Solve for x |3x+4|<8

$| 3 x + 4 | < 8$

Step 1

Write $| 3 x + 4 | < 8$ as a piecewise.

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

$3 x + 4 \geq 0$

Solve the inequality.

Subtract $4$ from both sides of the inequality.

$3 x \geq - 4$

Divide each term in $3 x \geq - 4$ by $3$ and simplify.

Divide each term in $3 x \geq - 4$ by $3$ .

$\frac{3 x}{3} \geq \frac{- 4}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} \geq \frac{- 4}{3}$

Divide $x$ by $1$ .

$x \geq \frac{- 4}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x \geq - \frac{4}{3}$

In the piece where $3 x + 4$ is non-negative, remove the absolute value.

$3 x + 4 < 8$

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

$3 x + 4 < 0$

Solve the inequality.

Subtract $4$ from both sides of the inequality.

$3 x < - 4$

Divide each term in $3 x < - 4$ by $3$ and simplify.

Divide each term in $3 x < - 4$ by $3$ .

$\frac{3 x}{3} < \frac{- 4}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} < \frac{- 4}{3}$

Divide $x$ by $1$ .

$x < \frac{- 4}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x < - \frac{4}{3}$

In the piece where $3 x + 4$ is negative, remove the absolute value and multiply by $- 1$ .

$- (3 x + 4) < 8$

Write as a piecewise.

${\begin{cases} 3 x + 4 < 8 & x \geq - \frac{4}{3} \\ - (3 x + 4) < 8 & x < - \frac{4}{3} \end{cases}$

Simplify $- (3 x + 4) < 8$ .

Apply the distributive property.

${\begin{cases} 3 x + 4 < 8 & x \geq - \frac{4}{3} \\ - (3 x) - 1 \cdot 4 < 8 & x < - \frac{4}{3} \end{cases}$

Multiply $3$ by $- 1$ .

${\begin{cases} 3 x + 4 < 8 & x \geq - \frac{4}{3} \\ - 3 x - 1 \cdot 4 < 8 & x < - \frac{4}{3} \end{cases}$

Multiply $- 1$ by $4$ .

${\begin{cases} 3 x + 4 < 8 & x \geq - \frac{4}{3} \\ - 3 x - 4 < 8 & x < - \frac{4}{3} \end{cases}$

Step 2

Solve $3 x + 4 < 8$ when $x \geq - \frac{4}{3}$ .

Solve $3 x + 4 < 8$ for $x$ .

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

Subtract $4$ from both sides of the inequality.

$3 x < 8 - 4$

Subtract $4$ from $8$ .

$3 x < 4$

Divide each term in $3 x < 4$ by $3$ and simplify.

Divide each term in $3 x < 4$ by $3$ .

$\frac{3 x}{3} < \frac{4}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} < \frac{4}{3}$

Divide $x$ by $1$ .

$x < \frac{4}{3}$

Find the intersection of $x < \frac{4}{3}$ and $x \geq - \frac{4}{3}$ .

$- \frac{4}{3} \leq x < \frac{4}{3}$

Step 3

Solve $- 3 x - 4 < 8$ when $x < - \frac{4}{3}$ .

Solve $- 3 x - 4 < 8$ for $x$ .

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

Add $4$ to both sides of the inequality.

$- 3 x < 8 + 4$

Add $8$ and $4$ .

$- 3 x < 12$

Divide each term in $- 3 x < 12$ by $- 3$ and simplify.

Divide each term in $- 3 x < 12$ by $- 3$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 3 x}{- 3} > \frac{12}{- 3}$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 x}{- 3} > \frac{12}{- 3}$

Divide $x$ by $1$ .

$x > \frac{12}{- 3}$

Simplify the right side.

Divide $12$ by $- 3$ .

$x > - 4$

Find the intersection of $x > - 4$ and $x < - \frac{4}{3}$ .

$- 4 < x < - \frac{4}{3}$

Step 4

Find the union of the solutions.

$- 4 < x < \frac{4}{3}$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$- 4 < x < \frac{4}{3}$

Interval Notation:

$(- 4, \frac{4}{3})$

Step 6

Solve for x |3x+4|<8

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