Solve for x |3x+2|>7

$| 3 x + 2 | > 7$

Step 1

Write $| 3 x + 2 | > 7$ as a piecewise.

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

$3 x + 2 \geq 0$

Solve the inequality.

Subtract $2$ from both sides of the inequality.

$3 x \geq - 2$

Divide each term in $3 x \geq - 2$ by $3$ and simplify.

Divide each term in $3 x \geq - 2$ by $3$ .

$\frac{3 x}{3} \geq \frac{- 2}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} \geq \frac{- 2}{3}$

Divide $x$ by $1$ .

$x \geq \frac{- 2}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x \geq - \frac{2}{3}$

In the piece where $3 x + 2$ is non-negative, remove the absolute value.

$3 x + 2 > 7$

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

$3 x + 2 < 0$

Solve the inequality.

Subtract $2$ from both sides of the inequality.

$3 x < - 2$

Divide each term in $3 x < - 2$ by $3$ and simplify.

Divide each term in $3 x < - 2$ by $3$ .

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

In the piece where $3 x + 2$ is negative, remove the absolute value and multiply by $- 1$ .

$- (3 x + 2) > 7$

Write as a piecewise.

${\begin{cases} 3 x + 2 > 7 & x \geq - \frac{2}{3} \\ - (3 x + 2) > 7 & x < - \frac{2}{3} \end{cases}$

Simplify $- (3 x + 2) > 7$ .

Apply the distributive property.

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

Multiply $3$ by $- 1$ .

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

Multiply $- 1$ by $2$ .

${\begin{cases} 3 x + 2 > 7 & x \geq - \frac{2}{3} \\ - 3 x - 2 > 7 & x < - \frac{2}{3} \end{cases}$

Step 2

Solve $3 x + 2 > 7$ for $x$ .

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

Subtract $2$ from both sides of the inequality.

$3 x > 7 - 2$

Subtract $2$ from $7$ .

$3 x > 5$

Divide each term in $3 x > 5$ by $3$ and simplify.

Divide each term in $3 x > 5$ by $3$ .

$\frac{3 x}{3} > \frac{5}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} > \frac{5}{3}$

Divide $x$ by $1$ .

$x > \frac{5}{3}$

Step 3

Solve $- 3 x - 2 > 7$ for $x$ .

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

Add $2$ to both sides of the inequality.

$- 3 x > 7 + 2$

Add $7$ and $2$ .

$- 3 x > 9$

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

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

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

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $9$ by $- 3$ .

$x < - 3$

Step 4

Find the union of the solutions.

$x < - 3$ or $x > \frac{5}{3}$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$x < - 3 or x > \frac{5}{3}$

Interval Notation:

$(- \infty, - 3) \cup (\frac{5}{3}, \infty)$

Step 6

Solve for x |3x+2|>7

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