.
