equationshorizontal alignmentmath-modesplit
I wrote an equation in LaTeX, and it is centered. How can I left align it but without losing the split? (I still want the equal signs one under each other).
Code:
\begin{equation*}
\begin{split}
a &=(3+6+9+12+...+147)+1125\\
&=3 \cdot (1+2+3+4+...+49)+1125\\
&=3 \cdot \displaystyle\frac{49 \cdot 50}{2} + 1125=49 \cdot 75 + 1125\\
&=3675 + 1125\\
&=4900\\
&=70^2 \Rightarrow \sqrt{a} = \sqrt{70^2}=70 \end{split}
\end{equation*}
use it this way:
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{align}
u(x+\Delta x, t) &= u(x,t) +\frac{\partial u(x,t)}{\partial x}\Delta x +\frac{\partial u(x,t)}{\partial t}(t-t) + \frac12\frac{\partial^2 u(x,t)}{\partial x^2}(\Delta x)^2 \\
& \qquad + \frac{\partial^2u(x,t)}{\partial x\partial y}(t-t)(\Delta x)^2 + \frac12\frac{\partial u(x,t)}{\partial^2t}(t-t)^2 \\
& \qquad +(...)\Delta x^3 +(...)\Delta x^4
\intertext{Seeing powers of t dropping in expansion around $x\pm\Delta x$}
u(x-\Delta x, t) &= u(x,t) -\frac{\partial u(x,t)}{\partial x}\Delta x + \frac12\frac{\partial^2 u(x,t)}{\partial x^2}(\Delta x)^2 -(...)\Delta x^3 +(...)\Delta x^4\\
& \left[\frac12\frac{\partial^2 u(x,t)}{\partial x^2}(\Delta x)^2+ \frac12\frac{\partial^2 u(x,t)}{\partial x^2}(\Delta x)^2\right] \\
& = u(x-\Delta x, t) + u(x+\Delta x, t) - u(x,t) - u(x,t) + O(\Delta x^4)
\intertext{Odd powers of $\Delta x$ cancelling eachother}
\frac{\partial^2 u(x,t)}{\partial x^2} &= \frac{u(x-\Delta x, t) + u(x+\Delta x, t) - 2u(x,t)}{\Delta x^2} + O(\Delta x^2)
\end{align}
\end{document}
A simple approach would be to duplicate the contents of the widest entry (either left or right) on the opposite side, accommodating for some overlap:
\documentclass{article}
\usepackage{lipsum}% http://ctan.org/pkg/lipsum
\usepackage{amsmath}% http://ctan.org/pkg/amsmath
\begin{document}
\lipsum[2]
\begin{align*}
-3z &= -9 \\
z &= 3 \\
-3y-3*3 &= 3 \\
y &= -4 \\
2x+8*(-4)+4*3 &= 2 \\
x &= 11
\end{align*}
\lipsum[2]
\begin{align*}
-3z &= -9 \\
z &= 3 \\
-3y-3*3 &= 3 \\
y &= -4 \\
2x+8*(-4)+4*3 &= \rlap{2}\phantom{2x+8*(-4)+4*3} \\
x &= 11
\end{align*}
\lipsum[2]
\end{document}
Alternatively, if you're not concerned about numbering the equations, then you can use the following construction:
\documentclass{article}
\usepackage{lipsum}% http://ctan.org/pkg/lipsum
\usepackage{amsmath}% http://ctan.org/pkg/amsmath
\usepackage{array}% http://ctan.org/pkg/array
\newsavebox{\mymathbox}
\newcolumntype{R}{>{\begin{lrbox}{\mymathbox}$}r<{$\end{lrbox}\llap{\usebox{\mymathbox}}}}%
\newcolumntype{L}{>{\begin{lrbox}{\mymathbox}$}l<{$\end{lrbox}\rlap{\usebox{\mymathbox}}}}%
\begin{document}
\lipsum[2]
\begin{align*}
-3z &= -9 \\
z &= 3 \\
-3y-3*3 &= 3 \\
y &= -4 \\
2x+8*(-4)+4*3 &= 2 \\
x &= 11
\end{align*}
\lipsum[2]
\[
\renewcommand{\arraystretch}{1.2}
\begin{array}{R@{{}={}}L}
-3z & -9 \\
z & 3 \\
-3y-3*3 & 3 \\
y & -4 \\
2x+8*(-4)+4*3 & 2 \\
x & 11
\end{array} \]
\lipsum[2]
\end{document}
This constructs an align* in the form of an array and places the contents on either side of the equal sign in a zero-width box. Consequently, then equations seem centred based on =.
Note that \rlap and \llap uses \makebox, which inserts its contents in text mode. Hence the switch back to math mode via $ in the automated construction of the L and R column types.
Best Answer
I find no reason for this display to be flush left. Anyway…