Solve the Inequality for x 10/(x+4)>3-x

$\frac{10}{x + 4} > 3 - x$

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

Subtract $3$ from both sides of the inequality.

$\frac{10}{x + 4} - 3 > - x$

Add $x$ to both sides of the inequality.

$\frac{10}{x + 4} - 3 + x > 0$

Simplify $\frac{10}{x + 4} - 3 + x$ .

To write $- 3$ as a fraction with a common denominator, multiply by $\frac{x + 4}{x + 4}$ .

$\frac{10}{x + 4} - 3 \cdot \frac{x + 4}{x + 4} + x > 0$

Combine $- 3$ and $\frac{x + 4}{x + 4}$ .

$\frac{10}{x + 4} + \frac{- 3 (x + 4)}{x + 4} + x > 0$

Combine the numerators over the common denominator.

$\frac{10 - 3 (x + 4)}{x + 4} + x > 0$

Simplify the numerator.

Apply the distributive property.

$\frac{10 - 3 x - 3 \cdot 4}{x + 4} + x > 0$

Multiply $- 3$ by $4$ .

$\frac{10 - 3 x - 12}{x + 4} + x > 0$

Subtract $12$ from $10$ .

$\frac{- 3 x - 2}{x + 4} + x > 0$

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

$\frac{- 3 x - 2}{x + 4} + \frac{x (x + 4)}{x + 4} > 0$

Combine the numerators over the common denominator.

$\frac{- 3 x - 2 + x (x + 4)}{x + 4} > 0$

Simplify the numerator.

Apply the distributive property.

$\frac{- 3 x - 2 + x \cdot x + x \cdot 4}{x + 4} > 0$

Multiply $x$ by $x$ .

$\frac{- 3 x - 2 + x^{2} + x \cdot 4}{x + 4} > 0$

Move $4$ to the left of $x$ .

$\frac{- 3 x - 2 + x^{2} + 4 \cdot x}{x + 4} > 0$

Add $- 3 x$ and $4 x$ .

$\frac{x - 2 + x^{2}}{x + 4} > 0$

Reorder terms.

$\frac{x^{2} + x - 2}{x + 4} > 0$

Factor $x^{2} + x - 2$ 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 $- 2$ and whose sum is $1$ .

$- 1, 2$

Write the factored form using these integers.

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

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

$x - 1 = 0$

$x + 2 = 0$

$x + 4 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Subtract $2$ from both sides of the equation.

$x = - 2$

Subtract $4$ from both sides of the equation.

$x = - 4$

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

$x = 1$

$x = - 2$

$x = - 4$

Consolidate the solutions.

$x = 1, - 2, - 4$

Find the domain of $\frac{(x - 1) (x + 2)}{x + 4}$ .

Set the denominator in $\frac{(x - 1) (x + 2)}{x + 4}$ equal to $0$ to find where the expression is undefined.

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

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

$(- \infty, - 4) \cup (- 4, \infty)$

Use each root to create test intervals.

$x < - 4$

$- 4 < x < - 2$

$- 2 < x < 1$

$x > 1$

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 $x < - 4$ 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.

$\frac{10}{(- 6) + 4} > 3 - (- 6)$

The left side $- 5$ is not greater than the right side $9$ , which means that the given statement is false.

False

Test a value on the interval $- 4 < x < - 2$ to see if it makes the inequality true.

Choose a value on the interval $- 4 < x < - 2$ and see if this value makes the original inequality true.

$x = - 3$

Replace $x$ with $- 3$ in the original inequality.

$\frac{10}{(- 3) + 4} > 3 - (- 3)$

The left side $10$ is greater than the right side $6$ , which means that the given statement is always true.

True

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.

$x = 0$

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

$\frac{10}{(0) + 4} > 3 - (0)$

The left side $2.5$ is not greater than the right side $3$ , which means that the given statement is false.

False

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

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

$x = 4$

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

$\frac{10}{(4) + 4} > 3 - (4)$

The left side $1.25$ is greater than the right side $- 1$ , which means that the given statement is always true.

True

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

$x < - 4$ False

$- 4 < x < - 2$ True

$- 2 < x < 1$ False

$x > 1$ True

$x < - 4$ False

$- 4 < x < - 2$ True

$- 2 < x < 1$ False

$x > 1$ True

The solution consists of all of the true intervals.

$- 4 < x < - 2$ or $x > 1$

The result can be shown in multiple forms.

Inequality Form:

$- 4 < x < - 2 or x > 1$

Interval Notation:

$(- 4, - 2) \cup (1, \infty)$

Solve the Inequality for x 10/(x+4)>3-x

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