Solve for p 3+|1+p|=12

$3 + | 1 + p | = 12$

Move all terms not containing $| 1 + p |$ to the right side of the equation.

Subtract $3$ from both sides of the equation.

$| 1 + p | = 12 - 3$

Subtract $3$ from $12$ .

$| 1 + p | = 9$

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$1 + p = \pm 9$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$1 + p = 9$

Move all terms not containing $p$ to the right side of the equation.

Subtract $1$ from both sides of the equation.

$p = 9 - 1$

Subtract $1$ from $9$ .

$p = 8$

Next, use the negative value of the $\pm$ to find the second solution.

$1 + p = - 9$

Move all terms not containing $p$ to the right side of the equation.

Subtract $1$ from both sides of the equation.

$p = - 9 - 1$

Subtract $1$ from $- 9$ .

$p = - 10$

The complete solution is the result of both the positive and negative portions of the solution.

$p = 8, - 10$