Задача

[Гост]

.

[ammornil]

[tex]\begin{array}{|l} 2{\Large\frac{x-3}{7}}+3{\Large\frac{y+2}{2}}=7 \\ \\ 3{\Large\frac{x-3}{7}}-5{\Large\frac{y+2}{2}}=1 \end{array}[/tex]

Полагаме [tex]{\Large\begin{array}{|l}\frac{x-3}{7} =u \\ \\ \frac{y+2}{2}=v \end{array}} \Rightarrow \begin{array}{|l}x=7u+3 \\ y=2v-2 \end{array}[/tex]

Заместваме с полагането в оригналното условие:
[tex]\begin{array}{|l} 2u+3v=7 |.(5) \\ 3u-5v=1 |.(3)\end{array} \Rightarrow \begin{array}{|l} 10u+15v=35 \\ 9u-15v=3 \end{array} \Rightarrow \begin{array}{|l} 10u+15v+9u-15v=35+3 \\ 2u+3v=7\end{array} \Rightarrow \begin{array}{|l} 19u=38 \\ 3v=7-2u\end{array} \Rightarrow[/tex]

[tex]\begin{array}{|l} u=2 \\ v=\frac{7-2.2}{3}=1\end{array}[/tex]

Връщаме се в полагането:
[tex]\begin{array}{|l}x=7u+3 \\ y=2v-2 \end{array} \Rightarrow \begin{array}{|l}x=7.2+3 \\ y=2.1-2 \end{array} \Rightarrow \begin{array}{|l}x=17 \\ y=0 \end{array}[/tex]