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

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

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 $5$ .

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

Subtract $16$ from $15$ .

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

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 $- 5$ .

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

Subtract $16$ from $- 15$ .

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

Solve $3 x - 1 \leq - 7$ when $x \geq - 5$ .

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

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

Add $1$ to both sides of the inequality.

$3 x \leq - 7 + 1$

Add $- 7$ and $1$ .

$3 x \leq - 6$

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

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

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $- 6$ by $3$ .

$x \leq - 2$

Find the intersection of $x \leq - 2$ and $x \geq - 5$ .

$- 5 \leq x \leq - 2$

Solve $- 3 x - 31 \leq - 7$ when $x < - 5$ .

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

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

Add $31$ to both sides of the inequality.

$- 3 x \leq - 7 + 31$

Add $- 7$ and $31$ .

$- 3 x \leq 24$

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

Divide each term in $- 3 x \leq 24$ by $- 3$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

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

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $24$ by $- 3$ .

$x \geq - 8$

Find the intersection of $x \geq - 8$ and $x < - 5$ .

$- 8 \leq x < - 5$

Find the union of the solutions.

$- 8 \leq x \leq - 2$

The result can be shown in multiple forms.

Inequality Form:

$- 8 \leq x \leq - 2$

Interval Notation:

$[- 8, - 2]$

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

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