I need help in writing this system in LaTeX:       EDIT: I need more help because I am still unable to use the provided solution. My system is:  \begin{empheq}[left=\empheqlbrace]{align} \begin{eqnarray}     \begin{rcases}     \frac{d\phi_{1}(t)}{dt}&amp;= -\phi_{1}(t)-c_{1}\phi_{3}(t),\\     \frac{d\phi_{2}(t)}{dt}&amp;= \alpha_{1}\phi_{3}(t)-(d_{1}+\mu_{1})\phi_{2}(t),\\ \nonumber     \frac{d\phi_{3}(t)}{dt}&amp;= \mu_{1}\phi_{2}(t)-d_1 \phi_{3}(t)-c_{2}\phi_3(t) \tilde{z_2}(t),\\ \nonumber     \frac{d\phi_{4}(t)}{dt}&amp;= \alpha_{2}\phi_{3}(t)\tilde{z_2}(t)-(d_2 + \mu_2) \phi_{4}(t),\\ \nonumber     \frac{d\phi_{5}(t)}{dt}&amp;= \mu_{2}\phi_{4}(t)-d_{2}\phi_{5}(t),\\ \nonumber     \end{rcases} \end{eqnarray} \\ \begin{eqnarray}     \begin{rcases}             \phi_{1}(t^+)&amp;=\phi_{1}(t),\nonumber \\             \phi_{2}(t^+)&amp;=\phi_{2}(t),\nonumber \\             \phi_{3}(t^+)&amp;=(1-\delta) \phi_{3}(t),\nonumber \\             \phi_{4}(t^+)&amp;=\phi_{4}(t),\nonumber \\             \phi_{5}(t^+)&amp;=\phi_{5}(t).\nonumber             \end{rcases} \end{eqnarray} \end{empheq}
Say I had;   \begin{equation}   \begin{cases}      x = 0 \\      y = 0 \\      z = 0     \end{cases}     \qquad \text{and} \qquad     \begin{cases}      x = a \\      y = a \\      z = b     \end{cases}   \end{equation}   How could I insert a brace ( } ) in the second 'cases' to say x and y are equal to a. (i.e. } = a ).   I hope this is clear, if not I can sketch it out - just let me know!  Thanks in advance!
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