Solve for x (6x+2)^-1<0

${(6 x + 2)}^{- 1} < 0$

Step 1

Simplify ${(6 x + 2)}^{- 1}$ .

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{6 x + 2} < 0$

Factor $2$ out of $6 x + 2$ .

Factor $2$ out of $6 x$ .

$\frac{1}{2 (3 x) + 2} < 0$

Factor $2$ out of $2$ .

$\frac{1}{2 (3 x) + 2 (1)} < 0$

Factor $2$ out of $2 (3 x) + 2 (1)$ .

$\frac{1}{2 (3 x + 1)} < 0$

Step 2

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

$3 x + 1 = 0$

Step 3

Subtract $1$ from both sides of the equation.

$3 x = - 1$

Step 4

Divide each term in $3 x = - 1$ by $3$ and simplify.

Divide each term in $3 x = - 1$ by $3$ .

$\frac{3 x}{3} = \frac{- 1}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 1}{3}$

Divide $x$ by $1$ .

$x = \frac{- 1}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{3}$

Step 5

The solution consists of all of the true intervals.

$x < - \frac{1}{3}$

Step 6

The result can be shown in multiple forms.

Inequality Form:

$x < - \frac{1}{3}$

Interval Notation:

$(- \infty, - \frac{1}{3})$

Step 7