Linear combination on $\mathbb R_{1} \left[ x \right ]$ Given the canonical basis of $\mathbb R_{1} [ x ]$,  $\{x,1\}$, I tryed to write each vector as a linear combination of the basis $B=\{3x+6,2x+10\}$.

So, I started by $x$.

$x=\lambda_{1}(3x+6)+\lambda_{2}(2x+10)=(3\lambda_{1}+2\lambda_{2})x+6\lambda_{1}+10\lambda_{2}$

Then I solved the system by a matrix,

$\begin{bmatrix}
 3&amp;2\\6&amp;10
\end{bmatrix}$
$\begin{bmatrix}
 \lambda_{1}\\\lambda_{2}
\end{bmatrix}$=
$\begin{bmatrix}
 1\\0
\end{bmatrix}$

The result was $\lambda_{1}=\frac{5}{9}$ and $\lambda_{2}=-\frac{1}{3}$.

But now I don't know how to define the coordinates of $x$ on the basis $B$. Thanks for the help
If I understand your question properly, then $x$ written in the basis $B$ simply has the form
$$ \begin{pmatrix} \frac{5}{9} &amp; -\frac{1}{3} \end{pmatrix}. $$
This is analogous to the fact that the basis vectors of $B$ have the form
$$ \begin{pmatrix} 3 &amp; 6 \end{pmatrix} \quad\text{and}\quad \begin{pmatrix} 2 &amp; 10 \end{pmatrix} $$
when written in the canonical base $(\,x,\, 1\,)$.
The coordinates of $x$ should be your $\lambda_1, \lambda_2$. You have already changed its basis to be written in terms of vectors in $B$ rather than vectors in the standard basis. 

$$
[x]_{\{x,1\}} = (1, 0) \\
[x]_{B} = (\lambda_1, \lambda_2)
$$