Solve for b 3+6|-2+b|=45

$3 + 6 | - 2 + b | = 45$

Move all terms not containing $| - 2 + b |$ to the right side of the equation.

Subtract $3$ from both sides of the equation.

$6 | - 2 + b | = 45 - 3$

Subtract $3$ from $45$ .

$6 | - 2 + b | = 42$

Divide each term in $6 | - 2 + b | = 42$ by $6$ and simplify.

Divide each term in $6 | - 2 + b | = 42$ by $6$ .

$\frac{6 | - 2 + b |}{6} = \frac{42}{6}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 | - 2 + b |}{6} = \frac{42}{6}$

Divide $| - 2 + b |$ by $1$ .

$| - 2 + b | = \frac{42}{6}$

Simplify the right side.

Divide $42$ by $6$ .

$| - 2 + b | = 7$

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

$- 2 + b = \pm 7$

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.

$- 2 + b = 7$

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

Add $2$ to both sides of the equation.

$b = 7 + 2$

Add $7$ and $2$ .

$b = 9$

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

$- 2 + b = - 7$

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

Add $2$ to both sides of the equation.

$b = - 7 + 2$

Add $- 7$ and $2$ .

$b = - 5$

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

$b = 9, - 5$