# how to align the following set of equations and have one equation number?

I have two equations. One is big and the other is small. I want to align them and want these two equations to have only one equation number. Another question is : Are there ways to make these equations look more appealing?

\documentclass{article}

\usepackage{amsmath,mathrsfs,xcolor,mathtools}

\begin{document}

$$\label{eq:lmlt} \begin{gathered} t_0 = 0, \quad s_0 = t_0+t_1, \quad t_{n+1} = s_n + \dfrac{M(1+M(s_n-t_n)) (s_n-t_n)^2}{2(1-M_0t_n)^5},\\ s_{n+m} = t_{n+m-1}+ \dfrac{1}{1-M_0t_{n+1}}\big[\dfrac{M(t_{n+1}-s_n)^2}{2} + \dfrac{13L(s_n-t_n)^4}{108} +\dfrac{{\color{red}N}M(s_n-t_n)^4}{9(1-M_0t_n)}\\ +\dfrac{M^3(s_n-t_n)^4}{3(1-M_0t_n)^2}\big] \end{gathered}$$

\end{document}

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which uses the aligned environment- note that using \big[ doesn't give the correct size of []; we'll fix that in the next solution below.

\subsection*{Original}
\label{eq:lmlt} \begin{gathered} t_0 = 0, \quad s_0 = t_0+t_1\\ \begin{aligned} t_{n+1} & = s_n + \dfrac{M(1+M(s_n-t_n)) (s_n-t_n)^2}{2(1-M_0t_n)^5}, \\ s_{n+m} & = t_{n+m-1}+ \dfrac{1}{1-M_0t_{n+1}}\big[\dfrac{M(t_{n+1}-s_n)^2}{2} + \dfrac{13L(s_n-t_n)^4}{108} \\ & \phantom{=} +\dfrac{{\color{red}N}M(s_n-t_n)^4}{9(1-M_0t_n)} +\dfrac{M^3(s_n-t_n)^4}{3(1-M_0t_n)^2}\big] \end{aligned} \end{gathered}


There are lots of different ways to present these equations- so this will be quite subjective. Here's one alternative- it's not drastically different, it just splits up the initial and subsequent iterations which can be a little easier to read.

Note that this solution uses \left[ ... \right. and \left. ... \right] to get the correct sizing of your [ ]; you should implement this in whichever solution you use.

\subsection*{Alternative}
The intial values of $t$ and $s$ are defined by
\begin{equation*}
t_0  = 0,   \qquad s_0  = t_0+t_1
\end{equation*}
with subsequent iterations following the formulas
\begin{align*}
t_{n+1} & = s_n + \dfrac{M(1+M(s_n-t_n)) (s_n-t_n)^2}{2(1-M_0t_n)^5},                                                    \\
s_{n+m} & = t_{n+m-1}+ \dfrac{1}{1-M_0t_{n+1}}\left[\dfrac{M(t_{n+1}-s_n)^2}{2} + \dfrac{13L(s_n-t_n)^4}{108}\right.     \\
& \phantom{=}+\left.\dfrac{{\color{red}N}M(s_n-t_n)^4}{9(1-M_0t_n)}+\dfrac{M^3(s_n-t_n)^4}{3(1-M_0t_n)^2}\right]
\end{align*}


Here's the complete MWE- note that the mathtools package loads amsmath so there's no need to load amsmath if you load mathtools

\documentclass{article}

\usepackage{xcolor,mathtools}

\begin{document}

\subsection*{Original}
\label{eq:lmlt} \begin{gathered} t_0 = 0, \quad s_0 = t_0+t_1\\ \begin{aligned} t_{n+1} & = s_n + \dfrac{M(1+M(s_n-t_n)) (s_n-t_n)^2}{2(1-M_0t_n)^5}, \\ s_{n+m} & = t_{n+m-1}+ \dfrac{1}{1-M_0t_{n+1}}\big[\dfrac{M(t_{n+1}-s_n)^2}{2} + \dfrac{13L(s_n-t_n)^4}{108} \\ & \phantom{=} +\dfrac{{\color{red}N}M(s_n-t_n)^4}{9(1-M_0t_n)} +\dfrac{M^3(s_n-t_n)^4}{3(1-M_0t_n)^2}\big] \end{aligned} \end{gathered}

\subsection*{Alternative}
The intial values of $t$ and $s$ are defined by
\begin{equation*}
t_0  = 0,   \qquad s_0  = t_0+t_1
\end{equation*}
with subsequent iterations following the formulas
\begin{align*}
t_{n+1} & = s_n + \dfrac{M(1+M(s_n-t_n)) (s_n-t_n)^2}{2(1-M_0t_n)^5},                                                    \\
s_{n+m} & = t_{n+m-1}+ \dfrac{1}{1-M_0t_{n+1}}\left[\dfrac{M(t_{n+1}-s_n)^2}{2} + \dfrac{13L(s_n-t_n)^4}{108}\right.     \\
& \phantom{=}+\left.\dfrac{{\color{red}N}M(s_n-t_n)^4}{9(1-M_0t_n)}+\dfrac{M^3(s_n-t_n)^4}{3(1-M_0t_n)^2}\right]
\end{align*}
\end{document}

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