Solve for a 1/3*|3a-3|-6>2

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

Step 1

Write $\frac{1}{3} \cdot | 3 a - 3 | - 6 > 2$ as a piecewise.

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

$3 a - 3 \geq 0$

Solve the inequality.

Add $3$ to both sides of the inequality.

$3 a \geq 3$

Divide each term in $3 a \geq 3$ by $3$ and simplify.

Divide each term in $3 a \geq 3$ by $3$ .

$\frac{3 a}{3} \geq \frac{3}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 a}{3} \geq \frac{3}{3}$

Divide $a$ by $1$ .

$a \geq \frac{3}{3}$

Simplify the right side.

Divide $3$ by $3$ .

$a \geq 1$

In the piece where $3 a - 3$ is non-negative, remove the absolute value.

$\frac{1}{3} \cdot (3 a - 3) - 6 > 2$

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

$3 a - 3 < 0$

Solve the inequality.

Add $3$ to both sides of the inequality.

$3 a < 3$

Divide each term in $3 a < 3$ by $3$ and simplify.

Divide each term in $3 a < 3$ by $3$ .

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $a$ by $1$ .

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

Simplify the right side.

Divide $3$ by $3$ .

$a < 1$

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

$\frac{1}{3} \cdot (- (3 a - 3)) - 6 > 2$

Write as a piecewise.

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

Simplify $\frac{1}{3} \cdot (3 a - 3) - 6 > 2$ .

Simplify each term.

Apply the distributive property.

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

Cancel the common factor of $3$ .

Factor $3$ out of $3 a$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $3$ .

Factor $3$ out of $- 3$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Subtract $6$ from $- 1$ .

${\begin{cases} a - 7 > 2 & a \geq 1 \\ \frac{1}{3} \cdot (- (3 a - 3)) - 6 > 2 & a < 1 \end{cases}$

Simplify $\frac{1}{3} \cdot (- (3 a - 3)) - 6 > 2$ .

Simplify each term.

Apply the distributive property.

${\begin{cases} a - 7 > 2 & a \geq 1 \\ \frac{1}{3} \cdot (- (3 a) - - 3) - 6 > 2 & a < 1 \end{cases}$

Multiply $3$ by $- 1$ .

${\begin{cases} a - 7 > 2 & a \geq 1 \\ \frac{1}{3} \cdot (- 3 a - - 3) - 6 > 2 & a < 1 \end{cases}$

Multiply $- 1$ by $- 3$ .

${\begin{cases} a - 7 > 2 & a \geq 1 \\ \frac{1}{3} \cdot (- 3 a + 3) - 6 > 2 & a < 1 \end{cases}$

Apply the distributive property.

${\begin{cases} a - 7 > 2 & a \geq 1 \\ \frac{1}{3} (- 3 a) + \frac{1}{3} \cdot 3 - 6 > 2 & a < 1 \end{cases}$

Cancel the common factor of $3$ .

Factor $3$ out of $- 3 a$ .

${\begin{cases} a - 7 > 2 & a \geq 1 \\ \frac{1}{3} (3 (- a)) + \frac{1}{3} \cdot 3 - 6 > 2 & a < 1 \end{cases}$

Cancel the common factor.

${\begin{cases} a - 7 > 2 & a \geq 1 \\ \frac{1}{3} (3 (- a)) + \frac{1}{3} \cdot 3 - 6 > 2 & a < 1 \end{cases}$

Rewrite the expression.

${\begin{cases} a - 7 > 2 & a \geq 1 \\ - a + \frac{1}{3} \cdot 3 - 6 > 2 & a < 1 \end{cases}$

Cancel the common factor of $3$ .

Cancel the common factor.

${\begin{cases} a - 7 > 2 & a \geq 1 \\ - a + \frac{1}{3} \cdot 3 - 6 > 2 & a < 1 \end{cases}$

Rewrite the expression.

${\begin{cases} a - 7 > 2 & a \geq 1 \\ - a + 1 - 6 > 2 & a < 1 \end{cases}$

Subtract $6$ from $1$ .

${\begin{cases} a - 7 > 2 & a \geq 1 \\ - a - 5 > 2 & a < 1 \end{cases}$

Step 2

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

Add $7$ to both sides of the inequality.

$a > 2 + 7$

Add $2$ and $7$ .

$a > 9$

Step 3

Solve $- a - 5 > 2$ for $a$ .

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

Add $5$ to both sides of the inequality.

$- a > 2 + 5$

Add $2$ and $5$ .

$- a > 7$

Divide each term in $- a > 7$ by $- 1$ and simplify.

Divide each term in $- a > 7$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- a}{- 1} < \frac{7}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{a}{1} < \frac{7}{- 1}$

Divide $a$ by $1$ .

$a < \frac{7}{- 1}$

Simplify the right side.

Divide $7$ by $- 1$ .

$a < - 7$

Step 4

Find the union of the solutions.

$a < - 7$ or $a > 9$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$a < - 7 or a > 9$

Interval Notation:

$(- \infty, - 7) \cup (9, \infty)$

Step 6

Solve for a 1/3*|3a-3|-6>2

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