Solve f:y=1/x^4+1

$$f : y = \frac { 1 } { x ^ { 4 } } + 1$$

Solve for f

$$f=y+\frac{y}{x^{4}}$$

$$y\neq 0\text{ and }x\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by yx^{4}, the least common multiple of y,x^{4}.

$$x^{4}f=y+yx^{4}$$

The equation is in standard form.

$$x^{4}f=yx^{4}+y$$

Divide both sides by x^{4}.

$$\frac{x^{4}f}{x^{4}}=\frac{yx^{4}+y}{x^{4}}$$

Dividing by x^{4} undoes the multiplication by x^{4}.

$$f=\frac{yx^{4}+y}{x^{4}}$$

Divide y+yx^{4} by x^{4}.

$$f=y+\frac{y}{x^{4}}$$

Solve for x (complex solution)

$$x=\sqrt{2}\left(-\frac{1}{2}+\frac{1}{2}i\right)\left(y-f\right)^{-\frac{1}{4}}\sqrt[4]{y}$$

$$x=\sqrt{2}\left(\frac{1}{2}+\frac{1}{2}i\right)\left(y-f\right)^{-\frac{1}{4}}\sqrt[4]{y}$$

$$x=\sqrt{2}\left(-\frac{1}{2}-\frac{1}{2}i\right)\left(y-f\right)^{-\frac{1}{4}}\sqrt[4]{y}$$

$$x=\sqrt{2}\left(\frac{1}{2}-\frac{1}{2}i\right)\left(y-f\right)^{-\frac{1}{4}}\sqrt[4]{y}\text{, }y\neq 0\text{ and }f\neq y$$

Solve for x

$$x=\sqrt[4]{-\frac{y}{y-f}}$$

$$x=-\sqrt[4]{-\frac{y}{y-f}}\text{, }\left(y>0\text{ and }f>y\right)\text{ or }\left(y<0\text{ and }f<y\right)$$