Wszystkie rozwiązałam metodą podstawiania
1.
[tex]\begin{cases}y=\cfrac{1}{3}(x-1)\\3(x+1)=-(y+1)\end{cases}\\\begin{cases}y=\cfrac{x}{3}-\cfrac{1}{3}\\3x+3=-y-1\ \leftarrow\text{podstawiam tutaj y}\end{cases}\\3x+3=-\left(\cfrac{x}{3}-\cfrac{1}{3}\right)-1\\3x+3=-\cfrac{x}{3}+\cfrac{1}{3}-1\\3\cfrac{1}{3}x=-3\cfrac{2}{3}\\10x=-11\\x=-\cfrac{11}{10}\\\text{a teraz y ze wzoru }y=\cfrac{x}{3}-\cfrac{1}{3}\\y=\cfrac{x}{3}-\cfrac{1}{3}\\y=\cfrac{-\frac{11}{10}}{3}-\cfrac{1}{3}\\y=-\cfrac{11}{30}-\cfrac{10}{30}=\cfrac{21}{30}\\y=-\cfrac{7}{10}[/tex]
[tex]\begin{cases}x=-\cfrac{11}{10}\\y=-\cfrac{7}{10}\end{cases}[/tex]
2.
[tex]\begin{cases}x-y=3x-3y+5\\x+y=3x+5y-2\end{cases}\\\begin{cases}2y=2x+5\\x+y=3x+5y-2\end{cases}\\\begin{cases}y=x+\cfrac{5}{2}\\2=2x+4y\end{cases}\\2=2x+4\left(x+\cfrac{5}{2}\right)\\2=2x+4x+10\\-8=6x\\x=-\cfrac{8}{6}=-\cfrac{4}{3}\\y=x+\cfrac{5}{2}\\y=-\cfrac{4}{3}+\cfrac{5}{2}\\y=-\cfrac{8}{6}+\cfrac{15}{6}=\cfrac{7}{6}\\\begin{cases}x=-\cfrac{4}{3}\\y=\cfrac{7}{6}\end{cases}[/tex]
3.
[tex]\begin{cases}2x=3(1-y)+2y\\3(y-x)=y-2\end{cases}\\\begin{cases}2x=3-3y+2y\\3y-3x=y-2\end{cases}\\\begin{cases}y=3-2x\\2y-3x=-2\end{cases}\\2(3-2x)-3x=-2\\6-4x-3x=-2\\-7x=-8\\x=\cfrac{8}{7}\\y=3-2x\\y=3-2*\cfrac{8}{7}\\y=\cfrac{21}{7}-\cfrac{16}{7}=\cfrac{5}{7}\\\begin{cases}x=\cfrac{8}{7}\\y=\cfrac{5}{7}\end{cases}[/tex]
4.
[tex]\begin{cases}\cfrac{2x-y}{3}=\cfrac{3x+y}{2}\\x=2(x+y+1)\end{cases}\\\begin{cases}4x-2y=9x+3y\\x=2x+2y+2\end{cases}\\\\\begin{cases}-5x=5y\\-x=2y+2\end{cases}\\\begin{cases}x=-y\\-x=2y+2\\\end{cases}\\-(-y)=2y+2\\y=2y+2\\-y=2\\y=-2\\x=-y\\x=-(-2)\\x=2\\\begin{cases}x=2\\y=-2\end{cases}[/tex]
