# Align equals inside case statement

I have the following code:

\begin{equation}
\frac {\partial C}{\partial {z_k}}= \begin{cases}
- \frac {1} {y_i} {y_i} (1-{y_i}) = y_{i} - 1 & k = i \\
\frac {1} {y_i} {y_{i}^{2}} = y_{i}  &  k \ne i  \\
\end{cases}
\end{equation}


It works fine, but what I want to achieve is to align the = from the first case and the second case. I did not manage to do that.

Any advice?

Thank you!

## 2 Answers

I can be done, using aligned instead, but I am not sure which exact alignment is desirable. Probably the best is to use a second cases for the simplification. \documentclass{article}

\usepackage{mathtools}

\begin{document}

\begin{equation}
\frac {\partial C}{\partial {z_k}}
=
\left\{
\begin{aligned}
- \frac {1} {y_i} {y_i} (1-{y_i})&= y_{i} - 1, &\qquad& k = i, \\
\frac {1} {y_i} {y_{i}^{2}}&= y_{i},  &&  k \ne i.
\end{aligned}
\right.
\end{equation}

\begin{equation}
\begin{split}
\frac {\partial C}{\partial {z_k}} &=
\begin{dcases}
- \frac {1} {y_i} {y_i} (1-{y_i}),& k = i, \\
\frac {1} {y_i} {y_{i}^{2}}, & k \ne i,
\end{dcases}\\
&=
\begin{cases}
y_{i} - 1, &k = i, \\
y_{i}, & k \ne i.
\end{cases}
\end{split}
\end{equation}

\end{document}

• Thank you for explaining and for giving me multiple options. I tried both (I was inclined to use the first one due in order to not take so much space) but I agree that the second one looks better, so I used that one. – elaRosca Jun 9 '14 at 10:56

It will suffice to insert \hspace{3.9em} at the beginning of your fourth line (the exact amount of space depends on the fontsize and the font you have chosen.

• Such manual adjustment is generally not advisable. – Andrew Swann Jun 9 '14 at 11:04