Solve the Inequality for x (4-x)/(x+3)>0

$\frac{4 - x}{x + 3} > 0$

Step 1

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

$4 - x = 0$

$x + 3 = 0$

Step 2

Subtract $4$ from both sides of the equation.

$- x = - 4$

Step 3

Divide each term in $- x = - 4$ by $- 1$ and simplify.

Divide each term in $- x = - 4$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 4}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 4}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 4}{- 1}$

Simplify the right side.

Divide $- 4$ by $- 1$ .

$x = 4$

Step 4

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 5

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

$x = 4$

$x = - 3$

Step 6

Consolidate the solutions.

$x = 4, - 3$

Step 7

Find the domain of $\frac{4 - x}{x + 3}$ .

Set the denominator in $\frac{4 - x}{x + 3}$ equal to $0$ to find where the expression is undefined.

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

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

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

Step 8

Use each root to create test intervals.

$x < - 3$

$- 3 < x < 4$

$x > 4$

Step 9

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.

$\frac{4 - (- 6)}{(- 6) + 3} > 0$

The left side $- 3.\overline{3}$ is not greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 3 < x < 4$ to see if it makes the inequality true.

Choose a value on the interval $- 3 < x < 4$ and see if this value makes the original inequality true.

$x = 0$

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

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

The left side $1.\overline{3}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > 4$ to see if it makes the inequality true.

Choose a value on the interval $x > 4$ and see if this value makes the original inequality true.

$x = 6$

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

$\frac{4 - (6)}{(6) + 3} > 0$

The left side $- 0.\overline{2}$ is not greater than the right side $0$ , which means that the given statement is false.

False

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

$x < - 3$ False

$- 3 < x < 4$ True

$x > 4$ False

$x < - 3$ False

$- 3 < x < 4$ True

$x > 4$ False

Step 10

The solution consists of all of the true intervals.

$- 3 < x < 4$

Step 11

The result can be shown in multiple forms.

Inequality Form:

$- 3 < x < 4$

Interval Notation:

$(- 3, 4)$

Step 12

Solve the Inequality for x (4-x)/(x+3)>0

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