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$\\Solution:\\\\ 2\left (\frac{x-1}{x+3} \right )-7\left (\frac{x+3}{x-1} \right )=5\\\\\\ Let\ \ a=\frac{x-1}{x+3}\ \ then\ \ \ \frac 1a=\frac{x+3}{x-1} \\\\\\ \Rightarrow 2a-7\times\frac 1a=5\\\\ \Rightarrow \frac{2a^2-7}{a}=5\\\\ \Rightarrow 2a^2-7=5a\\\\ \Rightarrow 2a^2-5a-7=0\\\\ \Rightarrow 2a^2+2a-7a-7=0\\\\ \Rightarrow 2a(a+1)-7(a+1)=0\\\\ \Rightarrow (a+1)(2a-7)=0\\\\ \Rightarrow a+1=0\ \ \ \ or\ \ \ \ 2a-7=0\\\\ \Rightarrow a=- 1\ \ \ \ \ or\ \ \ \ a=\frac{7}{2}\\\\$

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