(2, -2)
Y = 4/x^2
No, to be linear, both the power of x and y need to be 1. Since x is squared, the power of x is 2.
X^2 + Y^2 = 36 Y^2 = 36 - X^2 Y = (+/-)sqrt(36 - X^2) A circle. X = (+/-) 6 Y = (+/-) 6
If you are solving for y, it is fine. If you are solving for x, divide both sides by x and the equation should be x = y/x
2
(x, y) = (-3, -3) or (3, 3)
89
(2, -2)
The solutions are: x = 4, y = 2 and x = -4, y = -2
y equals x plus 4 when y equals 0 then x equals 2i i is the square root of negative 1
The two rational solutions are (0,0,0) and (1,1,1). There are no other real solutions.
Y = 4/x^2
They touch each other at (0, 100) on the x and y axis.
Y = X2 forms a parabola
If x = sin θ and y = cos θ then: sin² θ + cos² θ = 1 → x² + y² = 1 → x² = 1 - y²
They are +/- 5*sqrt(2)