# array equation with arrow

Please Help me in writing this equation

• Could you please show us what you've got so far. This is not a we-do-it-for-you-site, show some own effort and everybody here will be glad to help. You might take a look at the the LaTeX-Wikibook if you have no idea how to start. – Skillmon Jun 29 '17 at 11:46
• Slashed letter can be obtained with the cancel package. – Bernard Jun 29 '17 at 12:12

## 3 Answers

Fixing also various glitches in the original (due to indiscriminate usage of \left and \right, mainly):

\documentclass{article}
\usepackage{amsmath}
\usepackage{slashed}

\begin{document}

\begin{equation*}
\left(
\begin{aligned}
-\slashed{g}
&= y''\Bigl(\frac{dx}{ds}\Bigr)^{\!2}v^2
= \frac{y''v^2}{(ds/dx)^2}
\\
&= \frac{y''v^2}{1+(y')^2}
= \frac{y''2\slashed{g}(y_0-y)}{1+(y')^2}
\end{aligned}
\right)
\implies
2y''(y-y_0) = 1+(y')^2
\end{equation*}

\end{document}


Using array

  \documentclass[a4paper]{article}
\usepackage[left=1cm,right=1cm]{geometry}
\usepackage{amsmath,cancel}

\begin{document}
\begin{center}

$\left( \begin{array}{llll} -\cancel{g} & = y''\left( \dfrac{dx}{ds} \right)^{2}v^{2}&=\dfrac{y''v_{2}}{(ds/dx)^{2}}\\ &&&\\ & = \dfrac{y''v_{2}}{1+(y')^{2}}&=\dfrac{y''2\cancel{g}(y_{0}-y)}{1+(y')^{2}}\\ \end{array} \right)$$\implies 2y''(y-y_{0})=1+(y')^{2}$.

\end{center}

\end{document}


Use the array from amsmath for alignment at = and the cancel package to cancel out some terms.

\documentclass[12pt,a4paper]{article}
\usepackage{amsmath,cancel}

\begin{document}
$\left( \begin{array}{@{}r@{{}={}}l@{}} -\cancel{g}& y''\Big(\dfrac{dx}{ds}\Big)^2 v^2=\dfrac{y''v_2}{(ds/dx)^2}\\ & \dfrac{y'' v^2}{1+(y^\prime)^2}=\dfrac{y''2\cancel{g}(y_0-y)}{1+(y')^2} \end{array}\right) \Longrightarrow 2y''(y-y_0)=1+(y')^2.$

\end{document}


This can also be represented in a slightly different way, since it is a single equation divided over two lines and not a matrix.

$\left. \begin{array}{r@{{}={}}l} -\cancel{g} & y''\Big(\dfrac{dx}{ds}\Big)^2 v^2=\dfrac{y''v_2}{(ds/dx)^2}\\ & \dfrac{y'' v^2}{1+(y')^2}=\dfrac{y''2\cancel{g}(y_0-y)}{1+(y')^2} \end{array}\right\} \Longrightarrow 2y''(y-y_0)=1+(y')^2.$