Решить систему уравнений:1){x^2+y^2=74,{x+y=122){x^2-y^2=32,{x-y=43){(x-1)(y-1)=2,{x+y=54){x+y=3,{xy=-405){x-y=7,{xy=18

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1) x^2 + y^2 = 74
x + y = 122

Substitute y = 122 - x into the first equation:
x^2 + (122 - x)^2 = 74
x^2 + 14884 - 244x + x^2 = 74
2x^2 - 244x + 14710 = 0

Divide by 2 to simplify:
x^2 - 122x + 7355 = 0
(x - 65)(x - 113) = 0

x = 65 or x = 113

If x = 65, then y = 122 - 65 = 57
If x = 113, then y = 122 - 113 = 9

Therefore, the solutions are (65, 57) and (113, 9).

2) x^2 - y^2 = 32
x - y = 43

Substitute y = x - 43 into the first equation:
x^2 - (x - 43)^2 = 32
x^2 - (x^2 - 86x + 1849) = 32
86x - 1849 = 32
86x = 1881
x = 1881 / 86
x = 21.91

Substitute x = 21.91 into x - y = 43:
21.91 - y = 43
y = -21.91

Solution: x = 21.91, y = -21.91

3) (x - 1)(y - 1) = 2
x + y = 54

Expand the first equation:
xy - x - y + 1 = 2
xy - x - y - 1 = 0

Substitute x + y = 54 into xy - x - y - 1 = 0:
54y - x - y - 1 = 0
53y - x - 1 = 0
53(54 - y) - x - 1 = 0
2862 - 53y - x - 1 = 0
2861 - 53y = x

Substitute x = 2861 - 53y into x + y = 54:
2861 - 53y + y = 54
-52y = -2807
y = 53.98

Substitute y = 53.98 into x + y = 54:
x = 54 - 53.98
x = 0.02

Solution: x = 0.02, y = 53.98

4) x + y = 3
xy = -405

Substitute y = 3 - x into xy = -405:
x(3 - x) = -405
3x - x^2 = -405
x^2 - 3x + 405 = 0

The above quadratic equation has no real solutions, which means there are no real solutions for this system of equations.

5) x - y = 7
xy = 18

Substitute y = x - 7 into xy = 18:
x(x - 7) = 18
x^2 - 7x - 18 = 0
(x - 9)(x + 2) = 0

x = 9 or x = -2

If x = 9, then y = 9 - 7 = 2
If x = -2, then y = -2 - 7 = -9

Therefore, the solutions are (9, 2) and (-2, -9).