do đó ta có hpt:\(\left\{{}\begin{matrix}x-2y=-5\\y=2.\left(x-1\right)+3\end{matrix}\right.< =>\left\{{}\begin{matrix}x-2y=-5\\2x-y=-1\end{matrix}\right.< =>\left\{{}\begin{matrix}2x-4y=-10\\2x-y=-1\end{matrix}\right.< =>\left\{{}\begin{matrix}x-2y=-5\\-3y=-9\end{matrix}\right.< =>\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)
TH2:
x<1 ==> x-1<0 ==>\(\left|x-1\right|=1-x\)
do đó ta có hpt\(\left\{{}\begin{matrix}x-2y=-5\\y=2-2x+3\end{matrix}\right.< =>}\left\{{}\begin{matrix}x-2y=-5\\y+2x=5\end{matrix}\right.< =>\left\{{}\begin{matrix}2x-4y=-10\\y+2x=5\end{matrix}\right.< =>\left\{{}\begin{matrix}x-2y=-5\\-5y=-15\end{matrix}\right.< =>\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)