Odpowiedź:
[tex]d)\\\\\begin{cases}\frac{x-1}{2}+\frac{y-1}{3}=1\ \ /*6\\\frac{x-2}{8}-\frac{y-1}{4}=2\ \ /*8\end{cases}\\\\\\\begin{cases}3(x-1)+2(y-1)=6\\x-2-2(y-1)=16\end{cases}\\\\\\\begin{cases}3x-3+2y-2=6\\x-2-2y+2=16\end{cases}\\\\\\\begin{cases}3x+2y=6+3+2\\x-2y=16\end{cases}\\\\\\+\begin{cases}3x+2y=11\\x-2y=16\end{cases}\\----------\\4x=27\ \ /:4\\\\x=\frac{27}{4}[/tex]
[tex]3*\frac{27}{4}+2y=11\\\\\frac{81}{4}+2y=11\\\\2y=11-\frac{81}{4}\\\\2y=\frac{44}{4}- \frac{81}{4}\\\\2y=-\frac{37}{4}\ \ /:2\\\\y=-\frac{37}{4}*\frac{1}{2}\\\\y=-\frac{37}{8}\\\\\\\begin{cases}x=\frac{27}{4}\\y=-\frac{37}{8}\end{cases}[/tex]
Rozwiązaniem układu równań nie jest para liczb całkowitych
