Помогите решить. Не понимаю.

Решите задачу:

$1)\; \; \left \{ {{y=x} \atop {y=2-x}} \right. \; \; \to \; \; \; y=y\; \; \; \to \; \; \; x=2-x\; ,\; \; 2x=2\; ,\; x=1\\\\ y=x=1\\\\Otvet:\; \; (1,1)\; .\\\\2)\; \; \left \{ {{y=2x} \atop {y=6-x}} \right. \; \; \left \{ {{y=2x} \atop {2x=6-x}} \right. \; \left \{ {{y=2x} \atop {3x=6}} \right. \; \left \{ {{y=2\cdot 2} \atop {x=2}} \right. \; \left \{ {{y=4} \atop {x=2}} \right. \\\\Otvet:\; \; (2,4)\; .$

$3)\; \; \left \{ {{x+2y=6} \atop {x-y=4}} \right. \; \left \{ {{x=6-2y} \atop {x=4+y}} \right. \; \left \{ {{6-2y=4+y} \atop {x=4+y}} \right. \; \left \{ {{2=3y} \atop {x=4+y}} \right. \; \left \{ {{y=\frac{2}{3}} \atop {x=4+\frac{2}{3}}} \right. \\\\\left \{ {{y=\frac{2}{3}} \atop {x=4\frac{2}{3}}} \right. \\\\Otvet:\; \; (4\frac{2}{3},\frac{2}{3})\; .$

$4)\; \; \left \{ {{2x-y=-1} \atop {x+y=-2}} \right. \; \left \{ {{y=2x+1} \atop {y=-x-2}} \right. \; \left \{ {{-x-2=2x+1} \atop {y=-x-2}} \right. \; \left \{ {{3x=-3} \atop {y=-x-2}} \right. \; \left \{ {{x=-1} \atop {y=1-2}} \right. \\\\ \left \{ {{x=-1} \atop {y=-1}} \right. \\\\Otvet:\; \; (-1,-1)\; .$