No.
x2 + 2y - 6x + 8y - 1 = x2 - y2 + 4x + 6y - 1 y2 - 10x + 4y = 0 y(y+4) = 10x It cannot be solved completely because with two variables (x and y) you need two independent equations for a full solution.
x2 + 2y = 52 + 2*2 = 25 + 4 = 29
Implicit: x2 + 2y = 5 Explicit : y = (5 - x2)/2
x2+4x+4 = 25 x2+4x+4-25 = 0 x2+4x-21 = 0 (x+7)(x-3) = 0 x = -7 or x = 3
x^2+4x+7
x2 + 2y - 6x + 8y - 1 = x2 - y2 + 4x + 6y - 1 y2 - 10x + 4y = 0 y(y+4) = 10x It cannot be solved completely because with two variables (x and y) you need two independent equations for a full solution.
y = 4x-3
x2 + 2y = 52 + 2*2 = 25 + 4 = 29
Implicit: x2 + 2y = 5 Explicit : y = (5 - x2)/2
x2+4x+4 = 25 x2+4x+4-25 = 0 x2+4x-21 = 0 (x+7)(x-3) = 0 x = -7 or x = 3
x^2+4x+7
2x = x2 + 4x - 3 x2 + 2x - 3 = 0 (x - 1)(x - 2) = 0
x2 + 4x + 3 = 0 (x + 1)(x + 3) = 0 x ∈ {-3, -1}
y=-x^2+4x-3
Y = x2 + 4x + 11OK. That's very interesting.Is there a question ?
x2 + 4x - 9 = 5x + 3 ∴ x2 - x - 12 = 0 ∴ (x + 3)(x - 4) = 0 ∴ x ∈ {-3, 4}
x2+y2+4x+2y+3=0(x+2)2 + (Y+2.5)2 = 3This is the equation of circle with center at (-2,-2.5) with radius 3.5