[tex]\it \begin{cases} \it \dfrac{5}{2x-y}+\dfrac{6}{x-y}=3 \ \ \ \ \ \ (1)\\ \\ \\ \it \dfrac{2}{2x-y}+\dfrac{3}{y-x}=-\dfrac{3}{5} \ \ \ \ \ \ (2)\end{cases}[/tex]
Vom transforma ecuația (2), după cum urmează:
[tex]\it \dfrac{2}{2x-y}+\dfrac{^{-1)}3}{\ y-x}=-\dfrac{3}{5}|_{\cdot5} \Rightarrow \dfrac{10}{2x-y}-\dfrac{15}{x-y}=-3\\ \\ \\ Sistemul\ \ devine:\\ \\ \\ \begin{cases} \it \dfrac{5}{2x-y}+\dfrac{6}{x-y}=3\\ \\ \\ \it \dfrac{10}{2x-y}-\dfrac{15}{x-y}=-3 \end{cases}[/tex]
[tex]\it Vom\ nota:\\ \\ \dfrac{1}{2x-y}=a,\ \ \ \dfrac{1}{x-y}=b ,\ \ iar\ sistemul\ se\ scrie:\\ \\ \\ \begin{cases} \it 5a+6b=3|_{\cdot(-2)}\\ \\ \it 10a-15b=-3\end{cases} \Rightarrow \begin{cases} \it -10a-12b=-6\\ \\ \it 10a-15b=-3\end{cases}\\ \rule{0.1}{0.1}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \rule{70}{0.4}\\ \rule{0.1}{0.1}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -27b=-9 \Rightarrow b=\dfrac{-9}{-27}=\dfrac{1}{3}\ \ \ \ \ (3)[/tex]
[tex]\it 5a+6b=3\ \stackrel{(3)}{\Longrightarrow}\ 5a+2=3|_{-2} \Rightarrow 5a=1 \Rightarrow a=\dfrac{1}{5}[/tex]
[tex]\it a=\dfrac{1}{5} \Rightarrow \dfrac{1}{2x-y}=\dfrac{1}{5} \Rightarrow 2x-y=5 \Rightarrow y=2x-5\ \ \ \ \ \ (4)\\ \\ \\ \dfrac{1}{x-y}=\dfrac{1}{3} \Rightarrow x-y=3\ \stackrel{(4)}{\Longrightarrow}\ x-2x+5=3|_{-5} \Rightarrow -x=-2|_{\cdot(-1)} \Rightarrow x=2\\ \\ \\ x=2\ \stackrel{(4)}{\Longrightarrow}\ y=4-5 \Rightarrow y=-1\\ \\ \\ S=\{(2,\ -1)\}[/tex]
