\[ \begin{align} \nabla \times \vec{\mathbf{B}} -\, \frac1c\, \frac{\partial\vec{\mathbf{E}}}{\partial t} & = \frac{4\pi}{c}\vec{\mathbf{j}}
\nabla \cdot \vec{\mathbf{E}} & = 4 \pi \rho
\nabla \times \vec{\mathbf{E}}\, +\, \frac1c\, \frac{\partial\vec{\mathbf{B}}}{\partial t} & = \vec{\mathbf{0}}
\nabla \cdot \vec{\mathbf{B}} & = 0 \end{align} \]

something inbetween

\[ \begin{align} \nabla \times \vec{\mathbf{B}} -\, \frac1c\, \frac{\partial\vec{\mathbf{E}}}{\partial t} & = \frac{4\pi}{c}\vec{\mathbf{j}} \ \newline \nabla \cdot \vec{\mathbf{E}} & = 4 \pi \rho \ \newline \nabla \times \vec{\mathbf{E}}\, +\, \frac1c\, \frac{\partial\vec{\mathbf{B}}}{\partial t} & = \vec{\mathbf{0}} \ \newline \nabla \cdot \vec{\mathbf{B}} & = 0 \end{align} \]

Just testing \$tex\$, let’s see if this \(works as \neq \alpha\) expected.

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\).

\[ \frac{1}{(\sqrt{\phi \sqrt{5}}-\phi) e^{\frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\ldots} } } } \]

One more: \(s = a \times t^2 + v \times t + s\)

end of \(\LaTeX\)-test.