# How to align terms in separate lines in this case?

I want to get another variant for the alignment points in my question answered by Hendrik Vogt, see How to construct a long equation that is split in LHS and RHS to occupy a narrow column?.

\documentclass[twocolumn]{article}
\usepackage[a4paper,margin=1cm]{geometry}
\usepackage{amsmath}
\begin{document}
\begin{enumerate}
\item The degree of (C) is 3.
\item The degree of (A) is 1.
\item%
\! \begin{aligned}[t] -3x(x+1)&-2x(x-1)\\ &+4(x^2-3x-1) &= \!\begin{aligned}[t] &-3x^2-3x-2x^2+2x+v \\ &+4x^2-12x-4 \end{aligned} \\ &= \!\begin{aligned}[t] -3x^2-&3x-2x^2+2x+v \\ &+4x^2-12x-4 \end{aligned} \\ &= -x^2-13x-4 \end{aligned}
\end{enumerate}
\end{document}


The code above does not produce what I want to achieve as follows.

Note: Feel free to edit the title to be better because I have difficulty to rephrase it.

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...and you're open for alternatives? – Werner Jan 30 '13 at 5:38
@Werner: Of course but being inside the enumerate is a must. – kiss my armpit Jan 30 '13 at 5:50
Alignment of the - and + on the first RHS is incorrect. The first - is a unary operator, while + is binary. – Werner Jan 30 '13 at 5:54

Since you have multiple alignment points, could use the alignedat:

## Notes:

• I used \rlap on the two lines so it won't effect the other alignment points.
• The \hphantom{{}={}} was used to ensure that space was left equivalent to the =. The additional {} on either side of the equal ensures that the = is treated as a relational operator.
• The alignedat (similar to alignat) requires information as to the number of columns. The rule of thumb is to use half the of 1 + maximum number of &s in any row (rounded up in case of fractional result). I usually just use a number that is large enough so that there are not syntax errors. Have not yet encountered a problem with using a number larger that required. But in this case it appears that 3 is enough.

## Code:

\documentclass[twocolumn]{article}
\usepackage[a4paper,margin=1cm]{geometry}
\usepackage{amsmath}
\begin{document}
\begin{enumerate}
\item The degree of (C) is 3.
\item The degree of (A) is 1.
\item%
\! \begin{alignedat}[t]{4} -3x(x+1)&-2x(x-1)\\ &+4(x^2-3x-1) &&=\rlap{-3x^2-3x-2x^2+2x+v$} \\ & &&\hphantom{{}={}}{+}4x^2&&-12x-4\\ & &&=-3x^2 &&-3x-2x^2+2x+v \\ & && &&+4x^2-12x-4 \\ & &&=\rlap{$-x^2-13x-4} \end{alignedat}
\end{enumerate}
\end{document}

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Note that the spacing in the first +4x^2 on the right is not correct. – Hendrik Vogt Jan 30 '13 at 9:20
@HendrikVogt: Good point. I should have known that a + is not usually a unary operator. However, having the correct binary operator spacing around the operator has other consequences based on the requirement that the + and - signs be aligned as per OP's request. – Peter Grill Jan 30 '13 at 18:45
@GarbageCollector: Have added some info to address your question. – Peter Grill Jan 31 '13 at 0:22

An array provides rudimentary control over the display and alignment:

\documentclass[twocolumn]{article}
%\usepackage[a4paper,margin=1cm]{geometry}% http://ctan.org/pkg/geometry
%\usepackage{amsmath}% http://ctan.org/pkg/amsmath
\begin{document}
\begin{enumerate}
\item The degree of (C) is 3.
\item The degree of (A) is 1.
\item $\renewcommand{\arraystretch}{1.2} \begin{array}[t]{@{}l@{}l} -3x(x+1)-2x(x-1) \\ \phantom{-3x(x+1)}+4(x^2-3x-1) &{}= -3x^2-3x-2x^2+2x+v \\ &\phantom{{}={}}{}+4x^2-12x-4 \\ &{}=-3x^2-3x-2x^2+2x+v \\ &\phantom{{}=-3x^2}+4x^2-12x-4 \\ &{}= -x^2-13x-4 \end{array}$
\end{enumerate}
\end{document}

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