Solve 1/y=1/2x^2-z

$$\frac{ 1 }{ y } = \frac{ 1 }{ 2 { x }^{ 2 } } -z$$

Solve for y (complex solution)

$$y=-\frac{2x^{2}}{2zx^{2}-1}$$

$$\left(x\neq 0\text{ and }x\neq -\left(2z\right)^{-\frac{1}{2}}\text{ and }x\neq \left(2z\right)^{-\frac{1}{2}}\right)\text{ or }\left(x\neq 0\text{ and }z=0\right)$$

Steps for Solving Linear Equation

Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 2yx^{2}, the least common multiple of y,2x^{2}.

$$2x^{2}=y-z\times 2yx^{2}$$

Multiply -1 and 2 to get -2.

$$2x^{2}=y-2zyx^{2}$$

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

$$y-2zyx^{2}=2x^{2}$$

Combine all terms containing y.

$$\left(1-2zx^{2}\right)y=2x^{2}$$

Divide both sides by 1-2zx^{2}.

$$\frac{\left(1-2zx^{2}\right)y}{1-2zx^{2}}=\frac{2x^{2}}{1-2zx^{2}}$$

Dividing by 1-2zx^{2} undoes the multiplication by 1-2zx^{2}.

$$y=\frac{2x^{2}}{1-2zx^{2}}$$

Variable y cannot be equal to 0.

$$y=\frac{2x^{2}}{1-2zx^{2}}\text{, }y\neq 0$$

Solve for y

$$y=-\frac{2x^{2}}{2zx^{2}-1}$$

$$\left(x\neq 0\text{ and }|x|\neq \frac{1}{\sqrt{2z}}\right)\text{ or }\left(x\neq 0\text{ and }z\leq 0\right)$$

Steps for Solving Linear Equation

Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 2yx^{2}, the least common multiple of y,2x^{2}.

$$2x^{2}=y-z\times 2yx^{2}$$

Multiply -1 and 2 to get -2.

$$2x^{2}=y-2zyx^{2}$$

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

$$y-2zyx^{2}=2x^{2}$$

Combine all terms containing y.

$$\left(1-2zx^{2}\right)y=2x^{2}$$

Divide both sides by 1-2zx^{2}.

$$\frac{\left(1-2zx^{2}\right)y}{1-2zx^{2}}=\frac{2x^{2}}{1-2zx^{2}}$$

Dividing by 1-2zx^{2} undoes the multiplication by 1-2zx^{2}.

$$y=\frac{2x^{2}}{1-2zx^{2}}$$

Variable y cannot be equal to 0.

$$y=\frac{2x^{2}}{1-2zx^{2}}\text{, }y\neq 0$$

Solve for x (complex solution)

$$x=-\frac{i\left(-yz-1\right)^{-\frac{1}{2}}\sqrt{2y}}{2}$$

$$x=\frac{i\left(-yz-1\right)^{-\frac{1}{2}}\sqrt{2y}}{2}\text{, }y\neq 0\text{ and }\left(z=0\text{ or }y\neq -\frac{1}{z}\right)$$

Solve for x

$$x=\frac{\sqrt{\frac{2y}{yz+1}}}{2}$$

$$x=-\frac{\sqrt{\frac{2y}{yz+1}}}{2}\text{, }\left(y<-\frac{1}{z}\text{ and }y<0\text{ and }z>0\right)\text{ or }\left(y>-\frac{1}{z}\text{ and }y<0\text{ and }z<0\right)\text{ or }\left(y>-\frac{1}{z}\text{ and }y>0\text{ and }z>0\right)\text{ or }\left(z=0\text{ and }y>0\right)\text{ or }\left(y<-\frac{1}{z}\text{ and }y>0\text{ and }z<0\right)$$