Tutorial_01.tex

\documentclass[12pt]{article}

\usepackage{amsmath}
\begin{document}
Hello! This is my first LaTex document.

A rectangle has side lengths of $(x+1)$ and $(x+3)$.
The equation $$A(x)=x^2+4x+3$$ gives the area of the rectangle.

$\frac{5}{6}$\\
$3/4$\\
$\displaystyle{\frac{3}{4+1}}$\\
$\alpha$\\
$\Delta$\\
$$\delta$$
$$\sin 3\theta$$

$$\tan \phi=\frac{\sin \phi}{\cos \phi}$$

$$\ln x$$

$$\log_2 x$$

%Environment in LaTex
\begin{flushright}
This sentence is flush right.
\end{flushright}

\begin{center}
This sentence is centred.
\end{center}

\begin{flushleft}
This sentence is flush left.
\end{flushleft}

\begin{large}
This sentence is large.
\end{large}

\begin{Large}
This sentence is large.
\end{Large}

\begin{small}
This sentence is small.
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\begin{equation}\label{0}
\sqrt[3]{4x}=10
\end{equation}

\begin{equation}\label{1}
x^4+y^3+z^2=100
\end{equation}
Equation \ref {1} uses power.

\begin{equation*}\tag{ EQ5}
x^5+y^5=90
\end{equation*}

\[3x+5y=z\]
\[2x-4y=10\]

\begin{align*}
3x+5y &= z		& 2x-4y &= 10\\
3x+5y &= z	 	& 2x-4y &= 10
\end{align*}

\begin{align*}
|x+y|^2 &=(x+y)^2\\
&=x^2+2xy+y^2\\
&=|x|^2+2xy+|y|^2\\
&\leq|x|^2+2|xy|+|y|^2\\
&=|x|^2+2|x||y|+|y|^2\\
&=|x+y|^2
\end{align*}

\begin{enumerate}
\item Reddy
		\begin{enumerate}
		\item Ellen
		\item Pauline
		\item May
		\end{enumerate}
\item Vivian
\item Peppa
\item George
\end{enumerate}

\begin{itemize}
\item Reddy
		\begin{itemize}
		\item Ellen
		\item Pauline
		\item May
		\end{itemize}
\item Vivian
\item Peppa
\item George
\end{itemize}

\begin{description}
\item [GV903] {Method}
\item [GV994] {Presentation}
\end{description}
\end{document}