Theo đề ra ta có:
\(\left\{{}\begin{matrix}x.\left(x-y\right)=\frac{3}{10}\\y.\left(x-y\right)=-\frac{3}{50}\left(1\right)\end{matrix}\right.\)
\(\Rightarrow\frac{x\left(x-y\right)}{y\left(x-y\right)}=\frac{x}{y}=\frac{3}{10}:-\frac{3}{50}=-5\)
\(\Rightarrow\frac{x}{y}=-5\Rightarrow x=-5y\left(2\right)\)
Thay (2) vào (1) ta có :
\(y\left(-5y-y\right)=-\frac{3}{50}\)
\(\Rightarrow-6y^2=-\frac{3}{50}\)
\(\Rightarrow y^2=-\frac{3}{50}:-6\)
\(\Rightarrow y^2=\frac{1}{100}\)
\(\Rightarrow y=\sqrt{\frac{1}{100}}\)
\(\Rightarrow y=\frac{1}{10}\)
Ta có : \(x=-5.\frac{1}{10}\)
\(x=-\frac{1}{2}\)
Vậy \(x=-\frac{1}{2};y=\frac{1}{10}\)
