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

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

Solve for x

$$x=0$$

Steps for Solving Linear Equation

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}.

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

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 0.

$$-6x+9=6x+9$$

Subtract 6x from both sides.

$$-6x+9-6x=9$$

Combine -6x and -6x to get -12x.

$$-12x+9=9$$

Subtract 9 from both sides.

$$-12x=9-9$$

Subtract 9 from 9 to get 0.

$$-12x=0$$

Product of two numbers is equal to 0 if at least one of them is 0. Since -12 is not equal to 0, x must be equal to 0.

$$x=0$$