3x + y = 2
x^2 - y^2 = -12
From the first equation, we can rewrite it as:
y = 2 - 3x
Substitute y in 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 - 16 = 0
4x^2 - 6x + 8 = 0
Using the quadratic formula, we get:
x = (-(-6) ± sqrt((-6)^2 - 448)) / 2*4
x = (6 ± sqrt(36 - 128)) / 8
x = (6 ± sqrt(-92)) / 8
x = (6 ± 2*sqrt(23)i) / 8
x = (3 ± sqrt(23)i) / 4
Therefore, the solutions for x are -0.5 and 2.
Substitute x back to find y:
When x = -0.5, y = 2 - 3*(-0.5) = 3.5
So, the solution is (-0.5, 3.5).
When x = 2, y = 2 - 3*2 = -4
So, the solution is (2, -4).
3x + y = 2
x^2 - y^2 = -12
From the first equation, we can rewrite it as:
y = 2 - 3x
Substitute y in 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 - 16 = 0
4x^2 - 6x + 8 = 0
Using the quadratic formula, we get:
x = (-(-6) ± sqrt((-6)^2 - 448)) / 2*4
x = (6 ± sqrt(36 - 128)) / 8
x = (6 ± sqrt(-92)) / 8
x = (6 ± 2*sqrt(23)i) / 8
x = (3 ± sqrt(23)i) / 4
Therefore, the solutions for x are -0.5 and 2.
Substitute x back to find y:
When x = -0.5, y = 2 - 3*(-0.5) = 3.5
So, the solution is (-0.5, 3.5).
When x = 2, y = 2 - 3*2 = -4
So, the solution is (2, -4).