# How to align my equation to left?

I want to display four equations which have different length. I want they align left like this.

But my code give this

The following are my code:

\begin{aligned} p(a) &=p(a) p(a \rightarrow a)+p(b) p(b \rightarrow a)+p(c) p(c \rightarrow a)+p(d) p(d \rightarrow a) \\ &={\scriptstyle \frac{1}{2} \frac{2}{3}+\frac{1}{4} \frac{1}{3}+\frac{1}{3}+\frac{1}{3}=\frac{1}{2}} \end{aligned}
\begin{aligned} p(b) &=p(a) p(a \rightarrow b)+p(b) p(b \rightarrow b)+p(c) p(c \rightarrow b) \\ &={\scriptstyle\frac{1}{2} \frac{1}{6}+\frac{1}{4} \frac{1}{2}+\frac{1}{3} \frac{1}{3}=\frac{1}{4}} \end{aligned}
\begin{aligned} p(c) &=p(a) p(a \rightarrow c)+p(b) p(b \rightarrow c)+p(c) p(c \rightarrow c)+p(d) p(d \rightarrow c) \\ &={\scriptstyle\frac{1}{2} \frac{1}{12}+\frac{1}{4} \frac{1}{6}+\frac{1}{8} 0+\frac{1}{8}=\frac{1}{8}} \end{aligned}
\begin{aligned} p(d) &=p(a) p(a \rightarrow d)+p(c) p(c \rightarrow d)+p(d) p(d \rightarrow d) \\ &={\scriptstyle\frac{1}{2} \frac{1}{12}+\frac{1}{8} \frac{1}{3}+\frac{1}{8} \frac{1}{3}=\frac{1}{3}} \end{aligned}


Can anyone tell me what I can do?

• never use  in latex, use \[ and then the fleqn option will left align equations. – David Carlisle Mar 22 at 8:45 ## 2 Answers Here is a proposal \documentclass{article} \usepackage{amsmath} \begin{document} \begin{align*} p(a)&=p(a) p(a \rightarrow a)+p(b) p(b \rightarrow a)+p(c) p(c \rightarrow a)+p(d) p(d \rightarrow a)\\ &\;={\scriptstyle \frac{1}{2} \frac{2}{3}+\frac{1}{4} \frac{1}{3}+\frac{1}{3}+\frac{1}{3}=\frac{1}{2}}\\ p(b)&=p(a) p(a \rightarrow b)+p(b) p(b \rightarrow b)+p(c) p(c \rightarrow b)\\ &\;={\scriptstyle\frac{1}{2} \frac{1}{6}+\frac{1}{4} \frac{1}{2}+\frac{1}{3} \frac{1}{3}=\frac{1}{4}}\\ p(c)&=p(a) p(a \rightarrow c)+p(b) p(b \rightarrow c)+p(c) p(c \rightarrow c)+p(d) p(d \rightarrow c)\\ &\;={\scriptstyle\frac{1}{2} \frac{1}{12}+\frac{1}{4} \frac{1}{6}+\frac{1}{8} 0+\frac{1}{8}=\frac{1}{8}}\\ p(d)&=p(a) p(a \rightarrow d)+p(c) p(c \rightarrow d)+p(d) p(d \rightarrow d)\\ &\;={\scriptstyle\frac{1}{2} \frac{1}{12}+\frac{1}{8} \frac{1}{3}+\frac{1}{8} \frac{1}{3}=\frac{1}{3}} \end{align*} \end{document}  Why do you use \scriptstyle? It is very hard to read the fractions! Use \tfrac instead! You should remove \; in the &\;={\scriptstyle... lines to align the =s. I prefer that way. I strongly recommend this one \documentclass{article} \usepackage{amsmath} \begin{document} \begin{align*} p(a)&=p(a) p(a \rightarrow a)+p(b) p(b \rightarrow a)+p(c) p(c \rightarrow a)+p(d) p(d \rightarrow a)\\ &=\tfrac{1}{2} \tfrac{2}{3}+\tfrac{1}{4} \tfrac{1}{3}+\tfrac{1}{3}+\tfrac{1}{3}=\tfrac{1}{2}\\ p(b)&=p(a) p(a \rightarrow b)+p(b) p(b \rightarrow b)+p(c) p(c \rightarrow b)\\ &=\tfrac{1}{2} \tfrac{1}{6}+\tfrac{1}{4} \tfrac{1}{2}+\tfrac{1}{3} \tfrac{1}{3}=\tfrac{1}{4}\\ p(c)&=p(a) p(a \rightarrow c)+p(b) p(b \rightarrow c)+p(c) p(c \rightarrow c)+p(d) p(d \rightarrow c)\\ &=\tfrac{1}{2} \tfrac{1}{12}+\tfrac{1}{4} \tfrac{1}{6}+\tfrac{1}{8} 0+\tfrac{1}{8}=\tfrac{1}{8}\\ p(d)&=p(a) p(a \rightarrow d)+p(c) p(c \rightarrow d)+p(d) p(d \rightarrow d)\\ &=\tfrac{1}{2} \tfrac{1}{12}+\tfrac{1}{8} \tfrac{1}{3}+\tfrac{1}{8} \tfrac{1}{3}=\tfrac{1}{3} \end{align*} \end{document}  • Thank you very much. I use '\scriptstyle' because I want the second line of the equation to be small. – chole Mar 22 at 8:55 • @chole Then you should use \tfrac instead of \scriptstyle\frac. – JouleV Mar 22 at 8:58 • +1 for recommending \tfrac in place of \scriptstyle\frac. – Mico Mar 22 at 9:00 • @chole If my answer helps you, please mark your question as resolved by clicking the checkmark on the left of my answer. – JouleV Mar 22 at 9:02 • +1 yes as I commented under the question the direct answer why there was no left alignment was due to using  but If I had answered I would also have suggested align* rather than \[\begin{aligned} :-) – David Carlisle Mar 22 at 9:08

I propose this layout, using the fleqn environment and, as in my opinion, medium sized fractions will look best, I also use the \medmath and \mfrac commands, each from nccmath:

\documentclass{article}
\usepackage{amsmath, nccmath}

\begin{document}

\begin{fleqn}
\begin{align*}
p(a)&=p(a) p(a \rightarrow a)+p(b) p(b \rightarrow a)+p(c) p(c \rightarrow a)+p(d) p(d \rightarrow a)\\
p(b)&=p(a) p(a \rightarrow b)+p(b) p(b \rightarrow b)+p(c) p(c \rightarrow b)\\
&\quad= \medmath{\frac{1}{2}\cdot \frac{1}{6}+\frac{1}{4}\cdot \frac{1}{2}+\frac{1}{3}\cdot \frac{1}{3}}=\mfrac{1}{4} \\[1ex]
p(c)&=p(a) p(a \rightarrow c)+p(b) p(b \rightarrow c)+p(c) p(c \rightarrow c)+p(d) p(d \rightarrow c)\\
&\quad= \medmath{\frac{1}{2}\cdot \frac{1}{12}+\frac{1}{4}\cdot \frac{1}{6}+\frac{1}{8}\,0 + \frac{1}{8}}=\mfrac{1}{8} \\[1ex]
p(d)&=p(a) p(a \rightarrow d)+p(c) p(c \rightarrow d)+p(d) p(d \rightarrow d)\\
&\quad= \medmath{\frac{1}{2}\cdot \frac{1}{12}+\frac{1}{8}\cdot \frac{1}{3}+\frac{1}{8\cdot } \frac{1}{3}}=\mfrac{1}{3}
\end{align*}
\end{fleqn}

\end{document}