Solve the Inequality for x 3/(x-3)<=2/(x+2)

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

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

$\frac{3}{x - 3} - \frac{2}{x + 2} \leq 0$

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

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

$\frac{3}{x - 3} \cdot \frac{x + 2}{x + 2} - \frac{2}{x + 2} \leq 0$

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

$\frac{3}{x - 3} \cdot \frac{x + 2}{x + 2} - \frac{2}{x + 2} \cdot \frac{x - 3}{x - 3} \leq 0$

Write each expression with a common denominator of $(x - 3) (x + 2)$ , by multiplying each by an appropriate factor of $1$ .

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

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

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

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

Reorder the factors of $(x - 3) (x + 2)$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Apply the distributive property.

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

Multiply $3$ by $2$ .

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

Apply the distributive property.

$\frac{3 x + 6 - 2 x - 2 \cdot - 3}{(x + 2) (x - 3)} \leq 0$

Multiply $- 2$ by $- 3$ .

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

Subtract $2 x$ from $3 x$ .

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

Add $6$ and $6$ .

$\frac{x + 12}{(x + 2) (x - 3)} \leq 0$

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

$x + 12 = 0$

$x + 2 = 0$

$x - 3 = 0$

Subtract $12$ from both sides of the equation.

$x = - 12$

Subtract $2$ from both sides of the equation.

$x = - 2$

Add $3$ to 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.

$x = - 12$

$x = - 2$

$x = 3$

Consolidate the solutions.

$x = - 12, - 2, 3$

Find the domain of $\frac{x + 12}{(x + 2) (x - 3)}$ .

Set the denominator in $\frac{x + 12}{(x + 2) (x - 3)}$ equal to $0$ to find where the expression is undefined.

$(x + 2) (x - 3) = 0$

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 $0$ .

$x + 2 = 0$

$x - 3 = 0$

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

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.

$x = 3$

The final solution is all the values that make $(x + 2) (x - 3) = 0$ true.

$x = - 2, 3$

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

$(- \infty, - 2) \cup (- 2, 3) \cup (3, \infty)$

Use each root to create test intervals.

$x < - 12$

$- 12 < x < - 2$

$- 2 < x < 3$

$x > 3$

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 $x < - 12$ to see if it makes the inequality true.

Choose a value on the interval $x < - 12$ and see if this value makes the original inequality true.

$x = - 14$

Replace $x$ with $- 14$ in the original inequality.

$\frac{3}{(- 14) - 3} \leq \frac{2}{(- 14) + 2}$

The left side $- 0.17647058$ is less than the right side $- 0.1\overline{6}$ , which means that the given statement is always true.

True

Test a value on the interval $- 12 < x < - 2$ to see if it makes the inequality true.

Choose a value on the interval $- 12 < x < - 2$ and see if this value makes the original inequality true.

$x = - 7$

Replace $x$ with $- 7$ in the original inequality.

$\frac{3}{(- 7) - 3} \leq \frac{2}{(- 7) + 2}$

The left side $- 0.3$ is greater than the right side $- 0.4$ , which means that the given statement is false.

False

Test a value on the interval $- 2 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $- 2 < x < 3$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

$\frac{3}{(0) - 3} \leq \frac{2}{(0) + 2}$

The left side $- 1$ is less than the right side $1$ , which means that the given statement is always true.

True

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.

$\frac{3}{(6) - 3} \leq \frac{2}{(6) + 2}$

The left side $1$ is greater than the right side $0.25$ , which means that the given statement is false.

False

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

$x < - 12$ True

$- 12 < x < - 2$ False

$- 2 < x < 3$ True

$x > 3$ False

$x < - 12$ True

$- 12 < x < - 2$ False

$- 2 < x < 3$ True

$x > 3$ False

The solution consists of all of the true intervals.

$x \leq - 12$ or $- 2 < x < 3$

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 12 or - 2 < x < 3$

Interval Notation:

$(- \infty, - 12] \cup (- 2, 3)$

Solve the Inequality for x 3/(x-3)<=2/(x+2)

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