Question

giải hệ phương trình bằng phương pháp thế

Answer 1

\(\left(2\right)\Leftrightarrow\left|x-1\right|=3-3y\)

Thay vào \(\left(1\right)\Leftrightarrow3-3y+\left|y-2\right|=1\Leftrightarrow\left|y-2\right|=3y-2\)

\(\Leftrightarrow\left[{}\begin{matrix}y-2=3y-2\left(y\ge2\right)\\2-y=3y-2\left(y< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=0\left(tkm\right)\\y=1\left(tm\right)\end{matrix}\right.\)

Với \(y=1\Leftrightarrow\left|x-1\right|=3-3=0\Leftrightarrow x=1\)

Vậy \(\left(x;y\right)=\left(1;1\right)\)