решить уравнение: [x-1]+[x-3]=2

[tex][x-1]+[x-3]=2\\\\oboznachim\; \; y=x-3\; \; \to \; \; x=y+3\\\\x-1=(y+3)-1=y+2\; \; \to \\\\\; [x-1]+[x-3]=[y+2]+[y]\\\\\; [y+2]+[y]=2\\\\svojstvo:\; [y+2]=[y]+2\; \; \to \\\\\; [y+2]+[y]=[y]+2+[y]=2[y]+2\\\\2[y]+2=2\\\\2[y]=0\\\\\; [y]=0\\\\0 \leq y\ \textless \ 1\\\\0 \leq x-3\ \textless \ 1\\\\3 \leq x\ \textless \ 4[/tex]