Solve by Factoring 1/(x+3)-2/(3-x)=4/(x^2-9)

$\frac{1}{x + 3} - \frac{2}{3 - x} = \frac{4}{x^{2} - 9}$

Step 1

Subtract $\frac{4}{x^{2} - 9}$ from both sides of the equation.

$\frac{1}{x + 3} - \frac{2}{3 - x} - \frac{4}{x^{2} - 9} = 0$

Step 2

Simplify $\frac{1}{x + 3} - \frac{2}{3 - x} - \frac{4}{x^{2} - 9}$ .

Simplify the denominator.

Rewrite $9$ as $3^{2}$ .

$\frac{1}{x + 3} - \frac{2}{3 - x} - \frac{4}{x^{2} - 3^{2}} = 0$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 3$ .

$\frac{1}{x + 3} - \frac{2}{3 - x} - \frac{4}{(x + 3) (x - 3)} = 0$

To write $\frac{1}{x + 3}$ as a fraction with a common denominator, multiply by $\frac{3 - x}{3 - x}$ .

$\frac{1}{x + 3} \cdot \frac{3 - x}{3 - x} - \frac{2}{3 - x} - \frac{4}{(x + 3) (x - 3)} = 0$

To write $- \frac{2}{3 - x}$ as a fraction with a common denominator, multiply by $\frac{x + 3}{x + 3}$ .

$\frac{1}{x + 3} \cdot \frac{3 - x}{3 - x} - \frac{2}{3 - x} \cdot \frac{x + 3}{x + 3} - \frac{4}{(x + 3) (x - 3)} = 0$

Write each expression with a common denominator of $(x + 3) (3 - x)$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{1}{x + 3}$ by $\frac{3 - x}{3 - x}$ .

$\frac{3 - x}{(x + 3) (3 - x)} - \frac{2}{3 - x} \cdot \frac{x + 3}{x + 3} - \frac{4}{(x + 3) (x - 3)} = 0$

Multiply $\frac{2}{3 - x}$ by $\frac{x + 3}{x + 3}$ .

$\frac{3 - x}{(x + 3) (3 - x)} - \frac{2 (x + 3)}{(3 - x) (x + 3)} - \frac{4}{(x + 3) (x - 3)} = 0$

Reorder the factors of $(x + 3) (3 - x)$ .

$\frac{3 - x}{(3 - x) (x + 3)} - \frac{2 (x + 3)}{(3 - x) (x + 3)} - \frac{4}{(x + 3) (x - 3)} = 0$

Combine the numerators over the common denominator.

$\frac{3 - x - 2 (x + 3)}{(3 - x) (x + 3)} - \frac{4}{(x + 3) (x - 3)} = 0$

Simplify the numerator.

Apply the distributive property.

$\frac{3 - x - 2 x - 2 \cdot 3}{(3 - x) (x + 3)} - \frac{4}{(x + 3) (x - 3)} = 0$

Multiply $- 2$ by $3$ .

$\frac{3 - x - 2 x - 6}{(3 - x) (x + 3)} - \frac{4}{(x + 3) (x - 3)} = 0$

Subtract $6$ from $3$ .

$\frac{- x - 2 x - 3}{(3 - x) (x + 3)} - \frac{4}{(x + 3) (x - 3)} = 0$

Subtract $2 x$ from $- x$ .

$\frac{- 3 x - 3}{(3 - x) (x + 3)} - \frac{4}{(x + 3) (x - 3)} = 0$

Factor $3$ out of $- 3 x - 3$ .

Factor $3$ out of $- 3 x$ .

$\frac{3 (- x) - 3}{(3 - x) (x + 3)} - \frac{4}{(x + 3) (x - 3)} = 0$

Factor $3$ out of $- 3$ .

$\frac{3 (- x) + 3 (- 1)}{(3 - x) (x + 3)} - \frac{4}{(x + 3) (x - 3)} = 0$

Factor $3$ out of $3 (- x) + 3 (- 1)$ .

$\frac{3 (- x - 1)}{(3 - x) (x + 3)} - \frac{4}{(x + 3) (x - 3)} = 0$

Factor $- 1$ out of $x$ .

$\frac{3 (- x - 1)}{(3 - x) (x + 3)} - \frac{4}{(x + 3) (- 1 (- x) - 3)} = 0$

Rewrite $- 3$ as $- 1 (3)$ .

$\frac{3 (- x - 1)}{(3 - x) (x + 3)} - \frac{4}{(x + 3) (- 1 (- x) - 1 (3))} = 0$

Factor $- 1$ out of $- 1 (- x) - 1 (3)$ .

