Solve the Inequality for x -3|x+5|+16<=7

$- 3 | x + 5 | + 16 \leq 7$

Step 1

Write $- 3 | x + 5 | + 16 \leq 7$ as a piecewise.

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

$x + 5 \geq 0$

Subtract $5$ from both sides of the inequality.

$x \geq - 5$

In the piece where $x + 5$ is non-negative, remove the absolute value.

$- 3 (x + 5) + 16 \leq 7$

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

$x + 5 < 0$

Subtract $5$ from both sides of the inequality.

$x < - 5$

In the piece where $x + 5$ is negative, remove the absolute value and multiply by $- 1$ .

$- 3 (- (x + 5)) + 16 \leq 7$

Write as a piecewise.

${\begin{cases} - 3 (x + 5) + 16 \leq 7 & x \geq - 5 \\ - 3 (- (x + 5)) + 16 \leq 7 & x < - 5 \end{cases}$

Simplify $- 3 (x + 5) + 16 \leq 7$ .

Simplify each term.

Apply the distributive property.

${\begin{cases} - 3 x - 3 \cdot 5 + 16 \leq 7 & x \geq - 5 \\ - 3 (- (x + 5)) + 16 \leq 7 & x < - 5 \end{cases}$

Multiply $- 3$ by $5$ .

${\begin{cases} - 3 x - 15 + 16 \leq 7 & x \geq - 5 \\ - 3 (- (x + 5)) + 16 \leq 7 & x < - 5 \end{cases}$

Add $- 15$ and $16$ .

${\begin{cases} - 3 x + 1 \leq 7 & x \geq - 5 \\ - 3 (- (x + 5)) + 16 \leq 7 & x < - 5 \end{cases}$

Simplify $- 3 (- (x + 5)) + 16 \leq 7$ .

Simplify each term.

Apply the distributive property.

${\begin{cases} - 3 x + 1 \leq 7 & x \geq - 5 \\ - 3 (- x - 1 \cdot 5) + 16 \leq 7 & x < - 5 \end{cases}$

Multiply $- 1$ by $5$ .

${\begin{cases} - 3 x + 1 \leq 7 & x \geq - 5 \\ - 3 (- x - 5) + 16 \leq 7 & x < - 5 \end{cases}$

Apply the distributive property.

${\begin{cases} - 3 x + 1 \leq 7 & x \geq - 5 \\ - 3 (- x) - 3 \cdot - 5 + 16 \leq 7 & x < - 5 \end{cases}$

Multiply $- 1$ by $- 3$ .

${\begin{cases} - 3 x + 1 \leq 7 & x \geq - 5 \\ 3 x - 3 \cdot - 5 + 16 \leq 7 & x < - 5 \end{cases}$

Multiply $- 3$ by $- 5$ .

${\begin{cases} - 3 x + 1 \leq 7 & x \geq - 5 \\ 3 x + 15 + 16 \leq 7 & x < - 5 \end{cases}$

Add $15$ and $16$ .

${\begin{cases} - 3 x + 1 \leq 7 & x \geq - 5 \\ 3 x + 31 \leq 7 & x < - 5 \end{cases}$

Step 2

Solve $- 3 x + 1 \leq 7$ for $x$ .

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

Subtract $1$ from both sides of the inequality.

$- 3 x \leq 7 - 1$

Subtract $1$ from $7$ .

$- 3 x \leq 6$

Divide each term in $- 3 x \leq 6$ by $- 3$ and simplify.

Divide each term in $- 3 x \leq 6$ by $- 3$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 3 x}{- 3} \geq \frac{6}{- 3}$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 x}{- 3} \geq \frac{6}{- 3}$

Divide $x$ by $1$ .

$x \geq \frac{6}{- 3}$

Simplify the right side.

Divide $6$ by $- 3$ .

$x \geq - 2$

Step 3

Solve $3 x + 31 \leq 7$ for $x$ .

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

Subtract $31$ from both sides of the inequality.

$3 x \leq 7 - 31$

Subtract $31$ from $7$ .

$3 x \leq - 24$

Divide each term in $3 x \leq - 24$ by $3$ and simplify.

Divide each term in $3 x \leq - 24$ by $3$ .

$\frac{3 x}{3} \leq \frac{- 24}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} \leq \frac{- 24}{3}$

Divide $x$ by $1$ .

$x \leq \frac{- 24}{3}$

Simplify the right side.

Divide $- 24$ by $3$ .

$x \leq - 8$

Step 4

Find the union of the solutions.

$x \leq - 8$ or $x \geq - 2$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 8 or x \geq - 2$

Interval Notation:

$(- \infty, - 8] \cup [- 2, \infty)$

Step 6

Solve the Inequality for x -3|x+5|+16<=7

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