Solve the Inequality for x x/2>=5/(x+1)+4

$\frac{x}{2} \geq \frac{5}{x + 1} + 4$

Step 1

Move all the expressions to the left side of the equation.

Subtract $\frac{5}{x + 1}$ from both sides of the inequality.

$\frac{x}{2} - \frac{5}{x + 1} \geq 4$

Subtract $4$ from both sides of the inequality.

$\frac{x}{2} - \frac{5}{x + 1} - 4 \geq 0$

Step 2

Simplify $\frac{x}{2} - \frac{5}{x + 1} - 4$ .

To write $\frac{x}{2}$ as a fraction with a common denominator, multiply by $\frac{x + 1}{x + 1}$ .

$\frac{x}{2} \cdot \frac{x + 1}{x + 1} - \frac{5}{x + 1} - 4 \geq 0$

To write $- \frac{5}{x + 1}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{x}{2} \cdot \frac{x + 1}{x + 1} - \frac{5}{x + 1} \cdot \frac{2}{2} - 4 \geq 0$

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

Multiply $\frac{x}{2}$ by $\frac{x + 1}{x + 1}$ .

$\frac{x (x + 1)}{2 (x + 1)} - \frac{5}{x + 1} \cdot \frac{2}{2} - 4 \geq 0$

Multiply $\frac{5}{x + 1}$ by $\frac{2}{2}$ .

$\frac{x (x + 1)}{2 (x + 1)} - \frac{5 \cdot 2}{(x + 1) \cdot 2} - 4 \geq 0$

Reorder the factors of $(x + 1) \cdot 2$ .

$\frac{x (x + 1)}{2 (x + 1)} - \frac{5 \cdot 2}{2 (x + 1)} - 4 \geq 0$

Combine the numerators over the common denominator.

$\frac{x (x + 1) - 5 \cdot 2}{2 (x + 1)} - 4 \geq 0$

Multiply $- 5$ by $2$ .

$\frac{x (x + 1) - 10}{2 (x + 1)} - 4 \geq 0$

Simplify the numerator.

Apply the distributive property.

$\frac{x \cdot x + x \cdot 1 - 10}{2 (x + 1)} - 4 \geq 0$

Multiply $x$ by $x$ .

$\frac{x^{2} + x \cdot 1 - 10}{2 (x + 1)} - 4 \geq 0$

Multiply $x$ by $1$ .

$\frac{x^{2} + x - 10}{2 (x + 1)} - 4 \geq 0$

To write $- 4$ as a fraction with a common denominator, multiply by $\frac{2 (x + 1)}{2 (x + 1)}$ .

$\frac{x^{2} + x - 10}{2 (x + 1)} - 4 \cdot \frac{2 (x + 1)}{2 (x + 1)} \geq 0$

Combine $- 4$ and $\frac{2 (x + 1)}{2 (x + 1)}$ .

$\frac{x^{2} + x - 10}{2 (x + 1)} + \frac{- 4 (2 (x + 1))}{2 (x + 1)} \geq 0$

Combine the numerators over the common denominator.

$\frac{x^{2} + x - 10 - 4 (2 (x + 1))}{2 (x + 1)} \geq 0$

Simplify the numerator.

Multiply $- 4$ by $2$ .

$\frac{x^{2} + x - 10 - 8 (x + 1)}{2 (x + 1)} \geq 0$

Apply the distributive property.

$\frac{x^{2} + x - 10 - 8 x - 8 \cdot 1}{2 (x + 1)} \geq 0$

Multiply $- 8$ by $1$ .

$\frac{x^{2} + x - 10 - 8 x - 8}{2 (x + 1)} \geq 0$

Subtract $8 x$ from $x$ .

$\frac{x^{2} - 7 x - 10 - 8}{2 (x + 1)} \geq 0$

Subtract $8$ from $- 10$ .

$\frac{x^{2} - 7 x - 18}{2 (x + 1)} \geq 0$

Factor $x^{2} - 7 x - 18$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 18$ and whose sum is $- 7$ .

$- 9, 2$

Write the factored form using these integers.

$\frac{(x - 9) (x + 2)}{2 (x + 1)} \geq 0$

Step 3

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

$x - 9 = 0$

$x + 2 = 0$

$x + 1 = 0$

Step 4

Add $9$ to both sides of the equation.

$x = 9$

Step 5

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 6

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 7

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

$x = 9$

$x = - 2$

$x = - 1$

Step 8

Consolidate the solutions.

$x = 9, - 2, - 1$

Step 9

Find the domain of $\frac{(x - 9) (x + 2)}{2 (x + 1)}$ .

Set the denominator in $\frac{(x - 9) (x + 2)}{2 (x + 1)}$ equal to $0$ to find where the expression is undefined.

$2 (x + 1) = 0$

Solve for $x$ .

Divide each term in $2 (x + 1) = 0$ by $2$ and simplify.

Divide each term in $2 (x + 1) = 0$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x + 1$ by $1$ .

$x + 1 = \frac{0}{2}$

Simplify the right side.

Divide $0$ by $2$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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

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

Step 10

Use each root to create test intervals.

$x < - 2$

$- 2 < x < - 1$

$- 1 < x < 9$

$x > 9$

Step 11

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

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

$x = - 4$

Replace $x$ with $- 4$ in the original inequality.

$\frac{- 4}{2} \geq \frac{5}{(- 4) + 1} + 4$

The left side $- 2$ is less than the right side $2.\overline{3}$ , which means that the given statement is false.

False

Test a value on the interval $- 2 < x < - 1$ to see if it makes the inequality true.

Choose a value on the interval $- 2 < x < - 1$ and see if this value makes the original inequality true.

$x = - 1.5$

Replace $x$ with $- 1.5$ in the original inequality.

$\frac{- 1.5}{2} \geq \frac{5}{(- 1.5) + 1} + 4$

The left side $- 0.75$ is greater than the right side $- 6$ , which means that the given statement is always true.

True

Test a value on the interval $- 1 < x < 9$ to see if it makes the inequality true.

Choose a value on the interval $- 1 < x < 9$ and see if this value makes the original inequality true.

$x = 0$

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

$\frac{0}{2} \geq \frac{5}{(0) + 1} + 4$

The left side $0$ is less than the right side $9$ , which means that the given statement is false.

False

Test a value on the interval $x > 9$ to see if it makes the inequality true.

Choose a value on the interval $x > 9$ and see if this value makes the original inequality true.

$x = 12$

Replace $x$ with $12$ in the original inequality.

$\frac{12}{2} \geq \frac{5}{(12) + 1} + 4$

The left side $6$ is greater than the right side $4.\overline{384615}$ , which means that the given statement is always true.

True

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

$x < - 2$ False

$- 2 < x < - 1$ True

$- 1 < x < 9$ False

$x > 9$ True

$x < - 2$ False

$- 2 < x < - 1$ True

$- 1 < x < 9$ False

$x > 9$ True

Step 12

The solution consists of all of the true intervals.

$- 2 \leq x < - 1$ or $x \geq 9$

Step 13

The result can be shown in multiple forms.

Inequality Form:

$- 2 \leq x < - 1 or x \geq 9$

Interval Notation:

$[- 2, - 1) \cup [9, \infty)$

Step 14

Solve the Inequality for x x/2>=5/(x+1)+4

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