Convert to Interval Notation (3x+8)/(x-4)>=0

$\frac{3 x + 8}{x - 4} \geq 0$

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

$3 x + 8 = 0$

$x - 4 = 0$

Subtract $8$ from both sides of the equation.

$3 x = - 8$

Divide each term in $3 x = - 8$ by $3$ and simplify.

Divide each term in $3 x = - 8$ by $3$ .

$\frac{3 x}{3} = \frac{- 8}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 8}{3}$

Divide $x$ by $1$ .

$x = \frac{- 8}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{8}{3}$

Add $4$ to 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.

$x = - \frac{8}{3}$

$x = 4$

Consolidate the solutions.

$x = - \frac{8}{3}, 4$

Find the domain of $\frac{3 x + 8}{x - 4}$ .

Set the denominator in $\frac{3 x + 8}{x - 4}$ equal to $0$ to find where the expression is undefined.

$x - 4 = 0$

Add $4$ to 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.

$x < - \frac{8}{3}$

$- \frac{8}{3} < x < 4$

$x > 4$

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 $x < - \frac{8}{3}$ to see if it makes the inequality true.

Choose a value on the interval $x < - \frac{8}{3}$ and see if this value makes the original inequality true.

$x = - 5$

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

$\frac{3 (- 5) + 8}{(- 5) - 4} \geq 0$

The left side $0.\overline{7}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- \frac{8}{3} < x < 4$ to see if it makes the inequality true.

Choose a value on the interval $- \frac{8}{3} < x < 4$ and see if this value makes the original inequality true.

$x = 0$

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

$\frac{3 (0) + 8}{(0) - 4} \geq 0$

The left side $- 2$ is less than the right side $0$ , which means that the given statement is false.

False

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.

$\frac{3 (6) + 8}{(6) - 4} \geq 0$

The left side $13$ is greater than the right side $0$ , which means that the given statement is always true.

True

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

$x < - \frac{8}{3}$ True

$- \frac{8}{3} < x < 4$ False

$x > 4$ True

$x < - \frac{8}{3}$ True

$- \frac{8}{3} < x < 4$ False

$x > 4$ True

The solution consists of all of the true intervals.

$x \leq - \frac{8}{3}$ or $x > 4$

Convert the inequality to interval notation.

$(- \infty, - \frac{8}{3}] \cup (4, \infty)$

Convert to Interval Notation (3x+8)/(x-4)>=0

Related Questions

Support LCMGCF.com by disabling your adblocker.