Solve for B 4=6B+16C

$4 = 6 B + 16 C$

Step 1

Rewrite the equation as $6 B + 16 C = 4$ .

$6 B + 16 C = 4$

Step 2

Subtract $16 C$ from both sides of the equation.

$6 B = 4 - 16 C$

Step 3

Divide each term in $6 B = 4 - 16 C$ by $6$ and simplify.

Divide each term in $6 B = 4 - 16 C$ by $6$ .

$\frac{6 B}{6} = \frac{4}{6} + \frac{- 16 C}{6}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 B}{6} = \frac{4}{6} + \frac{- 16 C}{6}$

Divide $B$ by $1$ .

$B = \frac{4}{6} + \frac{- 16 C}{6}$

Simplify the right side.

Simplify each term.

Cancel the common factor of $4$ and $6$ .

Factor $2$ out of $4$ .

$B = \frac{2 (2)}{6} + \frac{- 16 C}{6}$

Cancel the common factors.

Factor $2$ out of $6$ .

$B = \frac{2 \cdot 2}{2 \cdot 3} + \frac{- 16 C}{6}$

Cancel the common factor.

$B = \frac{2 \cdot 2}{2 \cdot 3} + \frac{- 16 C}{6}$

Rewrite the expression.

$B = \frac{2}{3} + \frac{- 16 C}{6}$

Cancel the common factor of $- 16$ and $6$ .

Factor $2$ out of $- 16 C$ .

$B = \frac{2}{3} + \frac{2 (- 8 C)}{6}$

Cancel the common factors.

Factor $2$ out of $6$ .

$B = \frac{2}{3} + \frac{2 (- 8 C)}{2 (3)}$

Cancel the common factor.

$B = \frac{2}{3} + \frac{2 (- 8 C)}{2 \cdot 3}$

Rewrite the expression.

$B = \frac{2}{3} + \frac{- 8 C}{3}$

Move the negative in front of the fraction.

$B = \frac{2}{3} - \frac{8 C}{3}$