Solve the Inequality for x 5/(8x+3)<0

$\frac{5}{8 x + 3} < 0$

Step 1

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

$8 x + 3 = 0$

Step 2

Subtract $3$ from both sides of the equation.

$8 x = - 3$

Step 3

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

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

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

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 4

Find the domain of $\frac{5}{8 x + 3}$ .

Set the denominator in $\frac{5}{8 x + 3}$ equal to $0$ to find where the expression is undefined.

$8 x + 3 = 0$

Solve for $x$ .

Subtract $3$ from both sides of the equation.

$8 x = - 3$

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

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

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

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

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

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

Step 5

The solution consists of all of the true intervals.

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

Step 6

The result can be shown in multiple forms.

Inequality Form:

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

Interval Notation:

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

Step 7

Solve the Inequality for x 5/(8x+3)<0

Related Questions

Support LCMGCF.com by disabling your adblocker.