ĐKXĐ: x $\neq$ -3 ; y $\neq$ 1
Đặt x + 3 = a ( a $\neq$ 0) ; y - 1 = b ( b $\neq$ 0)
HPT trở thành: $\left \{ {{\frac{1}{a}-\frac{2}{b}=9} \atop {\frac{3}{a}+\frac{1}{b}=6}} \right.$
<=> $\left \{ {{\frac{3}{a}-\frac{6}{b}=27} \atop {\frac{3}{a}+\frac{1}{b}=6}} \right.$
<=> $\left \{ {{\frac{7}{b}=-21} \atop {\frac{3}{a}+\frac{1}{b}=6}} \right.$
<=> $\left \{ {{b=-\frac{1}{3}} \atop {\frac{3}{a}+\frac{1}{b}=6}} \right.$
<=> $\left \{ {{b=-\frac{1}{3}} \atop {\frac{3}{a}-3=6}} \right.$
<=> $\left \{ {{b=-\frac{1}{3}} \atop {a=\frac{1}{3}}} \right.$ (TM)
Với b = -$\frac{1}{3}$ <=> y - 1 = -$\frac{1}{3}$ <=> y = $\frac{2}{3}$ ( TM)
Với a = $\frac{1}{3}$ <=> x + 3 = $\frac{1}{3}$ <=> x = -$\frac{8}{3}$ ( TM)
Vậy x = -$\frac{8}{3}$ và y = $\frac{2}{3}$
