[A2/H3/E1] Solved

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diff --git a/Year 2/Algebra/Homework3/Exercise1.pdf b/Year 2/Algebra/Homework3/Exercise1.pdf
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diff --git a/Year 2/Algebra/Homework3/Exercise1.tex b/Year 2/Algebra/Homework3/Exercise1.tex
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+% Document MUST be built with XeLaTeX!
+\documentclass[11pt,a4paper]{article}
+
+\usepackage[margin=1.5cm]{geometry}
+
+% Поддръжка на български език
+% Thanks to https://tex.stackexchange.com/a/136926
+\usepackage{fontspec}
+\defaultfontfeatures{Ligatures=TeX}
+\setmainfont{CMU Serif}
+\setsansfont{CMU Sans Serif}
+
+\usepackage[bulgarian,english]{babel}
+
+\usepackage{amsmath}            % Математически символи
+\usepackage{amssymb}
+\usepackage{mathtools}
+\usepackage{amsfonts}           % Още математически символи
+\usepackage{cancel}             % Зачертаване
+\usepackage[dvipsnames]{xcolor} % Цветни кутийки + допълнителни цветове
+\usepackage{indentfirst}        % Индентация на параграф след секция
+
+\usepackage{enumitem}
+\setlist[enumerate]{label=\arabic*.}
+
+\renewcommand{\thesection}{}
+
+\let\oldforall\forall
+\renewcommand{\forall}{\; \oldforall}
+
+\let\oldexist\exists
+\renewcommand{\exists}{\; \oldexist}
+
+\usepackage[pdfauthor={Камен Димитров Младенов, КН Летен семестър, 2023},
+		    pdftitle={Домашна работа 3 по А2, Зад 1},
+		    pdfsubject={},
+		    pdfkeywords={},
+		    colorlinks=true,
+		    urlcolor=blue]{hyperref}
+
+\title{\vspace{-1.5cm}Домашно №2, Задача 1 по Алгебра 2 на Компютърни Науки}
+\author{Камен Димитров Младенов, Група 6}
+\date{31 Май 2023}
+
+\begin{document}
+
+\maketitle
+
+\section{Решение}
+
+\begin{align*}
+	\frac{f(x)}{g(x)} & = \frac{x^3+x^2-x+\bar{2}}{x^3-\bar{2}x^2-\bar{2}x+\bar{2}} \quad (x^3+x^2-x+\bar{2}) - (x^3-\bar{2}x^2-\bar{2}x+\bar{2}) = \bar{3}x^2+x = r_1 \quad f(x) = g(x) + r_1 \\
+	\frac{g(x)}{r_1} & = \frac{x^3-\bar{2}x^2-\bar{2}x+\bar{2}}{\bar{3}x^2+x} \quad (x^3-\bar{2}x^2-\bar{2}x+\bar{2}) - \bar{2}x(\bar{3}x^2+x) = -\bar{4}x^2 - \bar{2}x + \bar{2} = x^2 + \bar{3}x + \bar{2} \\
+	& \qquad\qquad\qquad\qquad\qquad\; (x^2 + \bar{3}x + \bar{2}) - \bar{2}(\bar{3}x^2+x) = x + \bar{2} = r_2 \quad g(x) = (\bar{2}x + \bar{2})r_1 + r_2 \\
+	\frac{r_1}{r_2} & = \frac{\bar{3}x^2+x}{x + \bar{2}} \quad \qquad\qquad\;\; (\bar{3}x^2+x) - (\bar{3}x)(x + \bar{2}) = 0 \quad r_1 = (\bar{3}x) r_2 \\
+\end{align*}
+\begin{align*}
+	& g(x) = \bar{2}(x + \bar{1})(\bar{3}x) r_2 + r_2 = (x^2 + x + \bar{1}) r_2 \qquad f(x) = g(x) + (\bar{3}x) r_2 = (x^2 + \bar{4} x + \bar{1})r_2 \implies (f, g) = r_2
+\end{align*}
+\begin{align*}
+	& f(x) = g(x) + r_1 \iff r_1 = f(x) - g(x) \\
+	& \implies g(x) = \bar{2}(x+\bar{1})r_1 + r_2 = \bar{2}(x+\bar{1})(f(x) - g(x)) + r_2 \\
+	& \iff r_2 = \bar{2}(x+\bar{1})(g(x) - f(x)) + g(x) \\
+	& \iff r_2 = (\bar{2}x + \bar{2} + \bar{1})g(x) + (-\bar{2})(x+\bar{1})f(x) \\
+	& \iff r_2 = \bar{3}(x+\bar{1})f(x) + (\bar{2}x + \bar{3})g(x) \\
+	& \implies u(x) = \bar{3}x + \bar{3} \quad v(x) = \bar{2}x + \bar{3}
+\end{align*}
+
+
+\end{document}
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