Вопрос в картинках...

Решите задачу:

$\left \{ {{3^{x+y}=27} \atop {(x-y)^2=1}} \right. \\ \left \{ {{3^{x+y}=3^3} \atop {(x-y)^2=1}} \right. \\ \left \{ {{x+y=3} \atop {(x-y)^2=1}} \right. \\ \left \{ {{x=3-y} \atop {(3-y-y)^2=1}} \right. \\ \left \{ {{x=3-y} \atop {3-2y=б1}} \right. \\ \left \{ {{x=3-y} \atop {y= \frac{3б1}{2} }} \right. \\ \\ \\ \left \{ {{x=3-y} \atop {y=1 }} \right. \\ \left \{ {{x=3-y} \atop {y=2 }} \right. \\ \\ \\ \left \{ {{x=2} \atop {y=1 }} \right. \\ \left \{ {{x=1} \atop {y=2 }} \right. \\ \\ OTBET:(1;2),(2;1)$