1){x3+y3=1 {x+y=1
2){x2+xy=12 {xy-y2=2
3)x3-y3=8 {x-y=2

Предыдущий
вопрос Следующий
вопрос

0 = 0
Since the result is 0 = 0, this system has infinitely many solutions.

2) Replace x in the second equation with (1 - y) from the first equation:
(1 - y)2 + (1 - y)y = 12
(1 - 2y + y2) + (y - y2) = 12
1 - 2y + y + y2 - y2 = 12
1 - y = 12
-y = 11
y = -11
substitute y in x = 1 - y
x = 1 - (-11)
x = 1 + 11
x = 12

Therefore, x = 12 and y = -11 is a solution to this system of equations.

3) Replace x in the first equation with (y + 2):
(y + 2)3 - y3 = 8
y3 + 6y2 + 12y + 8 - y3 = 8
6y2 + 12y = 0
6y(y + 2) = 0
y = 0 or y = -2

Substitute y = 0 into the first equation:
x3 - 0 = 8
x3 = 8
x = 2

Substitute y = -2 into the first equation:
x3 - (-2)3 = 8
x3 + 8 = 8
x3 = 0
x = 0

Therefore, x = 2 and y = 0 or x = 0 and y = -2 are solutions to this system of equations.