Solve the Inequality for x x/(x+3)-3/(3-x)>(x+9)/(x^2-9)

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

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

$\frac{x}{x + 3} - \frac{3}{3 - x} - \frac{x + 9}{x^{2} - 9} > 0$

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

Simplify the denominator.

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

$\frac{x}{x + 3} - \frac{3}{3 - x} - \frac{x + 9}{x^{2} - 3^{2}} > 0$

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 $b = 3$ .

$\frac{x}{x + 3} - \frac{3}{3 - x} - \frac{x + 9}{(x + 3) (x - 3)} > 0$

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

$\frac{x}{x + 3} \cdot \frac{3 - x}{3 - x} - \frac{3}{3 - x} - \frac{x + 9}{(x + 3) (x - 3)} > 0$

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

$\frac{x}{x + 3} \cdot \frac{3 - x}{3 - x} - \frac{3}{3 - x} \cdot \frac{x + 3}{x + 3} - \frac{x + 9}{(x + 3) (x - 3)} > 0$

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

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

$\frac{x (3 - x)}{(x + 3) (3 - x)} - \frac{3}{3 - x} \cdot \frac{x + 3}{x + 3} - \frac{x + 9}{(x + 3) (x - 3)} > 0$

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

$\frac{x (3 - x)}{(x + 3) (3 - x)} - \frac{3 (x + 3)}{(3 - x) (x + 3)} - \frac{x + 9}{(x + 3) (x - 3)} > 0$

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

$\frac{x (3 - x)}{(3 - x) (x + 3)} - \frac{3 (x + 3)}{(3 - x) (x + 3)} - \frac{x + 9}{(x + 3) (x - 3)} > 0$

Combine the numerators over the common denominator.

$\frac{x (3 - x) - 3 (x + 3)}{(3 - x) (x + 3)} - \frac{x + 9}{(x + 3) (x - 3)} > 0$

Simplify the numerator.

Apply the distributive property.

$\frac{x \cdot 3 + x (- x) - 3 (x + 3)}{(3 - x) (x + 3)} - \frac{x + 9}{(x + 3) (x - 3)} > 0$

Move $3$ to the left of $x$ .

$\frac{3 \cdot x + x (- x) - 3 (x + 3)}{(3 - x) (x + 3)} - \frac{x + 9}{(x + 3) (x - 3)} > 0$

Rewrite using the commutative property of multiplication.

$\frac{3 \cdot x - x \cdot x - 3 (x + 3)}{(3 - x) (x + 3)} - \frac{x + 9}{(x + 3) (x - 3)} > 0$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$\frac{3 x - (x \cdot x) - 3 (x + 3)}{(3 - x) (x + 3)} - \frac{x + 9}{(x + 3) (x - 3)} > 0$

Multiply $x$ by $x$ .

$\frac{3 x - x^{2} - 3 (x + 3)}{(3 - x) (x + 3)} - \frac{x + 9}{(x + 3) (x - 3)} > 0$

Apply the distributive property.

$\frac{3 x - x^{2} - 3 x - 3 \cdot 3}{(3 - x) (x + 3)} - \frac{x + 9}{(x + 3) (x - 3)} > 0$

Multiply $- 3$ by $3$ .

$\frac{3 x - x^{2} - 3 x - 9}{(3 - x) (x + 3)} - \frac{x + 9}{(x + 3) (x - 3)} > 0$

Subtract $3 x$ from $3 x$ .

$\frac{- x^{2} + 0 - 9}{(3 - x) (x + 3)} - \frac{x + 9}{(x + 3) (x - 3)} > 0$

Add $- x^{2}$ and $0$ .

$\frac{- x^{2} - 9}{(3 - x) (x + 3)} - \frac{x + 9}{(x + 3) (x - 3)} > 0$

Factor $- 1$ out of $x$ .

