Solve -x-3/x+3=3+x/x-3

$$- \frac{ x-3 }{ x+3 } = \frac{ 3+x }{ x-3 }$$

Solve for x (complex solution)

$$x=-3i$$

$$x=3i$$

Steps by Finding Square Root

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 x+3,x-3.

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

Multiply x-3 and x-3 to get \left(x-3\right)^{2}.

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

Multiply x+3 and 3+x to get \left(x+3\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$-\left(x^{2}-6x+9\right)=\left(x+3\right)^{2}$$

To find the opposite of x^{2}-6x+9, find the opposite of each term.

$$-x^{2}+6x-9=\left(x+3\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$-x^{2}+6x-9=x^{2}+6x+9$$

Subtract x^{2} from both sides.

$$-x^{2}+6x-9-x^{2}=6x+9$$

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

$$-2x^{2}+6x-9=6x+9$$

Subtract 6x from both sides.

$$-2x^{2}+6x-9-6x=9$$

Combine 6x and -6x to get 0.

$$-2x^{2}-9=9$$

Add 9 to both sides.

$$-2x^{2}=9+9$$

Add 9 and 9 to get 18.

$$-2x^{2}=18$$

Divide both sides by -2.

$$x^{2}=\frac{18}{-2}$$

Divide 18 by -2 to get -9.

$$x^{2}=-9$$

The equation is now solved.

$$x=3i x=-3i$$

Steps Using the Quadratic Formula

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 x+3,x-3.

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

Multiply x-3 and x-3 to get \left(x-3\right)^{2}.

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

Multiply x+3 and 3+x to get \left(x+3\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$-\left(x^{2}-6x+9\right)=\left(x+3\right)^{2}$$

To find the opposite of x^{2}-6x+9, find the opposite of each term.

$$-x^{2}+6x-9=\left(x+3\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$-x^{2}+6x-9=x^{2}+6x+9$$

Subtract x^{2} from both sides.

$$-x^{2}+6x-9-x^{2}=6x+9$$

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

$$-2x^{2}+6x-9=6x+9$$

Subtract 6x from both sides.

$$-2x^{2}+6x-9-6x=9$$

Combine 6x and -6x to get 0.

$$-2x^{2}-9=9$$

Subtract 9 from both sides.

$$-2x^{2}-9-9=0$$

Subtract 9 from -9 to get -18.

$$-2x^{2}-18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 0 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-2\right)\left(-18\right)}}{2\left(-2\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-2\right)\left(-18\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{0±\sqrt{8\left(-18\right)}}{2\left(-2\right)}$$

Multiply 8 times -18.

$$x=\frac{0±\sqrt{-144}}{2\left(-2\right)}$$

Take the square root of -144.

$$x=\frac{0±12i}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{0±12i}{-4}$$

Now solve the equation x=\frac{0±12i}{-4} when ± is plus.

$$x=-3i$$

Now solve the equation x=\frac{0±12i}{-4} when ± is minus.

$$x=3i$$

The equation is now solved.

$$x=-3i x=3i$$