Solve 3x^2+9999996z^2=x^2+y

$$3 { x }^{ 2 } +9999996 { z }^{ 2 } = { x }^{ 2 } +y$$

Solve for y

$$y=2\left(x^{2}+4999998z^{2}\right)$$

Solution Steps

Swap sides so that all variable terms are on the left hand side.

$$x^{2}+y=3x^{2}+9999996z^{2}$$

Subtract x^{2} from both sides.

$$y=3x^{2}+9999996z^{2}-x^{2}$$

Combine 3x^{2} and -x^{2} to get 2x^{2}.

$$y=2x^{2}+9999996z^{2}$$

Solve for x (complex solution)

$$x=-\frac{\sqrt{2\left(y-9999996z^{2}\right)}}{2}$$

$$x=\frac{\sqrt{2\left(y-9999996z^{2}\right)}}{2}$$

Solve for x

$$x=\frac{\sqrt{2\left(y-9999996z^{2}\right)}}{2}$$

$$x=-\frac{\sqrt{2\left(y-9999996z^{2}\right)}}{2}\text{, }y\geq 9999996z^{2}$$