Convert to Interval Notation 8/(6-x)>=x

$\frac{8}{6 - x} \geq x$

Subtract $x$ from both sides of the inequality.

$\frac{8}{6 - x} - x \geq 0$

Simplify $\frac{8}{6 - x} - x$ .

To write $- x$ as a fraction with a common denominator, multiply by $\frac{6 - x}{6 - x}$ .

$\frac{8}{6 - x} - x \cdot \frac{6 - x}{6 - x} \geq 0$

Combine $- x$ and $\frac{6 - x}{6 - x}$ .

$\frac{8}{6 - x} + \frac{- x (6 - x)}{6 - x} \geq 0$

Combine the numerators over the common denominator.

$\frac{8 - x (6 - x)}{6 - x} \geq 0$

Simplify the numerator.

Apply the distributive property.

$\frac{8 - x \cdot 6 - x (- x)}{6 - x} \geq 0$

Multiply $6$ by $- 1$ .

$\frac{8 - 6 x - x (- x)}{6 - x} \geq 0$

Rewrite using the commutative property of multiplication.

$\frac{8 - 6 x - 1 \cdot - 1 x \cdot x}{6 - x} \geq 0$

Simplify each term.

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$\frac{8 - 6 x - 1 \cdot - 1 (x \cdot x)}{6 - x} \geq 0$

Multiply $x$ by $x$ .

$\frac{8 - 6 x - 1 \cdot - 1 x^{2}}{6 - x} \geq 0$

Multiply $- 1$ by $- 1$ .

$\frac{8 - 6 x + 1 x^{2}}{6 - x} \geq 0$

Multiply $x^{2}$ by $1$ .

$\frac{8 - 6 x + x^{2}}{6 - x} \geq 0$

Factor $8 - 6 x + x^{2}$ 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 $8$ and whose sum is $- 6$ .

$- 4, - 2$

Write the factored form using these integers.

$\frac{(x - 4) (x - 2)}{6 - x} \geq 0$

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

$x - 4 = 0$

$x - 2 = 0$

$6 - x = 0$

Add $4$ to both sides of the equation.

$x = 4$

Add $2$ to both sides of the equation.

$x = 2$

Subtract $6$ from both sides of the equation.

$- x = - 6$

Divide each term in $- x = - 6$ by $- 1$ and simplify.

Divide each term in $- x = - 6$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 6}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 6}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 6}{- 1}$

Simplify the right side.

Divide $- 6$ by $- 1$ .

$x = 6$

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

$x = 4$

$x = 2$

$x = 6$

Consolidate the solutions.

$x = 4, 2, 6$

Find the domain of $\frac{(x - 4) (x - 2)}{6 - x}$ .

Set the denominator in $\frac{(x - 4) (x - 2)}{6 - x}$ equal to $0$ to find where the expression is undefined.

$6 - x = 0$

Solve for $x$ .

Subtract $6$ from both sides of the equation.

$- x = - 6$

Divide each term in $- x = - 6$ by $- 1$ and simplify.

Divide each term in $- x = - 6$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 6}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 6}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 6}{- 1}$

Simplify the right side.

Divide $- 6$ by $- 1$ .

$x = 6$

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

$(- \infty, 6) \cup (6, \infty)$

Use each root to create test intervals.

$x < 2$

$2 < x < 4$

$4 < x < 6$

$x > 6$

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.

$x = 0$

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

$\frac{8}{6 - (0)} \geq 0$

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 $2 < x < 4$ to see if it makes the inequality true.

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

$x = 3$

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

$\frac{8}{6 - (3)} \geq 3$

The left side $2.\overline{6}$ is less than the right side $3$ , which means that the given statement is false.

False

Test a value on the interval $4 < x < 6$ to see if it makes the inequality true.

Choose a value on the interval $4 < x < 6$ and see if this value makes the original inequality true.

$x = 5$

Replace $x$ with $5$ in the original inequality.

$\frac{8}{6 - (5)} \geq 5$

The left side $8$ is greater than the right side $5$ , which means that the given statement is always true.

True

Test a value on the interval $x > 6$ to see if it makes the inequality true.

Choose a value on the interval $x > 6$ and see if this value makes the original inequality true.

$x = 8$

Replace $x$ with $8$ in the original inequality.

$\frac{8}{6 - (8)} \geq 8$

The left side $- 4$ is less than the right side $8$ , which means that the given statement is false.

False

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

$x < 2$ True

$2 < x < 4$ False

$4 < x < 6$ True

$x > 6$ False

$x < 2$ True

$2 < x < 4$ False

$4 < x < 6$ True

$x > 6$ False

The solution consists of all of the true intervals.

$x \leq 2$ or $4 \leq x < 6$

Convert the inequality to interval notation.

$(- \infty, 2] \cup [4, 6)$

Convert to Interval Notation 8/(6-x)>=x

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