$\frac{3 (- x - 1)}{(3 - x) (x + 3)} - \frac{4}{(x + 3) (- 1 (- x + 3))} = 0$

Reorder terms.

$\frac{3 (- x - 1)}{(- x + 3) (x + 3)} - \frac{4}{(x + 3) (- 1 (- x + 3))} = 0$

To write $\frac{3 (- x - 1)}{(- x + 3) (x + 3)}$ as a fraction with a common denominator, multiply by $\frac{- 1}{- 1}$ .

$\frac{3 (- x - 1)}{(- x + 3) (x + 3)} \cdot \frac{- 1}{- 1} - \frac{4}{(x + 3) (- 1 (- x + 3))} = 0$

Write each expression with a common denominator of $(- x + 3) (x + 3) \cdot - 1$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{3 (- x - 1)}{(- x + 3) (x + 3)}$ by $\frac{- 1}{- 1}$ .

$\frac{3 (- x - 1) \cdot - 1}{(- x + 3) (x + 3) \cdot - 1} - \frac{4}{(x + 3) (- 1 (- x + 3))} = 0$

Reorder the factors of $(- x + 3) (x + 3) \cdot - 1$ .

$\frac{3 (- x - 1) \cdot - 1}{- (- x + 3) (x + 3)} - \frac{4}{(x + 3) (- 1 (- x + 3))} = 0$

Reorder the factors of $(x + 3) (- 1 (- x + 3))$ .

$\frac{3 (- x - 1) \cdot - 1}{- (- x + 3) (x + 3)} - \frac{4}{- (- x + 3) (x + 3)} = 0$

Combine the numerators over the common denominator.

$\frac{3 (- x - 1) \cdot - 1 - 4}{- (- x + 3) (x + 3)} = 0$

Simplify the numerator.

Apply the distributive property.

$\frac{(3 (- x) + 3 \cdot - 1) \cdot - 1 - 4}{- (- x + 3) (x + 3)} = 0$

Multiply $- 1$ by $3$ .

$\frac{(- 3 x + 3 \cdot - 1) \cdot - 1 - 4}{- (- x + 3) (x + 3)} = 0$

Multiply $3$ by $- 1$ .

$\frac{(- 3 x - 3) \cdot - 1 - 4}{- (- x + 3) (x + 3)} = 0$

Apply the distributive property.

$\frac{- 3 x \cdot - 1 - 3 \cdot - 1 - 4}{- (- x + 3) (x + 3)} = 0$

Multiply $- 1$ by $- 3$ .

$\frac{3 x - 3 \cdot - 1 - 4}{- (- x + 3) (x + 3)} = 0$

Multiply $- 3$ by $- 1$ .

$\frac{3 x + 3 - 4}{- (- x + 3) (x + 3)} = 0$

Subtract $4$ from $3$ .

$\frac{3 x - 1}{- (- x + 3) (x + 3)} = 0$

Move the negative in front of the fraction.

$- \frac{3 x - 1}{(- x + 3) (x + 3)} = 0$

Factor $- 1$ out of $- x$ .

$- \frac{3 x - 1}{(- (x) + 3) (x + 3)} = 0$

Rewrite $3$ as $- 1 (- 3)$ .

$- \frac{3 x - 1}{(- (x) - 1 (- 3)) (x + 3)} = 0$

Factor $- 1$ out of $- (x) - 1 (- 3)$ .

$- \frac{3 x - 1}{- (x - 3) (x + 3)} = 0$

Rewrite $- (x - 3)$ as $- 1 (x - 3)$ .

$- \frac{3 x - 1}{- 1 (x - 3) (x + 3)} = 0$

Move the negative in front of the fraction.

$- - \frac{3 x - 1}{(x - 3) (x + 3)} = 0$

Multiply $- 1$ by $- 1$ .

$1 \frac{3 x - 1}{(x - 3) (x + 3)} = 0$

Multiply $\frac{3 x - 1}{(x - 3) (x + 3)}$ by $1$ .

$\frac{3 x - 1}{(x - 3) (x + 3)} = 0$

Step 3

Set the numerator equal to zero.

$3 x - 1 = 0$

Step 4

Solve the equation for $x$ .

Add $1$ to both sides of the equation.

$3 x = 1$

Divide each term in $3 x = 1$ by $3$ and simplify.

Divide each term in $3 x = 1$ by $3$ .

$\frac{3 x}{3} = \frac{1}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{1}{3}$

Divide $x$ by $1$ .

$x = \frac{1}{3}$

Solve by Factoring 1/(x+3)-2/(3-x)=4/(x^2-9)

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