Solve for b 15=|-3b+10|

$15 = | - 3 b + 10 |$

Rewrite the equation as $| - 3 b + 10 | = 15$ .

$| - 3 b + 10 | = 15$

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

$- 3 b + 10 = \pm 15$

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.

$- 3 b + 10 = 15$

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

Subtract $10$ from both sides of the equation.

$- 3 b = 15 - 10$

Subtract $10$ from $15$ .

$- 3 b = 5$

Divide each term in $- 3 b = 5$ by $- 3$ and simplify.

Divide each term in $- 3 b = 5$ by $- 3$ .

$\frac{- 3 b}{- 3} = \frac{5}{- 3}$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 b}{- 3} = \frac{5}{- 3}$

Divide $b$ by $1$ .

$b = \frac{5}{- 3}$

Simplify the right side.

Move the negative in front of the fraction.

$b = - \frac{5}{3}$

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

$- 3 b + 10 = - 15$

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

Subtract $10$ from both sides of the equation.

$- 3 b = - 15 - 10$

Subtract $10$ from $- 15$ .

$- 3 b = - 25$

Divide each term in $- 3 b = - 25$ by $- 3$ and simplify.

Divide each term in $- 3 b = - 25$ by $- 3$ .

$\frac{- 3 b}{- 3} = \frac{- 25}{- 3}$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 b}{- 3} = \frac{- 25}{- 3}$

Divide $b$ by $1$ .

$b = \frac{- 25}{- 3}$

Simplify the right side.

Dividing two negative values results in a positive value.

$b = \frac{25}{3}$

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

$b = - \frac{5}{3}, \frac{25}{3}$

The result can be shown in multiple forms.

Exact Form:

$b = - \frac{5}{3}, \frac{25}{3}$

Decimal Form:

$b = - 1.\overline{6}, 8.\overline{3}$

Mixed Number Form:

$b = - 1 \frac{2}{3}, 8 \frac{1}{3}$

Solve for b 15=|-3b+10|

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