$\frac{- x^{2} - 9}{(3 - x) (x + 3)} - \frac{x + 9}{(x + 3) (- 1 (- x) - 3)} > 0$

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

$\frac{- x^{2} - 9}{(3 - x) (x + 3)} - \frac{x + 9}{(x + 3) (- 1 (- x) - 1 (3))} > 0$

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

$\frac{- x^{2} - 9}{(3 - x) (x + 3)} - \frac{x + 9}{(x + 3) (- 1 (- x + 3))} > 0$

Reorder terms.

$\frac{- x^{2} - 9}{(- x + 3) (x + 3)} - \frac{x + 9}{(x + 3) (- 1 (- x + 3))} > 0$

To write $\frac{- x^{2} - 9}{(- x + 3) (x + 3)}$ as a fraction with a common denominator, multiply by $\frac{- 1}{- 1}$ .

$\frac{- x^{2} - 9}{(- x + 3) (x + 3)} \cdot \frac{- 1}{- 1} - \frac{x + 9}{(x + 3) (- 1 (- x + 3))} > 0$

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

Multiply $\frac{- x^{2} - 9}{(- x + 3) (x + 3)}$ by $\frac{- 1}{- 1}$ .

$\frac{(- x^{2} - 9) \cdot - 1}{(- x + 3) (x + 3) \cdot - 1} - \frac{x + 9}{(x + 3) (- 1 (- x + 3))} > 0$

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

$\frac{(- x^{2} - 9) \cdot - 1}{- (- x + 3) (x + 3)} - \frac{x + 9}{(x + 3) (- 1 (- x + 3))} > 0$

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

$\frac{(- x^{2} - 9) \cdot - 1}{- (- x + 3) (x + 3)} - \frac{x + 9}{- (- x + 3) (x + 3)} > 0$

Combine the numerators over the common denominator.

$\frac{(- x^{2} - 9) \cdot - 1 - (x + 9)}{- (- x + 3) (x + 3)} > 0$

Simplify the numerator.

Apply the distributive property.

$\frac{- x^{2} \cdot - 1 - 9 \cdot - 1 - (x + 9)}{- (- x + 3) (x + 3)} > 0$

Multiply $- x^{2} \cdot - 1$ .

Multiply $- 1$ by $- 1$ .

$\frac{1 x^{2} - 9 \cdot - 1 - (x + 9)}{- (- x + 3) (x + 3)} > 0$

Multiply $x^{2}$ by $1$ .

$\frac{x^{2} - 9 \cdot - 1 - (x + 9)}{- (- x + 3) (x + 3)} > 0$

Multiply $- 9$ by $- 1$ .

$\frac{x^{2} + 9 - (x + 9)}{- (- x + 3) (x + 3)} > 0$

Apply the distributive property.

$\frac{x^{2} + 9 - x - 1 \cdot 9}{- (- x + 3) (x + 3)} > 0$

Multiply $- 1$ by $9$ .

$\frac{x^{2} + 9 - x - 9}{- (- x + 3) (x + 3)} > 0$

Subtract $9$ from $9$ .

$\frac{x^{2} - x + 0}{- (- x + 3) (x + 3)} > 0$

Add $x^{2} - x$ and $0$ .

$\frac{x^{2} - x}{- (- x + 3) (x + 3)} > 0$

Factor $x$ out of $x^{2} - x$ .

Factor $x$ out of $x^{2}$ .

$\frac{x \cdot x - x}{- (- x + 3) (x + 3)} > 0$

Factor $x$ out of $- x$ .

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

Factor $x$ out of $x \cdot x + x \cdot - 1$ .

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

Move the negative in front of the fraction.

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

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

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

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

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

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

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

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

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

Move the negative in front of the fraction.

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

Multiply $- 1$ by $- 1$ .

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

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

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

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$x = 0$

$x - 1 = 0$

$x - 3 = 0$

$x + 3 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Add $3$ to both sides of the equation.

$x = 3$

Subtract $3$ from both sides of the equation.

$x = - 3$

Solve for each factor to find the values where the absolute value expression goes from negative to positive.

