Solve x+3/x-3-2/x+3=1

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

Solve for x

$$x=-6$$

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.

$$\left(x+3\right)\left(x+3\right)-\left(x-3\right)\times 2=\left(x-3\right)\left(x+3\right)$$

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

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

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

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

Use the distributive property to multiply x-3 by 2.

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

To find the opposite of 2x-6, find the opposite of each term.

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

Combine 6x and -2x to get 4x.

$$x^{2}+4x+9+6=\left(x-3\right)\left(x+3\right)$$

Add 9 and 6 to get 15.

$$x^{2}+4x+15=\left(x-3\right)\left(x+3\right)$$

Consider \left(x-3\right)\left(x+3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$x^{2}+4x+15=x^{2}-9$$

Subtract x^{2} from both sides.

$$x^{2}+4x+15-x^{2}=-9$$

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

$$4x+15=-9$$

Subtract 15 from both sides.

$$4x=-9-15$$

Subtract 15 from -9 to get -24.

$$4x=-24$$

Divide both sides by 4.

$$x=\frac{-24}{4}$$

Divide -24 by 4 to get -6.

$$x=-6$$