3х+у=2, х^(2)-у^(2)=-12

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First, let's rearrange the first equation to solve for y:

3x + y = 2
y = 2 - 3x

Now we can substitute this expression for y into the second equation:

x^2 - (2 - 3x)^2 = -12
x^2 - (4 - 12x + 9x^2) = -12
x^2 - 4 + 12x - 9x^2 = -12
-8x^2 + 12x - 4 = -12
-8x^2 + 12x + 8 = 0

Now we have a quadratic equation in terms of x. We can solve for x using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a
x = (-12 ± √(12^2 - 4(-8)(8))) / 2(-8)
x = (-12 ± √(144 + 256)) / -16
x = (-12 ± √400) / -16
x = (-12 ± 20) / -16

This gives two possible values for x:

x = (8/4) or x = (-32/4)

Therefore, x = 2 or x = -8

Now that we have the values of x, we can substitute them back into the equation y = 2 - 3x to find the corresponding values of y:

If x = 2:
y = 2 - 3(2)
y = 2 - 6
y = -4

If x = -8:
y = 2 - 3(-8)
y = 2 + 24
y = 26

Therefore, the solutions to the system of equations are:
x = 2, y = -4
x = -8, y = 26