Solve 1/x*2x^4-1/y*2x=1/2

$$\frac { 1 } { x } \cdot 2 x ^ { 4 } - \frac { 1 } { y } \cdot 2 x = \frac { 1 } { 2 }$$

Solve for y

$$y=-\frac{4x}{1-4x^{3}}$$

$$x\neq 0\text{ and }x\neq \frac{4^{\frac{2}{3}}}{4}$$

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 2xy, the least common multiple of x,y,2.

$$2y\times 1\times 2x^{4}-2x\times 1\times 2x=xy$$

Multiply 2 and 1 to get 2.

$$2y\times 2x^{4}-2x\times 1\times 2x=xy$$

Multiply 2 and 2 to get 4.

$$4yx^{4}-2x\times 1\times 2x=xy$$

Multiply x and x to get x^{2}.

$$4yx^{4}-2x^{2}\times 1\times 2=xy$$

Multiply 2 and 1 to get 2.

$$4yx^{4}-2x^{2}\times 2=xy$$

Multiply 2 and 2 to get 4.

$$4yx^{4}-4x^{2}=xy$$

Subtract xy from both sides.

$$4yx^{4}-4x^{2}-xy=0$$

Add 4x^{2} to both sides. Anything plus zero gives itself.

$$4yx^{4}-xy=4x^{2}$$

Combine all terms containing y.

$$\left(4x^{4}-x\right)y=4x^{2}$$

Divide both sides by 4x^{4}-x.

$$\frac{\left(4x^{4}-x\right)y}{4x^{4}-x}=\frac{4x^{2}}{4x^{4}-x}$$

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

$$y=\frac{4x^{2}}{4x^{4}-x}$$

Divide 4x^{2} by 4x^{4}-x.

$$y=\frac{4x}{4x^{3}-1}$$

Variable y cannot be equal to 0.

$$y=\frac{4x}{4x^{3}-1}\text{, }y\neq 0$$