Solve 3x^2+8/x=8/x+3x

$$\frac { 3 x ^ { 2 } + 8 } { x } = \frac { 8 } { x } + 3 x$$

Solve for x

$$x\neq 0$$

Solution Steps

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$3x^{2}+8=8+3xx$$

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

$$3x^{2}+8=8+3x^{2}$$

Subtract 8 from both sides.

$$3x^{2}+8-8=3x^{2}$$

Subtract 8 from 8 to get 0.

$$3x^{2}=3x^{2}$$

Subtract 3x^{2} from both sides.

$$3x^{2}-3x^{2}=0$$

Combine 3x^{2} and -3x^{2} to get 0.

$$0=0$$

Compare 0 and 0.

$$\text{true}$$

This is true for any x.

$$x\in \mathrm{R}$$

Variable x cannot be equal to 0.

$$x\in \mathrm{R}\setminus 0$$