12/у-12/х=1 12/(у+1)-12/(х+2)=1 решите пожалуйста систему подробно

$\left \{ {{\frac{12}{y}-\frac{12}{x}=1} \atop { \frac{12}{y+1} }-\frac{12}{x+2}=1} \right. \; \; \; t=\frac{12}{y}\; ,\; p=\frac{12}{x}\; \to \; y=\frac{12}{t}\; ,\; x=\frac{12}{p}\\\\\\x\ne-2,y\ne-1\\\\\frac{12}{y+1}=\frac{12}{\frac{12}{t}+1}= \frac{12t}{12+t}\; \; ,\; \; \frac{12}{x+2} =\frac{12}{\frac{12}{p}+2}=\frac{12p}{12+2p}\\\\ \left \{ {{t-p=1} \atop {\frac{12t}{12+t}-\frac{12p}{12+2p}=1}} \right. \; \left \{ {{t=p+2} \atop {\frac{12+12p}{13+p}-\frac{12p}{12+2p}-1=0}} \right.$

$\left \{ {{t=1+p} \atop {\frac{(12+12p)(12+2p)-12p(13+p)-(13+p)(12+2p)}{(12+2p)(13+p)}}=0} \right. \; \left \{ {{\frac{2(5p^2-13p-6)}{(12+2p)(13+p)}=0} \atop {t=1+p}} \right. \\\\\\5p^2-13p-6=0\; ;\; \; D=289\\\\p_1=\frac{13-17}{10}=-\frac{2}{5}=-0,4\; ;\; \; p_2=3\\\\t_1=1+p_1=1-0,4=0,6\; ;\; \; t_2=1+p_2=1+3=4\\\\y_1=\frac{12}{p_1}=\frac{12}{0,6}=20\; ;\; \; x_1=\frac{12}{p_1}=\frac{12}{-0,4}=-30\\\\y_2=\frac{12}{t_2}=\frac{12}{4}=3\; ;\; \; x_2=\frac{12}{p_2}=\frac{12}{3}=4$

Ответ: (-30,20) , (4,3) .