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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}
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\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}
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\renewcommand{\thesection}{}
\let\oldforall\forall
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\let\oldexist\exists
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\usepackage[pdfauthor={Камен Димитров Младенов, КН Летен семестър, 2023},
pdftitle={Домашна работа 3 по А2, Зад 4},
pdfsubject={},
pdfkeywords={},
colorlinks=true,
urlcolor=blue]{hyperref}
\title{\vspace{-1.5cm}Домашно №3, Задача 4 по Алгебра 2 на Компютърни Науки}
\author{Камен Димитров Младенов, Група 6}
\date{31 Май 2023}
\begin{document}
\maketitle
\section{Решение}
\begin{align*}
& \left| \begin{array}{ll}
x_1 + x_2 + x_3 + x_4 + x_5 & = 3 \\
x_1 x_2 + x_1 x_3 + x_1 x_4 + x_1 x_5 + x_2 x_3 + x_2 x_4 + x_2 x_5 + x_3 x_4 + x_3 x_5 + x_4 x_5 & = -5 \\
x_1 x_2 x_3 + x_1 x_2 x_4 + x_1 x_2 x_5 + x_1 x_3 x_4 + x_1 x_3 x_5 + x_1 x_4 x_5 + x_2 x_3 x_4 + x_2 x_3 x_5 + x_2 x_4 x_5 + x_3 x_4 x_5 & = -p \\
x_1 x_2 x_3 x_4 + x_1 x_2 x_3 x_5 + x_1 x_2 x_4 x_5 + x_1 x_3 x_4 x_5 + x_2 x_3 x_4 x_5 & = q \\
x_1 x_2 x_3 x_4 x_5 & = -r \\
x_1 + x_2 + x_3 = 2(x_4 + x_5) \\
x_4 x_5 = 6(x_1 x_2 + x_1 x_3 + x_2 x_3) = 3 x_1 x_2 x_3
\end{array} \right. \\
& \text{От равенства 1 и 6} \implies 2(x_4 + x_5) + x_4 + x_5 = 3 \iff x_4 + x_5 = 1 \\
& \left| \begin{array}{ll}
x_4 + x_5 & = 1 \\
(x_4 + x_5)(x_1 + x_2 + x_3) + x_1 x_2 + x_1 x_3 + x_2 x_3 + x_4 x_5 & = -5 \\
(x_4 + x_5)(x_1 x_2 + x_1 x_3 + x_2 x_3) + x_1 x_2 x_3 + x_1 x_4 x_5 + x_2 x_4 x_5 + x_3 x_4 x_5 & = -p \\
(x_4 + x_5) x_1 x_2 x_3 + x_1 x_2 x_4 x_5 + x_1 x_3 x_4 x_5 + x_2 x_3 x_4 x_5 & = q \\
x_1 x_2 x_3 x_4 x_5 & = -r \\
x_1 + x_2 + x_3 = 2 \\
x_4 x_5 = 6(x_1 x_2 + x_1 x_3 + x_2 x_3) = 3 x_1 x_2 x_3
\end{array} \right. \\
& \text{От равенства 2 и 6} \implies x_1 x_2 + x_1 x_3 + x_2 x_3 + x_4 x_5 = -7 \quad \text{ от равенство 7 } \implies \frac{1}{6} x_4 x_5 + x_4 x_5 = -7 \\
& \iff \frac{7}{6}x_4 x_5 = -7 \iff x_4 x_5 = -6 \\
& \left| \begin{array}{ll}
x_4 + x_5 & = 1 \\
x_4 x_5 & = -6 \\
(x_4 x_5)(x_1 + x_2 + x_3) + x_1 x_2 + x_1 x_3 + x_2 x_3 + x_1 x_2 x_3 & = -p \\
(x_4 x_5)(x_1 x_2 + x_1 x_3 + x_2 x_3) + x_1 x_2 x_3 & = q \\
(x_4 x_5) x_1 x_2 x_3 & = -r \\
x_1 + x_2 + x_3 = 2 \\
-6 = 6(x_1 x_2 + x_1 x_3 + x_2 x_3) = 3 x_1 x_2 x_3
\end{array} \right. \\
& \text{От равенства 3 и 2 и 6} \implies -6 . 2 + (x_1 x_2 + x_1 x_3 + x_2 x_3) + x_1 x_2 x_3 = -p \quad \text{ от рав. 7 } \\
& \implies -12 - 1 + x_1 x_2 x_3 = -p \iff x_1 x_2 x_3 = 13 - p \quad \text{ от рав 7 } \implies -2 = 13 - p \implies p = 15 \\
& \text{От равенства 4 и 2 и 7} \implies -6 . (-1) + x_1 x_2 x_3 = q \quad \text{ от рав 7 } \implies 6 + (-2) = q \implies q = 4 \\
& \text{От равенства 5 и 2 и 7} \implies (-6) . (-2) = -r \implies r = -12
\end{align*}
\end{document}
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