Solve 1/3-x+1/3+x=6/9-x^2

$$\frac { 1 } { 3 - x } + \frac { 1 } { 3 + x } = \frac { 6 } { 9 - x ^ { 2 } }$$

Solve for x (complex solution)

$$x\in \mathrm{C}\setminus 3,-3$$

Solution Steps

Variable x cannot be equal to any of the values -3,3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)\left(x+3\right), the least common multiple of 3-x,3+x,9-x^{2}.

$$-\left(3+x\right)+x-3=-6$$

To find the opposite of 3+x, find the opposite of each term.

$$-3-x+x-3=-6$$

Combine -x and x to get 0.

$$-3-3=-6$$

Subtract 3 from -3 to get -6.

$$-6=-6$$

Compare -6 and -6.

$$\text{true}$$

This is true for any x.

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

Variable x cannot be equal to any of the values -3,3.

$$x\in \mathrm{C}\setminus -3,3$$

Solve for x

$$x\in \mathrm{R}\setminus 3,-3$$

Solution Steps

Variable x cannot be equal to any of the values -3,3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)\left(x+3\right), the least common multiple of 3-x,3+x,9-x^{2}.

$$-\left(3+x\right)+x-3=-6$$

To find the opposite of 3+x, find the opposite of each term.

$$-3-x+x-3=-6$$

Combine -x and x to get 0.

$$-3-3=-6$$

Subtract 3 from -3 to get -6.

$$-6=-6$$

Compare -6 and -6.

$$\text{true}$$

This is true for any x.

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

Variable x cannot be equal to any of the values -3,3.

$$x\in \mathrm{R}\setminus -3,3$$