Solve for x 4|x+3|>=8

$4 | x + 3 | \geq 8$

Step 1

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

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

$x + 3 \geq 0$

Subtract $3$ from both sides of the inequality.

$x \geq - 3$

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

$4 (x + 3) \geq 8$

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

$x + 3 < 0$

Subtract $3$ from both sides of the inequality.

$x < - 3$

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

$4 (- (x + 3)) \geq 8$

Write as a piecewise.

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

Simplify $4 (x + 3) \geq 8$ .

Apply the distributive property.

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

Multiply $4$ by $3$ .

${\begin{cases} 4 x + 12 \geq 8 & x \geq - 3 \\ 4 (- (x + 3)) \geq 8 & x < - 3 \end{cases}$

Simplify $4 (- (x + 3)) \geq 8$ .

Apply the distributive property.

${\begin{cases} 4 x + 12 \geq 8 & x \geq - 3 \\ 4 (- x - 1 \cdot 3) \geq 8 & x < - 3 \end{cases}$

Multiply $- 1$ by $3$ .

${\begin{cases} 4 x + 12 \geq 8 & x \geq - 3 \\ 4 (- x - 3) \geq 8 & x < - 3 \end{cases}$

Apply the distributive property.

${\begin{cases} 4 x + 12 \geq 8 & x \geq - 3 \\ 4 (- x) + 4 \cdot - 3 \geq 8 & x < - 3 \end{cases}$

Multiply $- 1$ by $4$ .

${\begin{cases} 4 x + 12 \geq 8 & x \geq - 3 \\ - 4 x + 4 \cdot - 3 \geq 8 & x < - 3 \end{cases}$

Multiply $4$ by $- 3$ .

${\begin{cases} 4 x + 12 \geq 8 & x \geq - 3 \\ - 4 x - 12 \geq 8 & x < - 3 \end{cases}$

Step 2

Solve $4 x + 12 \geq 8$ for $x$ .

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

Subtract $12$ from both sides of the inequality.

$4 x \geq 8 - 12$

Subtract $12$ from $8$ .

$4 x \geq - 4$

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

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

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

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $- 4$ by $4$ .

$x \geq - 1$

Step 3

Solve $- 4 x - 12 \geq 8$ for $x$ .

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

Add $12$ to both sides of the inequality.

$- 4 x \geq 8 + 12$

Add $8$ and $12$ .

$- 4 x \geq 20$

Divide each term in $- 4 x \geq 20$ by $- 4$ and simplify.

Divide each term in $- 4 x \geq 20$ by $- 4$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 4 x}{- 4} \leq \frac{20}{- 4}$

Simplify the left side.

Cancel the common factor of $- 4$ .

Cancel the common factor.

$\frac{- 4 x}{- 4} \leq \frac{20}{- 4}$

Divide $x$ by $1$ .

$x \leq \frac{20}{- 4}$

Simplify the right side.

Divide $20$ by $- 4$ .

$x \leq - 5$

Step 4

Find the union of the solutions.

$x \leq - 5$ or $x \geq - 1$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 5 or x \geq - 1$

Interval Notation:

$(- \infty, - 5] \cup [- 1, \infty)$

Step 6

Solve for x 4|x+3|>=8

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