[Tex/LaTex] increase vertical space between lines

arraysspacing

I want to increase the vertical space between each of the lines. I know you can use \arraystretch but i am struggling with the package.
here is my source:

\documentclass{article}
\usepackage[fleqn]{amsmath}
\usepackage{stix}
\usepackage[shortlabels]{enumitem}
\everymath{\displaystyle}
\begin{document}
\begin{enumerate}
  \item Express $0.\dot{3}$ as a fraction. \\
        Let $x=0.\dot{3}$ \\
        then $10x=3.\dot{3}$ \\
        $10x-x=3.\dot{3}-0.\dot{3}$ \\
        $9x=3$ \\
        $9x=\frac{3}{9}$
\end{enumerate}
\end{document

and the output:

1

Best Answer

You should use the setspace package and, e.g. its doublespace environment, or if you want something more customised, the spacing environment, which has an argument, the value of \baselinestretch. Beware that doublespacing is not the same as choosing \baseline stretch=2., as the example below demonstrates.

The setspace package defines 4 commands : \singlespacing, \onehalfspacing, doublespacing and \setstretch{#1}, to be used in the preamble, and 4 environments, to be used in the body of the document: \singlespace, \onehalfspace, doublespace and spacing{#1}.

\documentclass{article}
\usepackage[fleqn]{amsmath}
\usepackage{stix}
\usepackage{setspace}
\usepackage[shortlabels]{enumitem}
\everymath{\displaystyle}
\begin{document}
\begin{enumerate}
  \item Express $0.\dot{3}$ as a fraction. \\
        Let $x=0.\dot{3}$ \\
        then $10x=3.\dot{3}$ \\
        $10x-x=3.\dot{3}-0.\dot{3}$ \\
        $9x=3$ \\
        $9x=\frac{3}{9}$
\end{enumerate}


\verb+doublespace+: \vspace{-1\baselineskip}
\begin{doublespace}
\begin{enumerate}
  \item Express $0.\dot{3}$ as a fraction. \\
        Let $x=0.\dot{3}$ \\
        then $10x=3.\dot{3}$ \\
        $10x-x=3.\dot{3}-0.\dot{3}$ \\
        $9x=3$ \\
        $9x=\frac{3}{9}$
\end{enumerate}
\end{doublespace}


\verb+spacing{2}+:\vspace{-1.333\baselineskip}
\begin{spacing}{2}
\begin{enumerate}
  \item Express $0.\dot{3}$ as a fraction. \\
        Let $x=0.\dot{3}$ \\
        then $10x=3.\dot{3}$ \\
        $10x-x=3.\dot{3}-0.\dot{3}$ \\
        $9x=3$ \\
        $9x=\frac{3}{9}$
\end{enumerate}
\end{spacing}

\end{document} 

enter image description here

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