Solve for d 3|d+1|-7>=-1

$3 | d + 1 | - 7 \geq - 1$

Step 1

Write $3 | d + 1 | - 7 \geq - 1$ as a piecewise.

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

$d + 1 \geq 0$

Subtract $1$ from both sides of the inequality.

$d \geq - 1$

In the piece where $d + 1$ is non-negative, remove the absolute value.

$3 (d + 1) - 7 \geq - 1$

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

$d + 1 < 0$

Subtract $1$ from both sides of the inequality.

$d < - 1$

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

$3 (- (d + 1)) - 7 \geq - 1$

Write as a piecewise.

${\begin{cases} 3 (d + 1) - 7 \geq - 1 & d \geq - 1 \\ 3 (- (d + 1)) - 7 \geq - 1 & d < - 1 \end{cases}$

Simplify $3 (d + 1) - 7 \geq - 1$ .

Simplify each term.

Apply the distributive property.

${\begin{cases} 3 d + 3 \cdot 1 - 7 \geq - 1 & d \geq - 1 \\ 3 (- (d + 1)) - 7 \geq - 1 & d < - 1 \end{cases}$

Multiply $3$ by $1$ .

${\begin{cases} 3 d + 3 - 7 \geq - 1 & d \geq - 1 \\ 3 (- (d + 1)) - 7 \geq - 1 & d < - 1 \end{cases}$

Subtract $7$ from $3$ .

${\begin{cases} 3 d - 4 \geq - 1 & d \geq - 1 \\ 3 (- (d + 1)) - 7 \geq - 1 & d < - 1 \end{cases}$

Simplify $3 (- (d + 1)) - 7 \geq - 1$ .

Simplify each term.

Apply the distributive property.

${\begin{cases} 3 d - 4 \geq - 1 & d \geq - 1 \\ 3 (- d - 1 \cdot 1) - 7 \geq - 1 & d < - 1 \end{cases}$

Multiply $- 1$ by $1$ .

${\begin{cases} 3 d - 4 \geq - 1 & d \geq - 1 \\ 3 (- d - 1) - 7 \geq - 1 & d < - 1 \end{cases}$

Apply the distributive property.

${\begin{cases} 3 d - 4 \geq - 1 & d \geq - 1 \\ 3 (- d) + 3 \cdot - 1 - 7 \geq - 1 & d < - 1 \end{cases}$

Multiply $- 1$ by $3$ .

${\begin{cases} 3 d - 4 \geq - 1 & d \geq - 1 \\ - 3 d + 3 \cdot - 1 - 7 \geq - 1 & d < - 1 \end{cases}$

Multiply $3$ by $- 1$ .

${\begin{cases} 3 d - 4 \geq - 1 & d \geq - 1 \\ - 3 d - 3 - 7 \geq - 1 & d < - 1 \end{cases}$

Subtract $7$ from $- 3$ .

${\begin{cases} 3 d - 4 \geq - 1 & d \geq - 1 \\ - 3 d - 10 \geq - 1 & d < - 1 \end{cases}$

Step 2

Solve $3 d - 4 \geq - 1$ for $d$ .

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

Add $4$ to both sides of the inequality.

$3 d \geq - 1 + 4$

Add $- 1$ and $4$ .

$3 d \geq 3$

Divide each term in $3 d \geq 3$ by $3$ and simplify.

Divide each term in $3 d \geq 3$ by $3$ .

$\frac{3 d}{3} \geq \frac{3}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 d}{3} \geq \frac{3}{3}$

Divide $d$ by $1$ .

$d \geq \frac{3}{3}$

Simplify the right side.

Divide $3$ by $3$ .

$d \geq 1$

Step 3

Solve $- 3 d - 10 \geq - 1$ for $d$ .

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

Add $10$ to both sides of the inequality.

$- 3 d \geq - 1 + 10$

Add $- 1$ and $10$ .

$- 3 d \geq 9$

Divide each term in $- 3 d \geq 9$ by $- 3$ and simplify.

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

$\frac{- 3 d}{- 3} \leq \frac{9}{- 3}$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 d}{- 3} \leq \frac{9}{- 3}$

Divide $d$ by $1$ .

$d \leq \frac{9}{- 3}$

Simplify the right side.

Divide $9$ by $- 3$ .

$d \leq - 3$

Step 4

Find the union of the solutions.

$d \leq - 3$ or $d \geq 1$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$d \leq - 3 or d \geq 1$

Interval Notation:

$(- \infty, - 3] \cup [1, \infty)$

Step 6

Solve for d 3|d+1|-7>=-1

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