Solve x/x-3+2x/x+3=18/x^2-9

$$\frac { x } { x - 3 } + \frac { 2 x } { x + 3 } = \frac { 18 } { x ^ { 2 } - 9 }$$

Solve for x

$$x=-2$$

Steps Using Factoring By Grouping

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,x^{2}-9.

$$\left(x+3\right)x+\left(x-3\right)\times 2x=18$$

Use the distributive property to multiply x+3 by x.

$$x^{2}+3x+\left(x-3\right)\times 2x=18$$

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

$$x^{2}+3x+\left(2x-6\right)x=18$$

Use the distributive property to multiply 2x-6 by x.

$$x^{2}+3x+2x^{2}-6x=18$$

Combine x^{2} and 2x^{2} to get 3x^{2}.

$$3x^{2}+3x-6x=18$$

Combine 3x and -6x to get -3x.

$$3x^{2}-3x=18$$

Subtract 18 from both sides.

$$3x^{2}-3x-18=0$$

Divide both sides by 3.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=-1 ab=1\left(-6\right)=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -1.

$$a=-3 b=2$$

Rewrite x^{2}-x-6 as \left(x^{2}-3x\right)+\left(2x-6\right).

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

Factor out x in the first and 2 in the second group.

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

Factor out common term x-3 by using distributive property.

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

To find equation solutions, solve x-3=0 and x+2=0.

$$x=3 x=-2$$

Variable x cannot be equal to 3.

$$x=-2$$

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,x^{2}-9.

$$\left(x+3\right)x+\left(x-3\right)\times 2x=18$$

Use the distributive property to multiply x+3 by x.

$$x^{2}+3x+\left(x-3\right)\times 2x=18$$

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

$$x^{2}+3x+\left(2x-6\right)x=18$$

Use the distributive property to multiply 2x-6 by x.

$$x^{2}+3x+2x^{2}-6x=18$$

Combine x^{2} and 2x^{2} to get 3x^{2}.

$$3x^{2}+3x-6x=18$$

Combine 3x and -6x to get -3x.

$$3x^{2}-3x=18$$

Subtract 18 from both sides.

$$3x^{2}-3x-18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -3 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 3\left(-18\right)}}{2\times 3}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 3\left(-18\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-3\right)±\sqrt{9-12\left(-18\right)}}{2\times 3}$$

Multiply -12 times -18.

$$x=\frac{-\left(-3\right)±\sqrt{9+216}}{2\times 3}$$

Add 9 to 216.

$$x=\frac{-\left(-3\right)±\sqrt{225}}{2\times 3}$$

Take the square root of 225.

$$x=\frac{-\left(-3\right)±15}{2\times 3}$$

The opposite of -3 is 3.

$$x=\frac{3±15}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{3±15}{6}$$

Now solve the equation x=\frac{3±15}{6} when ± is plus. Add 3 to 15.

$$x=\frac{18}{6}$$

Divide 18 by 6.

$$x=3$$

Now solve the equation x=\frac{3±15}{6} when ± is minus. Subtract 15 from 3.

$$x=-\frac{12}{6}$$

Divide -12 by 6.

$$x=-2$$

The equation is now solved.

$$x=3 x=-2$$

Variable x cannot be equal to 3.

$$x=-2$$

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,x^{2}-9.

$$\left(x+3\right)x+\left(x-3\right)\times 2x=18$$

Use the distributive property to multiply x+3 by x.

$$x^{2}+3x+\left(x-3\right)\times 2x=18$$

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

$$x^{2}+3x+\left(2x-6\right)x=18$$

Use the distributive property to multiply 2x-6 by x.

$$x^{2}+3x+2x^{2}-6x=18$$

Combine x^{2} and 2x^{2} to get 3x^{2}.

$$3x^{2}+3x-6x=18$$

Combine 3x and -6x to get -3x.

$$3x^{2}-3x=18$$

Divide both sides by 3.

$$\frac{3x^{2}-3x}{3}=\frac{18}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\left(-\frac{3}{3}\right)x=\frac{18}{3}$$

Divide -3 by 3.

$$x^{2}-x=\frac{18}{3}$$

Divide 18 by 3.

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

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=6+\frac{1}{4}$$

Add 6 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{25}{4}$$

Factor x^{2}-x+\frac{1}{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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{5}{2} x-\frac{1}{2}=-\frac{5}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$x=3 x=-2$$

Variable x cannot be equal to 3.

$$x=-2$$