Х^2+y^2=3 в системе x-y=2

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To solve this system of equations, we can substitute the value of y from the second equation into the first equation:

x - y = 2
x = y + 2

Substitute x = y + 2 into the first equation:

(y + 2)^2 + y^2 = 3
y^2 + 4y + 4 + y^2 = 3
2y^2 + 4y + 4 = 3
2y^2 + 4y + 1 = 0

Now we need to solve this quadratic equation for y. We can use the quadratic formula:

y = (-b ± √(b^2 - 4ac)) / 2a

In this case, a = 2, b = 4, and c = 1. Plugging the values into the formula:

y = (-4 ± √(4^2 - 421)) / 2*2
y = (-4 ± √(16 - 8)) / 4
y = (-4 ± √8) / 4
y = (-4 ± 2√2) / 4
y = (-2 ± √2) / 2

Now that we have the values for y, we can find the corresponding values for x using the equation x = y + 2:

For y = (-2 + √2) / 2:
x = (-2 + √2) / 2 + 2
x = (-2 + √2 + 4) / 2
x = (2 + √2) / 2

For y = (-2 - √2) / 2:
x = (-2 - √2) / 2 + 2
x = (-2 - √2 + 4) / 2
x = (2 - √2) / 2

Therefore, the solutions to the system of equations are:
(x, y) = ((2 + √2) / 2, (-2 + √2) / 2) and ((2 - √2) / 2, (-2 - √2) / 2)