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

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

Solve for x

$$x=-6$$

$$x=-1$$

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$$-12x+9-9=\left(x-3\right)\left(x-2\right)$$

Subtract 9 from 9 to get 0.

$$-12x=\left(x-3\right)\left(x-2\right)$$

Use the distributive property to multiply x-3 by x-2 and combine like terms.

$$-12x=x^{2}-5x+6$$

Subtract x^{2} from both sides.

$$-12x-x^{2}=-5x+6$$

Add 5x to both sides.

$$-12x-x^{2}+5x=6$$

Combine -12x and 5x to get -7x.

$$-7x-x^{2}=6$$

Subtract 6 from both sides.

$$-7x-x^{2}-6=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-x^{2}-7x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -7 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\left(-1\right)\left(-6\right)}}{2\left(-1\right)}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\left(-1\right)\left(-6\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-7\right)±\sqrt{49+4\left(-6\right)}}{2\left(-1\right)}$$

Multiply 4 times -6.

$$x=\frac{-\left(-7\right)±\sqrt{49-24}}{2\left(-1\right)}$$

Add 49 to -24.

$$x=\frac{-\left(-7\right)±\sqrt{25}}{2\left(-1\right)}$$

Take the square root of 25.

$$x=\frac{-\left(-7\right)±5}{2\left(-1\right)}$$

The opposite of -7 is 7.

$$x=\frac{7±5}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{7±5}{-2}$$

Now solve the equation x=\frac{7±5}{-2} when ± is plus. Add 7 to 5.

$$x=\frac{12}{-2}$$

Divide 12 by -2.

$$x=-6$$

Now solve the equation x=\frac{7±5}{-2} when ± is minus. Subtract 5 from 7.

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

Divide 2 by -2.

$$x=-1$$

The equation is now solved.

$$x=-6 x=-1$$

Steps for Completing the Square

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$$-12x+9-9=\left(x-3\right)\left(x-2\right)$$

Subtract 9 from 9 to get 0.

$$-12x=\left(x-3\right)\left(x-2\right)$$

Use the distributive property to multiply x-3 by x-2 and combine like terms.

$$-12x=x^{2}-5x+6$$

Subtract x^{2} from both sides.

$$-12x-x^{2}=-5x+6$$

Add 5x to both sides.

$$-12x-x^{2}+5x=6$$

Combine -12x and 5x to get -7x.

$$-7x-x^{2}=6$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-x^{2}-7x=6$$

Divide both sides by -1.

$$\frac{-x^{2}-7x}{-1}=\frac{6}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{7}{-1}\right)x=\frac{6}{-1}$$

Divide -7 by -1.

$$x^{2}+7x=\frac{6}{-1}$$

Divide 6 by -1.

$$x^{2}+7x=-6$$

Divide 7, the coefficient of the x term, by 2 to get \frac{7}{2}. Then add the square of \frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+7x+\left(\frac{7}{2}\right)^{2}=-6+\left(\frac{7}{2}\right)^{2}$$

Square \frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+7x+\frac{49}{4}=-6+\frac{49}{4}$$

Add -6 to \frac{49}{4}.

$$x^{2}+7x+\frac{49}{4}=\frac{25}{4}$$

Factor x^{2}+7x+\frac{49}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{7}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x+\frac{7}{2}=\frac{5}{2} x+\frac{7}{2}=-\frac{5}{2}$$

Subtract \frac{7}{2} from both sides of the equation.

$$x=-1 x=-6$$