от ammornil » 14 Дек 2021, 00:40
Пренареждаме уравненията, изразяваме [tex]x[/tex] и [tex]z[/tex] чрез [tex]y[/tex] и намираме [tex]y[/tex].
[tex]\begin{array}{|l} 2x+y=9 \\ 3y-2z=1 \\ x-y+5z=8 \end{array} \Leftrightarrow \begin{array}{|l} 2x=9-y \\ 2z=3y-1 \\ x-y+5z=8 \end{array} \Leftrightarrow \begin{array}{|l} x={\Large \frac{9-y}{2}} \\ z={\Large \frac{3y-1}{2}} \\ {\Large \frac{9-y}{2}}-y+5\left( {\Large \frac{3y-1}{2}} \right)=8 \end{array} \Leftrightarrow \begin{array}{|l} x={\Large \frac{9-y}{2}} \\ z={\Large \frac{3y-1}{2}} \\ 9-y-2y+5(3y-1)=2.8 \end{array} \Leftrightarrow[/tex]
[tex]\begin{array}{|l} x={\Large \frac{9-y}{2}} \\ z={\Large \frac{3y-1}{2}} \\ 9-3y+15y-5=16 \end{array} \Leftrightarrow \begin{array}{|l} x={\Large \frac{9-y}{2}} \\ z={\Large \frac{3y-1}{2}} \\ 12y=12 \end{array} \Leftrightarrow \begin{array}{|l} x={\Large \frac{9-1}{2}} \\ z={\Large \frac{3.1-1}{2}} \\ y=1 \end{array} \Leftrightarrow \begin{array}{|l} x=4 \\ z=1 \\ y=1 \end{array} \Leftrightarrow \begin{array}{|l} x= 4 \\ y=1 \\ z=1 \end{array}[/tex]
[tex]\color{lightseagreen}\text{''Който никога не е правил грешка, никога не е опитвал нещо ново.''} \\
\hspace{21em}\text{(Алберт Айнщайн)}[/tex]