Solve the Inequality for t |3t+1|>8

$| 3 t + 1 | > 8$

Write $| 3 t + 1 | > 8$ as a piecewise.

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$3 t + 1 \geq 0$

Solve the inequality.

Subtract $1$ from both sides of the inequality.

$3 t \geq - 1$

Divide each term in $3 t \geq - 1$ by $3$ and simplify.

Divide each term in $3 t \geq - 1$ by $3$ .

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $t$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

In the piece where $3 t + 1$ is non-negative, remove the absolute value.

$3 t + 1 > 8$

To find the interval for the second piece, find where the inside of the absolute value is negative.

$3 t + 1 < 0$

Solve the inequality.

Subtract $1$ from both sides of the inequality.

$3 t < - 1$

Divide each term in $3 t < - 1$ by $3$ and simplify.

Divide each term in $3 t < - 1$ by $3$ .

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $t$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

In the piece where $3 t + 1$ is negative, remove the absolute value and multiply by $- 1$ .

$- (3 t + 1) > 8$

Write as a piecewise.

${\begin{cases} 3 t + 1 > 8 & t \geq - \frac{1}{3} \\ - (3 t + 1) > 8 & t < - \frac{1}{3} \end{cases}$

Simplify $- (3 t + 1) > 8$ .

Apply the distributive property.

${\begin{cases} 3 t + 1 > 8 & t \geq - \frac{1}{3} \\ - (3 t) - 1 \cdot 1 > 8 & t < - \frac{1}{3} \end{cases}$

Multiply $3$ by $- 1$ .

${\begin{cases} 3 t + 1 > 8 & t \geq - \frac{1}{3} \\ - 3 t - 1 \cdot 1 > 8 & t < - \frac{1}{3} \end{cases}$

Multiply $- 1$ by $1$ .

${\begin{cases} 3 t + 1 > 8 & t \geq - \frac{1}{3} \\ - 3 t - 1 > 8 & t < - \frac{1}{3} \end{cases}$

Solve $3 t + 1 > 8$ for $t$ .

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

Subtract $1$ from both sides of the inequality.

$3 t > 8 - 1$

Subtract $1$ from $8$ .

$3 t > 7$

Divide each term in $3 t > 7$ by $3$ and simplify.

Divide each term in $3 t > 7$ by $3$ .

$\frac{3 t}{3} > \frac{7}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 t}{3} > \frac{7}{3}$

Divide $t$ by $1$ .

$t > \frac{7}{3}$

Solve $- 3 t - 1 > 8$ for $t$ .

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

Add $1$ to both sides of the inequality.

$- 3 t > 8 + 1$

Add $8$ and $1$ .

$- 3 t > 9$

Divide each term in $- 3 t > 9$ by $- 3$ and simplify.

Divide each term in $- 3 t > 9$ by $- 3$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 3 t}{- 3} < \frac{9}{- 3}$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 t}{- 3} < \frac{9}{- 3}$

Divide $t$ by $1$ .

$t < \frac{9}{- 3}$

Simplify the right side.

Divide $9$ by $- 3$ .

$t < - 3$

Find the union of the solutions.

$t < - 3$ or $t > \frac{7}{3}$

The result can be shown in multiple forms.

Inequality Form:

$t < - 3 or t > \frac{7}{3}$

Interval Notation:

$(- \infty, - 3) \cup (\frac{7}{3}, \infty)$

Solve the Inequality for t |3t+1|>8

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