$x = 0$

$x = 1$

$x = 3$

$x = - 3$

Consolidate the solutions.

$x = 0, 1, 3, - 3$

Find the domain of $\frac{x (x - 1)}{(x - 3) (x + 3)}$ .

Set the denominator in $\frac{x (x - 1)}{(x - 3) (x + 3)}$ equal to $0$ to find where the expression is undefined.

$(x - 3) (x + 3) = 0$

Solve for $x$ .

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 3 = 0$

$x + 3 = 0$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

The final solution is all the values that make $(x - 3) (x + 3) = 0$ true.

$x = 3, - 3$

The domain is all values of $x$ that make the expression defined.

$(- \infty, - 3) \cup (- 3, 3) \cup (3, \infty)$

Use each root to create test intervals.

$x < - 3$

$- 3 < x < 0$

$0 < x < 1$

$1 < x < 3$

$x > 3$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < - 3$ to see if it makes the inequality true.

Choose a value on the interval $x < - 3$ and see if this value makes the original inequality true.

$x = - 6$

Replace $x$ with $- 6$ in the original inequality.

$\frac{- 6}{(- 6) + 3} - \frac{3}{3 - (- 6)} > \frac{(- 6) + 9}{{(- 6)}^{2} - 9}$

The left side $1.\overline{6}$ is greater than the right side $0.\overline{1}$ , which means that the given statement is always true.

True

Test a value on the interval $- 3 < x < 0$ to see if it makes the inequality true.

Choose a value on the interval $- 3 < x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Replace $x$ with $- 2$ in the original inequality.

$\frac{- 2}{(- 2) + 3} - \frac{3}{3 - (- 2)} > \frac{(- 2) + 9}{{(- 2)}^{2} - 9}$

The left side $- 2.6$ is not greater than the right side $- 1.4$ , which means that the given statement is false.

False

Test a value on the interval $0 < x < 1$ to see if it makes the inequality true.

Choose a value on the interval $0 < x < 1$ and see if this value makes the original inequality true.

$x = 0.5$

Replace $x$ with $0.5$ in the original inequality.

$\frac{0.5}{(0.5) + 3} - \frac{3}{3 - (0.5)} > \frac{(0.5) + 9}{{(0.5)}^{2} - 9}$

The left side $- 1.0\overline{571428}$ is greater than the right side $- 1.0\overline{857142}$ , which means that the given statement is always true.

True

Test a value on the interval $1 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $1 < x < 3$ and see if this value makes the original inequality true.

$x = 2$

Replace $x$ with $2$ in the original inequality.

$\frac{2}{(2) + 3} - \frac{3}{3 - (2)} > \frac{(2) + 9}{{(2)}^{2} - 9}$

The left side $- 2.6$ is not greater than the right side $- 2.2$ , which means that the given statement is false.

False

Test a value on the interval $x > 3$ to see if it makes the inequality true.

Choose a value on the interval $x > 3$ and see if this value makes the original inequality true.

$x = 6$

Replace $x$ with $6$ in the original inequality.

$\frac{6}{(6) + 3} - \frac{3}{3 - (6)} > \frac{(6) + 9}{{(6)}^{2} - 9}$

The left side $1.\overline{6}$ is greater than the right side $0.\overline{5}$ , which means that the given statement is always true.

True

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 3$ True

$- 3 < x < 0$ False

$0 < x < 1$ True

$1 < x < 3$ False

$x > 3$ True

$x < - 3$ True

$- 3 < x < 0$ False

$0 < x < 1$ True

$1 < x < 3$ False

$x > 3$ True

The solution consists of all of the true intervals.

$x < - 3$ or $0 < x < 1$ or $x > 3$

The result can be shown in multiple forms.

Inequality Form:

$x < - 3 or 0 < x < 1 or x > 3$

Interval Notation:

$(- \infty, - 3) \cup (0, 1) \cup (3, \infty)$

Solve the Inequality for x x/(x+3)-3/(3-x)>(x+9)/(x^2-9)

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