Solve the Inequality for a (|6a+3|)/3>2

$\frac{| 6 a + 3 |}{3} > 2$

Multiply both sides by $3$ .

$\frac{| 6 a + 3 |}{3} \cdot 3 > 2 \cdot 3$

Simplify.

Simplify the left side.

Simplify $\frac{| 6 a + 3 |}{3} \cdot 3$ .

Simplify the numerator.

Factor $3$ out of $6 a + 3$ .

Factor $3$ out of $6 a$ .

$\frac{| 3 (2 a) + 3 |}{3} \cdot 3 > 2 \cdot 3$

Factor $3$ out of $3$ .

$\frac{| 3 (2 a) + 3 (1) |}{3} \cdot 3 > 2 \cdot 3$

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

$\frac{| 3 (2 a + 1) |}{3} \cdot 3 > 2 \cdot 3$

Apply the distributive property.

$\frac{| 3 (2 a) + 3 \cdot 1 |}{3} \cdot 3 > 2 \cdot 3$

Multiply $2$ by $3$ .

$\frac{| 6 a + 3 \cdot 1 |}{3} \cdot 3 > 2 \cdot 3$

Multiply $3$ by $1$ .

$\frac{| 6 a + 3 |}{3} \cdot 3 > 2 \cdot 3$

Factor $3$ out of $6 a + 3$ .

Factor $3$ out of $6 a$ .

$\frac{| 3 (2 a) + 3 |}{3} \cdot 3 > 2 \cdot 3$

Factor $3$ out of $3$ .

$\frac{| 3 (2 a) + 3 (1) |}{3} \cdot 3 > 2 \cdot 3$

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

$\frac{| 3 (2 a + 1) |}{3} \cdot 3 > 2 \cdot 3$

Remove non-negative terms from the absolute value.

$\frac{3 | 2 a + 1 |}{3} \cdot 3 > 2 \cdot 3$

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 | 2 a + 1 |}{3} \cdot 3 > 2 \cdot 3$

Rewrite the expression.

$3 | 2 a + 1 | > 2 \cdot 3$

Simplify the right side.

Multiply $2$ by $3$ .

$3 | 2 a + 1 | > 6$

Solve for $a$ .

Write $3 | 2 a + 1 | > 6$ as a piecewise.

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

$2 a + 1 \geq 0$

Solve the inequality.

Subtract $1$ from both sides of the inequality.

$2 a \geq - 1$

Divide each term in $2 a \geq - 1$ by $2$ and simplify.

Divide each term in $2 a \geq - 1$ by $2$ .

$\frac{2 a}{2} \geq \frac{- 1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 a}{2} \geq \frac{- 1}{2}$

Divide $a$ by $1$ .

$a \geq \frac{- 1}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$a \geq - \frac{1}{2}$

In the piece where $2 a + 1$ is non-negative, remove the absolute value.

$3 (2 a + 1) > 6$

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

$2 a + 1 < 0$

Solve the inequality.

Subtract $1$ from both sides of the inequality.

$2 a < - 1$

Divide each term in $2 a < - 1$ by $2$ and simplify.

Divide each term in $2 a < - 1$ by $2$ .

$\frac{2 a}{2} < \frac{- 1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 a}{2} < \frac{- 1}{2}$

Divide $a$ by $1$ .

$a < \frac{- 1}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$a < - \frac{1}{2}$

In the piece where $2 a + 1$ is negative, remove the absolute value and multiply by $- 1$ .

$3 (- (2 a + 1)) > 6$

Write as a piecewise.

${\begin{cases} 3 (2 a + 1) > 6 & a \geq - \frac{1}{2} \\ 3 (- (2 a + 1)) > 6 & a < - \frac{1}{2} \end{cases}$

Simplify $3 (2 a + 1) > 6$ .

Apply the distributive property.

${\begin{cases} 3 (2 a) + 3 \cdot 1 > 6 & a \geq - \frac{1}{2} \\ 3 (- (2 a + 1)) > 6 & a < - \frac{1}{2} \end{cases}$

Multiply $2$ by $3$ .

${\begin{cases} 6 a + 3 \cdot 1 > 6 & a \geq - \frac{1}{2} \\ 3 (- (2 a + 1)) > 6 & a < - \frac{1}{2} \end{cases}$

Multiply $3$ by $1$ .

