# global align parameters across items

I've read questions asking about align across multiple environments, but still am stumped about how I might address this situation.

The following code:

\section*{Solutions to Practice Problems}

\begin{enumerate}
\item Write down the meaning of the Commutative, Associative and Distributive properties.  Create an example
that shows how each one works.\-2em] \begin{align*} \intertext{ \begin{description}[noitemsep, topsep=0pt] \item[Commutative Property:] The order in which terms appear can be changed, but the result remains unchanged. \begin{example*} \end{example*} \end{description} }\\[-2em] &&& \text{(Addition)} & 1,000 + x \ &= \ x + 1,000 &&&\\ &&& \text{(Multiplication)} & ab \ &= \ ba &&& \intertext{ \begin{description}[noitemsep, topsep=0pt] \item[Associative Property:] The order in which terms are \emph{grouped together} can be changed, but the result remains unchanged. \begin{example*} \end{example*} \end{description} }\\[-2em] &&& \text{(Addition)} & (1,000 + 500) + y \ &= \ 1,000 + (500 + y) &&& \\ &&& \text{(Multiplication)} & a(bc) \ &= \ (ab)c &&& \intertext{ \begin{description}[noitemsep, topsep=0pt] \item[Distributive Property:] The product of the sums equals the sum of the products. \begin{example*} \end{example*} \end{description} }\\[-2em] &&& \text{(Numbers only)} & 3(40 - 2) \ &= \ 3 \cdot 40 - 3 \cdot 2 &&& \\ &&& \text{(With variables)} & 5(x - 12) \ &= \ 5x - 60 &&& \\ \end{align*} \item Write down the meaning of the Additive and Multiplicative Identity properties. Create an example that shows how each one works. \begin{align*} \intertext{ \begin{description}[noitemsep, topsep=0pt] \item[Additive Identity:] There exists a number 0 such that addition with 0 and another term leaves the term unchanged. \begin{example*} \end{example*} \end{description} }\\[-2em] &&& \text{(Numbers)} & 1,000 + 0 \ &= \ 1,000 &&&\\ &&& \text{(Variables)} & x + 0 \ &= \ x &&& \intertext{ \begin{description}[noitemsep, topsep=0pt] \item[Multiplicative Identity:] There exists a number 1 such that multiplication with 1 and another term leaves the term unchanged. \begin{example*} \end{example*} \end{description} }\\[-2em] &&& \text{(Numbers)} & 12 \cdot 1 \ &= \ 12 &&& \\ &&& \text{(Variables)} & x \cdot 1 \ &= \ x &&& \end{align*}  produces this result: I can fudge this enough to align equations within each item, but not across items. What I'm looking for is uniformity with left aligning the (Numbers) text across every item, and ideally aligning every equation at the = sign across items. The closest question I found to this is here. I am looking for only a slight modification to what I have here, if possible. If not, then are there ways to set "global" align parameters throughout a document that can be across multiple align environments? ## 1 Answer For code readability you may be better off to set the content using a specially-formed array: \documentclass{article} \usepackage[margin=1in]{geometry}% Just for this example \usepackage{amsmath,array,enumitem} \newlength{\LHS} \newlength{\RHS} \newenvironment{bigalign} {\[% http://tex.stackexchange.com/q/31672/5764 %\renewcommand{\arraystretch}{1.5}% Stretch array vertically \begin{array}{p{7em}>{\raggedleft}p{\LHS}<{}@{}>{{}}p{\RHS}<{}}} {\end{array}}
\begin{document}

\begin{enumerate}
\settowidth{\LHS}{$(1\,000 + 500) + y$}% Longest left-hand side
\settowidth{\RHS}{${}= 1\,000 + (500 + y)$}% Longest right-hand side
\item Write down the meaning of the Commutative, Associative and Distributive properties.  Create an example
that shows how each one works.

\begin{description}[noitemsep, topsep=0pt]
\item[Commutative Property:] The order in which terms appear can be changed, but the
result remains unchanged.

\textbf{Example.}
\begin{bigalign}
\text{(Addition)}       &         1\,000 + x &= x + 1\,000 \\
\text{(Multiplication)} &                 ab &= ba
\end{bigalign}

\item[Associative Property:] The order in which terms are \emph{grouped together} can be changed,
but the result remains unchanged.

\textbf{Example.}
\begin{bigalign}
\text{(Addition)}       & (1\,000 + 500) + y &= 1\,000 + (500 + y) \\
\text{(Multiplication)} &              a(bc) &= (ab)c
\end{bigalign}

\item[Distributive Property:] The product of the sums equals the sum of the products.

\textbf{Example.}
\begin{bigalign}
\text{(Numbers only)}   &          3(40 - 2) &= 3 \cdot 40 - 3 \cdot 2 \\
\text{(With variables)} &          5(x - 12) &= 5x - 60 \\
\end{bigalign}

\end{description}

\item Write down the meaning of the Additive and Multiplicative Identity properties. Create an example that
shows how each one works.

\begin{description}[noitemsep, topsep=0pt]
\item[Additive Identity:]  There exists a number 0 such that addition with 0 and another term leaves the
term unchanged.

\textbf{Example.}
\begin{bigalign}
\text{(Numbers)}        &         1\,000 + 0 &= 1\,000 \\
\text{(Variables)}      &              x + 0 &= x
\end{bigalign}

\item[Multiplicative Identity:] There exists a number 1 such that multiplication with 1 and another term
leaves the term unchanged.

\textbf{Example.}
\begin{bigalign}
\text{(Numbers)}        &         12 \cdot 1 &= 12 \\
\text{(Variables)}      &          x \cdot 1 &= x
\end{bigalign}

\end{description}
\end{enumerate}

\end{document}


The array (hidden inside bigalign) uses a fixed-width columns for each of the components. The first column is the left-most notation. I chose 7em since it provided a nice gap between the notation and the other content. The second column is also a fixed-width column, just like the third, but measured to the widest element across the alignments (the \settowidth commands). On the left-hand side, the widest element is (1\,000 + 500) + y, while the widest element on the right-hand side is = 1\,000 + (500 + y).

You may want to increase \arraystretch to suit your needs. I've kept it at 1 for the example.

• Thank you this is perfect! Incidentally, I was using 'array' at first, which had much better vertical spacing, but then ran across alignment problems between items. Thank you so much again for the clarification. – Abbas Jaffary Feb 23 '15 at 17:48