System of equation involving tensor I have to solve a system of equation involving tensor:
\begin{align}
\underline{\underline{a_1}}\cdot\underline{x} + \underline{\underline{\underline{\underline{b_1}}}} \therefore \underline{\underline{\underline{y}}} = \underline{c_1} \\
\underline{\underline{a_2}}\cdot\underline{x} + \underline{\underline{\underline{\underline{b_2}}}} \therefore \underline{\underline{\underline{y}}} = \underline{c_2}
\end{align}
where $\underline{\cdot}, \underline{\underline{\cdot}}, \underline{\underline{\underline{\cdot}}}, \underline{\underline{\underline{\underline{\cdot}}}}$ hold for first, second, third and fourth-rank tensor. The $(\cdot)$ and $(\therefore)$ hold for simple and triple tensor contraction. I need to find out $\underline{x}$ and $\underline{\underline{\underline{y}}}$. Suppose that $\underline{\underline{\underline{y}}}$ has $n$ independent components.
What are common technique to solve this system of equation?
 A: You say $\underline{\underline{\underline y}}$ has $n$ independent components. If its dependence on these components is linear then we can write $\underline{\underline{\underline y}}=\underline{\underline{\underline{\underline d}}}\cdot\underline{y'}$ for some $\underline{\underline{\underline{\underline d}}}$ and  $n$ component vector $\underline{y'}$.
Then note $\underline{\underline{\underline{\underline{b_i}}}} \therefore \underline{\underline{\underline{y}}} =\underline{\underline{\underline{\underline{b_i}}}} \therefore \left(\underline{\underline{\underline{\underline d}}}\cdot\underline{y'}\right) =\left( \underline{\underline{\underline{\underline{b_i}}}} \therefore \underline{\underline{\underline{\underline d}}}\right)\cdot\underline{y'}$, and so let $\underline{\underline{b_i'}}=\underline{\underline{\underline{\underline{b_i}}}} \therefore \underline{\underline{\underline{\underline d}}}$ for $i=1,2$.
Then our system of equations becomes
$$
\begin{align}
\underline{\underline{a_1}}\cdot\underline{x} + \underline{\underline{b_1'}} \cdot\underline{y'} = \underline{c_1} \\
\underline{\underline{a_2}}\cdot\underline{x} + \underline{\underline{b_2'}} \cdot\underline{y'} = \underline{c_2}
\end{align}
$$
which we can write in block matrices as
$$
\begin{pmatrix}
a_1&b_1'\\
a_2&b_2'\\
\end{pmatrix}
\begin{pmatrix}
x\\
y'\\
\end{pmatrix}
=
\begin{pmatrix}
c_1\\
c_2\\
\end{pmatrix}
$$
and so this can be solved by standard matrix inversion.
We can use a similar solution if $\underline{\underline{\underline y}}$ depends affinely on its parameters, i.e. if $\underline{\underline{\underline y}}=\underline{\underline{\underline{\underline d}}}\cdot\underline{y'}+\underline{\underline{\underline {y_0}}}$. But if $\underline{\underline{\underline y}}$ depends on its parameters in a completely non-linear way then we shouldn't hope for a solution from linear algebra.
N.B. It seems to me that index notation is better than this ridiculous lines-and-dots thing, and you should learn it if you haven't already.
