Ответ:

[tex]\bigg (1\dfrac{4}{13} \ ; \ 12\dfrac{1}{13} \bigg )[/tex]

Объяснение:

[tex]\displaystyle \left \{ {{\dfrac{1}{4}(y-1)-\dfrac{1}{3}(x+1)=2} \atop {\dfrac{1}{4}(x+3)-\dfrac{1}{3}(y+1)=-4}} \right. \Leftrightarrow \left \{ {{3(y-1)-4(x+1)=24} \atop {x+3-4(y+1)=-48}} \right. \Leftrightarrow[/tex]

[tex]\displaystyle \Leftrightarrow \left \{ {{3y-3-4x-4=24} \atop {x+3-4y-4=-48}} \right. \Leftrightarrow \left \{ {{3y-4x-7=24} \atop {x-4y-1=-48}} \right. \Leftrightarrow \left \{ {{3y-4x=24+7} \atop {x-4y=-48+1}} \right. \Leftrightarrow[/tex]

[tex]\displaystyle \Leftrightarrow \left \{ {{3y-4x=31} \atop {x-4y=-47}} \right. \Leftrightarrow \left \{ {{3y-4x=31} \atop {4x-16y=-188}} \right. \bigg |+[/tex]

[tex]3y+4x-4x-16y=31-188;[/tex]

[tex]-13y=-157; \quad y=\dfrac{-157}{-13}=\dfrac{157}{13}=12\dfrac{1}{13};[/tex]

[tex]x=4y-47; \quad x=4 \cdot 12\dfrac{1}{13}-47=48\dfrac{4}{13}-47=1\dfrac{4}{13};[/tex]

[tex]\bigg (1\dfrac{4}{13} \ ; \ 12\dfrac{1}{13} \bigg );[/tex]