4 (x-y)=-2,
{ 3x-7y=-2,5-2(x+y);
1) Поработаем со вторым уравнением, упростим его.
3x-7y=-2,5-2(x+y)
3x-7y=-2,5-2x-2y
Перенесем выражения с переменными в левую сторону, свободные члены в правую.
3x-7y+2x+2y=-2,5
5x-5y=-2,5 |:5 (поделим все уравнение на пять)
x-y=-0.5
2. Запишем получившуюся систему:
{4 (x-y)=-2
{x-y=-0.5
Раскроем скобки в первом уравнении, получим:
{4x- 4y=-2
{x-y=-0.5
3. Выразим из второго уравнения x.
x-y=-0.5
x=y-0.5
4. Подставим получившийся х в первое уравнение:
4*(у-0.5)-4у=-2
4у-2-4у=-2
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Выражение( а в квадрате целое) при рациональных а только, если а целое.
Значит цеое число должно быть (n-3)/n=1 -3/n
Но 3 делится нацело только на 1 (-1) и 3 (-3)
ответ: n может принимать значения: 1,-1,3,-3