# Adding labels to the equations when they are broken

\begin{align}
\begin{split}
\pd{EU}{\alpha} &= d(\theta) U'(\pi_{1}) [G_{\alpha}(\alpha, \theta, \epsilon) - \widebar{P} C_{\alpha}(\alpha, \theta)] \\
&+ (1-d(\theta)) U'(\pi_{0}) [G_{\alpha}(\alpha, \theta, 0) -  C_{\alpha}(\alpha, \theta)] = 0,\\
\pd{EU}{\theta} &= h'(\theta)[U(\pi_{0}) - U(\pi_{1})] \\
&+ d(\theta) U'(\pi_{1})[G_{\theta}(\alpha, \theta, \epsilon) - \widebar{P} C_{\theta}(\alpha, \theta)] \\
&+(1-d(\theta))U'(\pi_{0})[G_{\theta}(\alpha, \theta, 0) -  1] = 0.\\
\end{split}
\end{align}


I want to label the equations 4 and 5 and then reference them. Here I only have the two equations labelled as one.

EDIT:

From the answer below, I was able to label the equations:

\begin{align}
\begin{split}
\pd{EU}{\alpha} &= d(\theta) U'(\pi_{1}) [G_{\alpha}(\alpha,   \theta, \epsilon) - \widebar{P} C_{\alpha}(\alpha, \theta)] \\
&+ (1-d(\theta)) U'(\pi_{0}) [G_{\alpha}(\alpha, \theta, 0) - C_{\alpha}(\alpha, \theta)]=0
\end{split}\label{eqn:4}\\
\begin{split}
\pd{EU}{\theta} &= h'(\theta)[U(\pi_{0}) - U(\pi_{1})] \\
&+ d(\theta) U'(\pi_{1})[G_{\theta}(\alpha, \theta, \epsilon) - \widebar{P} C_{\theta}(\alpha, \theta)] \\
&+(1-d(\theta))U'(\pi_{0})[G_{\theta}(\alpha, \theta, 0) -  1] = 0.\\
\end{split}\label{eqn:5}\
\end{align}


However, now the two equations are not aligned with each other, and I want them to be aligned. Any suggestions?

Use \nonumber (or \notag) to avoid numbering a specific equation inside an align, otherwise it will be numbered (and you can \label-\ref it):

\documentclass{article}
\usepackage{amsmath,mathabx}
\newcommand{\pd}[2]{\frac{\partial #1}{\partial #2}}
\begin{document}

\begin{align}
\pd{EU}{\alpha} &= d(\theta) U'(\pi_{1}) \bigl[G_{\alpha}(\alpha, \theta, \epsilon) - \widebar{P} C_{\alpha}(\alpha, \theta)\bigr] \nonumber \\
&\hphantom{=} + (1-d(\theta)) U'(\pi_{0}) [G_{\alpha}(\alpha, \theta, 0) -  C_{\alpha}(\alpha, \theta)] = 0, \label{eq:first} \\
\pd{EU}{\theta} &= h'(\theta)\bigl[ U(\pi_{0}) - U(\pi_{1})\bigr] \nonumber \\
&\hphantom{=} + d(\theta) U'(\pi_{1})\bigl[G_{\theta}(\alpha, \theta, \epsilon) - \widebar{P} C_{\theta}(\alpha, \theta)\bigr] \nonumber \\
&\hphantom{=} + (1-d(\theta))U'(\pi_{0})\bigl[G_{\theta}(\alpha, \theta, 0) -  1\bigr] = 0. \label{eq:second}
\end{align}
Consider reviewing~\eqref{eq:first} and~\eqref{eq:second}.

\end{document}

• To avoid the slightly disconcerting look of having a wider space between the first and second line than between the second and third line of each subequation, you might want to consider "wrapping" the large partial derivative fractional terms at the start of each subequation in a \smash[b]{...} directive. – Mico May 25 '15 at 9:15

here's an approach that should give you the result you're looking for.

\documentclass{article}
\usepackage{amsmath}
\newcommand{\pd}[2]{\frac{\partial {#1}}{\partial {#2}}}
\begin{document}
\setcounter{equation}{3}
\begin{align}
\begin{split}\smash[b]{\pd{EU}{\alpha}}
&= d(\theta) U'(\pi_{1})
[G_{\alpha}(\alpha, \theta, \epsilon) - \bar{P} C_{\alpha}(\alpha, \theta)] \\
[G_{\alpha}(\alpha, \theta, 0) - C_{\alpha}(\alpha, \theta)] = 0,
\end{split}\label{aa}\\
\begin{split}\smash[b]{\pd{EU}{\theta}}
&= h'(\theta)[U(\pi_{0}) - U(\pi_{1})] \\
[G_{\theta}(\alpha, \theta, \epsilon) -  \bar{P} C_{\theta}(\alpha, \theta)] \\
[G_{\theta}(\alpha, \theta, 0) - 1] = 0.
\end{split}\label{bb}
\end{align}

some text \eqref{aa} some more text \eqref{bb}

\end{document}


there are a couple of things to note:

• the fractions would have spread the space between the first and second line of each "split" group, so i applied \smash to the bottom [b]; the wider space between the two groups is appropriate, but since there's nothing at the right with a height and depth of more than a single line, the result (to my eyes) looks better this way.

• the fraction is the only part of each group that is to the left of the = sign. all the succeeding lines are indented by a \quad to make the structure clearer.

Is this what you want:

\documentclass{article}
\usepackage{amsmath}
\newcommand{\pd}[2]{\frac{\partial {#1}}{\partial {#2}}}
\begin{document}

\begin{align}
\begin{split} \pd{EU}{\alpha} &= d(\theta) U'(\pi_{1}) [G_{\alpha}(\alpha,   \theta, \epsilon) - \bar{P} C_{\alpha}(\alpha, \theta)] \\
&\quad+ (1-d(\theta)) U'(\pi_{0}) [G_{\alpha}(\alpha, \theta, 0) - C_{\alpha}(\alpha, \theta)] = 0,\end{split}\label{aa}\\
\begin{split}\pd{EU}{\theta} &= h'(\theta)[U(\pi_{0}) - U(\pi_{1})] \\
&\quad+ d(\theta) U'(\pi_{1})[G_{\theta}(\alpha, \theta, \epsilon) -  \bar{P} C_{\theta}(\alpha, \theta)] \\ &\quad+(1-d(\theta))U'(\pi_{0})[G_{\theta}(\alpha, \theta, 0) - 1] = 0. \end{split}\label{bb}
\end{align}

some text \eqref{aa} some more text \eqref{bb}

\end{document}