${\begin{cases} 6 a + 3 > 6 & a \geq - \frac{1}{2} \\ 3 (- (2 a + 1)) > 6 & a < - \frac{1}{2} \end{cases}$

Simplify $3 (- (2 a + 1)) > 6$ .

Apply the distributive property.

${\begin{cases} 6 a + 3 > 6 & a \geq - \frac{1}{2} \\ 3 (- (2 a) - 1 \cdot 1) > 6 & a < - \frac{1}{2} \end{cases}$

Multiply $2$ by $- 1$ .

${\begin{cases} 6 a + 3 > 6 & a \geq - \frac{1}{2} \\ 3 (- 2 a - 1 \cdot 1) > 6 & a < - \frac{1}{2} \end{cases}$

Multiply $- 1$ by $1$ .

${\begin{cases} 6 a + 3 > 6 & a \geq - \frac{1}{2} \\ 3 (- 2 a - 1) > 6 & a < - \frac{1}{2} \end{cases}$

Apply the distributive property.

${\begin{cases} 6 a + 3 > 6 & a \geq - \frac{1}{2} \\ 3 (- 2 a) + 3 \cdot - 1 > 6 & a < - \frac{1}{2} \end{cases}$

Multiply $- 2$ by $3$ .

${\begin{cases} 6 a + 3 > 6 & a \geq - \frac{1}{2} \\ - 6 a + 3 \cdot - 1 > 6 & a < - \frac{1}{2} \end{cases}$

Multiply $3$ by $- 1$ .

${\begin{cases} 6 a + 3 > 6 & a \geq - \frac{1}{2} \\ - 6 a - 3 > 6 & a < - \frac{1}{2} \end{cases}$

Solve $6 a + 3 > 6$ for $a$ .

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

Subtract $3$ from both sides of the inequality.

$6 a > 6 - 3$

Subtract $3$ from $6$ .

$6 a > 3$

Divide each term in $6 a > 3$ by $6$ and simplify.

Divide each term in $6 a > 3$ by $6$ .

$\frac{6 a}{6} > \frac{3}{6}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 a}{6} > \frac{3}{6}$

Divide $a$ by $1$ .

$a > \frac{3}{6}$

Simplify the right side.

Cancel the common factor of $3$ and $6$ .

Factor $3$ out of $3$ .

$a > \frac{3 (1)}{6}$

Cancel the common factors.

Factor $3$ out of $6$ .

$a > \frac{3 \cdot 1}{3 \cdot 2}$

Cancel the common factor.

$a > \frac{3 \cdot 1}{3 \cdot 2}$

Rewrite the expression.

$a > \frac{1}{2}$

Solve $- 6 a - 3 > 6$ for $a$ .

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

Add $3$ to both sides of the inequality.

$- 6 a > 6 + 3$

Add $6$ and $3$ .

$- 6 a > 9$

Divide each term in $- 6 a > 9$ by $- 6$ and simplify.

Divide each term in $- 6 a > 9$ by $- 6$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 6 a}{- 6} < \frac{9}{- 6}$

Simplify the left side.

Cancel the common factor of $- 6$ .

Cancel the common factor.

$\frac{- 6 a}{- 6} < \frac{9}{- 6}$

Divide $a$ by $1$ .

$a < \frac{9}{- 6}$

Simplify the right side.

Cancel the common factor of $9$ and $- 6$ .

Factor $3$ out of $9$ .

$a < \frac{3 (3)}{- 6}$

Cancel the common factors.

Factor $3$ out of $- 6$ .

$a < \frac{3 \cdot 3}{3 \cdot - 2}$

Cancel the common factor.

$a < \frac{3 \cdot 3}{3 \cdot - 2}$

Rewrite the expression.

$a < \frac{3}{- 2}$

Move the negative in front of the fraction.

$a < - \frac{3}{2}$

Find the union of the solutions.

$a < - \frac{3}{2}$ or $a > \frac{1}{2}$

The result can be shown in multiple forms.

Inequality Form:

$a < - \frac{3}{2} or a > \frac{1}{2}$

Interval Notation:

$(- \infty, - \frac{3}{2}) \cup (\frac{1}{2}, \infty)$

Solve the Inequality for a (|6a+3|)/3>2

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