Solve frac{sqrt{3}}{x^2-1}=frac{sqrt{3}}{2x-2}-frac{sqrt{3}}{2x+2}?

$$\frac { \sqrt { 3 } } { x ^ { 2 } - 1 } = \frac { \sqrt { 3 } } { 2 x - 2 } - \frac { \sqrt { 3 } } { 2 x + 2 } ?$$

Solve for x (complex solution)

$$x\in \mathrm{C}\setminus -1,1$$

Solution Steps

Variable x cannot be equal to any of the values -1,1 since division by zero is not defined. Multiply both sides of the equation by 2\left(x-1\right)\left(x+1\right), the least common multiple of x^{2}-1,2x-2,2x+2.

$$2\sqrt{3}=\left(x+1\right)\sqrt{3}-\left(x-1\right)\sqrt{3}$$

Use the distributive property to multiply x+1 by \sqrt{3}.

$$2\sqrt{3}=x\sqrt{3}+\sqrt{3}-\left(x-1\right)\sqrt{3}$$

Use the distributive property to multiply x-1 by \sqrt{3}.

$$2\sqrt{3}=x\sqrt{3}+\sqrt{3}-\left(x\sqrt{3}-\sqrt{3}\right)$$

To find the opposite of x\sqrt{3}-\sqrt{3}, find the opposite of each term.

$$2\sqrt{3}=x\sqrt{3}+\sqrt{3}-x\sqrt{3}+\sqrt{3}$$

Combine x\sqrt{3} and -x\sqrt{3} to get 0.

$$2\sqrt{3}=\sqrt{3}+\sqrt{3}$$

Combine \sqrt{3} and \sqrt{3} to get 2\sqrt{3}.

$$2\sqrt{3}=2\sqrt{3}$$

Subtract 2\sqrt{3} from both sides.

$$2\sqrt{3}-2\sqrt{3}=0$$

Combine 2\sqrt{3} and -2\sqrt{3} to get 0.

$$0=0$$

Compare 0 and 0.

$$\text{true}$$

This is true for any x.

$$x\in \mathrm{C}$$

Variable x cannot be equal to any of the values -1,1.

$$x\in \mathrm{C}\setminus -1,1$$

Solve for x

$$x\in \mathrm{R}\setminus 1,-1$$

Solution Steps

Variable x cannot be equal to any of the values -1,1 since division by zero is not defined. Multiply both sides of the equation by 2\left(x-1\right)\left(x+1\right), the least common multiple of x^{2}-1,2x-2,2x+2.

$$2\sqrt{3}=\left(x+1\right)\sqrt{3}-\left(x-1\right)\sqrt{3}$$

Use the distributive property to multiply x+1 by \sqrt{3}.

$$2\sqrt{3}=x\sqrt{3}+\sqrt{3}-\left(x-1\right)\sqrt{3}$$

Use the distributive property to multiply x-1 by \sqrt{3}.

$$2\sqrt{3}=x\sqrt{3}+\sqrt{3}-\left(x\sqrt{3}-\sqrt{3}\right)$$

To find the opposite of x\sqrt{3}-\sqrt{3}, find the opposite of each term.

$$2\sqrt{3}=x\sqrt{3}+\sqrt{3}-x\sqrt{3}+\sqrt{3}$$

Combine x\sqrt{3} and -x\sqrt{3} to get 0.

$$2\sqrt{3}=\sqrt{3}+\sqrt{3}$$

Combine \sqrt{3} and \sqrt{3} to get 2\sqrt{3}.

$$2\sqrt{3}=2\sqrt{3}$$

Subtract 2\sqrt{3} from both sides.

$$2\sqrt{3}-2\sqrt{3}=0$$

Combine 2\sqrt{3} and -2\sqrt{3} 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 any of the values -1,1.

$$x\in \mathrm{R}\setminus -1,1$$