Solve the Inequality for a (a+1)/(a-2)>3

$\frac{a + 1}{a - 2} > 3$

Subtract $3$ from both sides of the inequality.

$\frac{a + 1}{a - 2} - 3 > 0$

Simplify $\frac{a + 1}{a - 2} - 3$ .

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

$\frac{a + 1}{a - 2} - 3 \cdot \frac{a - 2}{a - 2} > 0$

Combine $- 3$ and $\frac{a - 2}{a - 2}$ .

$\frac{a + 1}{a - 2} + \frac{- 3 (a - 2)}{a - 2} > 0$

Combine the numerators over the common denominator.

$\frac{a + 1 - 3 (a - 2)}{a - 2} > 0$

Simplify the numerator.

Apply the distributive property.

$\frac{a + 1 - 3 a - 3 \cdot - 2}{a - 2} > 0$

Multiply $- 3$ by $- 2$ .

$\frac{a + 1 - 3 a + 6}{a - 2} > 0$

Subtract $3 a$ from $a$ .

$\frac{- 2 a + 1 + 6}{a - 2} > 0$

Add $1$ and $6$ .

$\frac{- 2 a + 7}{a - 2} > 0$

Factor $- 1$ out of $- 2 a$ .

$\frac{- (2 a) + 7}{a - 2} > 0$

Rewrite $7$ as $- 1 (- 7)$ .

$\frac{- (2 a) - 1 (- 7)}{a - 2} > 0$

Factor $- 1$ out of $- (2 a) - 1 (- 7)$ .

$\frac{- (2 a - 7)}{a - 2} > 0$

Rewrite $- (2 a - 7)$ as $- 1 (2 a - 7)$ .

$\frac{- 1 (2 a - 7)}{a - 2} > 0$

Move the negative in front of the fraction.

$- \frac{2 a - 7}{a - 2} > 0$

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

$2 a - 7 = 0$

$a - 2 = 0$

Add $7$ to both sides of the equation.

$2 a = 7$

Divide each term in $2 a = 7$ by $2$ and simplify.

Divide each term in $2 a = 7$ by $2$ .

$\frac{2 a}{2} = \frac{7}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 a}{2} = \frac{7}{2}$

Divide $a$ by $1$ .

$a = \frac{7}{2}$

Add $2$ to both sides of the equation.

$a = 2$

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

$a = \frac{7}{2}$

$a = 2$

Consolidate the solutions.

$a = \frac{7}{2}, 2$

Find the domain of $- \frac{2 a - 7}{a - 2}$ .

Set the denominator in $\frac{2 a - 7}{a - 2}$ equal to $0$ to find where the expression is undefined.

$a - 2 = 0$

Add $2$ to both sides of the equation.

$a = 2$

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

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

Use each root to create test intervals.

$a < 2$

$2 < a < \frac{7}{2}$

$a > \frac{7}{2}$

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 $a < 2$ to see if it makes the inequality true.

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

$a = 0$

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

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

The left side $- 0.5$ is not greater than the right side $3$ , which means that the given statement is false.

False

Test a value on the interval $2 < a < \frac{7}{2}$ to see if it makes the inequality true.

Choose a value on the interval $2 < a < \frac{7}{2}$ and see if this value makes the original inequality true.

$a = 3$

Replace $a$ with $3$ in the original inequality.

$\frac{(3) + 1}{(3) - 2} > 3$

The left side $4$ is greater than the right side $3$ , which means that the given statement is always true.

True

Test a value on the interval $a > \frac{7}{2}$ to see if it makes the inequality true.

Choose a value on the interval $a > \frac{7}{2}$ and see if this value makes the original inequality true.

$a = 6$

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

$\frac{(6) + 1}{(6) - 2} > 3$

The left side $1.75$ is not greater than the right side $3$ , which means that the given statement is false.

False

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

$a < 2$ False

$2 < a < \frac{7}{2}$ True

$a > \frac{7}{2}$ False

$a < 2$ False

$2 < a < \frac{7}{2}$ True

$a > \frac{7}{2}$ False

The solution consists of all of the true intervals.

$2 < a < \frac{7}{2}$

The result can be shown in multiple forms.

Inequality Form:

$2 < a < \frac{7}{2}$

Interval Notation:

$(2, \frac{7}{2})$

Solve the Inequality for a (a+1)/(a-2)>3

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