Solve 2y^3/2y^4+16y^3=1/y+8

$$\frac { 2 y ^ { 3 } } { 2 y ^ { 4 } + 16 y ^ { 3 } } = \frac { 1 } { y + 8 }$$

Solve for y (complex solution)

$$y\in \mathrm{C}\setminus 0,-8$$

Solution Steps

Variable y cannot be equal to any of the values -8,0 since division by zero is not defined. Multiply both sides of the equation by 2\left(y+8\right)y^{3}, the least common multiple of 2y^{4}+16y^{3},y+8.

$$2y^{3}=2y^{3}$$

This is true for any y.

$$y\in \mathrm{C}$$

Variable y cannot be equal to any of the values 0,-8.

$$y\in \mathrm{C}\setminus -8,0$$

Solve for y

$$y\in \mathrm{R}\setminus -8,0$$

Solution Steps

Variable y cannot be equal to any of the values -8,0 since division by zero is not defined. Multiply both sides of the equation by 2\left(y+8\right)y^{3}, the least common multiple of 2y^{4}+16y^{3},y+8.

$$2y^{3}=2y^{3}$$

This is true for any y.

$$y\in \mathrm{R}$$

Variable y cannot be equal to any of the values 0,-8.

$$y\in \mathrm{R}\setminus -8,0$$