Solve the Inequality for t 5t+1>=-t-1

$5 t + 1 \geq - t - 1$

Move all terms containing $t$ to the left side of the inequality.

Add $t$ to both sides of the inequality.

$5 t + 1 + t \geq - 1$

Add $5 t$ and $t$ .

$6 t + 1 \geq - 1$

Move all terms not containing $t$ to the right side of the inequality.

Subtract $1$ from both sides of the inequality.

$6 t \geq - 1 - 1$

Subtract $1$ from $- 1$ .

$6 t \geq - 2$

Divide each term in $6 t \geq - 2$ by $6$ and simplify.

Divide each term in $6 t \geq - 2$ by $6$ .

$\frac{6 t}{6} \geq \frac{- 2}{6}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 t}{6} \geq \frac{- 2}{6}$

Divide $t$ by $1$ .

$t \geq \frac{- 2}{6}$

Simplify the right side.

Cancel the common factor of $- 2$ and $6$ .

Factor $2$ out of $- 2$ .

$t \geq \frac{2 (- 1)}{6}$

Cancel the common factors.

Factor $2$ out of $6$ .

$t \geq \frac{2 \cdot - 1}{2 \cdot 3}$

Cancel the common factor.

$t \geq \frac{2 \cdot - 1}{2 \cdot 3}$

Rewrite the expression.

$t \geq \frac{- 1}{3}$

Move the negative in front of the fraction.

$t \geq - \frac{1}{3}$

The result can be shown in multiple forms.

Inequality Form:

$t \geq - \frac{1}{3}$

Interval Notation:

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