Solve for b b/2=c+(ab)/3

$\frac{b}{2} = c + \frac{a b}{3}$

Step 1

Multiply both sides by $2$ .

$\frac{b}{2} \cdot 2 = (c + \frac{a b}{3}) \cdot 2$

Step 2

Simplify.

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{b}{2} \cdot 2 = (c + \frac{a b}{3}) \cdot 2$

Rewrite the expression.

$b = (c + \frac{a b}{3}) \cdot 2$

Simplify the right side.

Simplify $(c + \frac{a b}{3}) \cdot 2$ .

Simplify terms.

Apply the distributive property.

$b = c \cdot 2 + \frac{a b}{3} \cdot 2$

Move $2$ to the left of $c$ .

$b = 2 \cdot c + \frac{a b}{3} \cdot 2$

Combine $\frac{a b}{3}$ and $2$ .

$b = 2 \cdot c + \frac{a b \cdot 2}{3}$

Move $2$ to the left of $a b$ .

$b = 2 c + \frac{2 a b}{3}$

Reorder $2 c$ and $\frac{2 a b}{3}$ .

$b = \frac{2 a b}{3} + 2 c$

Step 3

Solve for $b$ .

Subtract $\frac{2 a b}{3}$ from both sides of the equation.

$b - \frac{2 a b}{3} = 2 c$

Factor $b$ out of $b - \frac{2 a b}{3}$ .

Factor $b$ out of $b^{1}$ .

$b \cdot 1 - \frac{2 a b}{3} = 2 c$

Factor $b$ out of $- \frac{2 a b}{3}$ .

$b \cdot 1 + b (- \frac{2 a}{3}) = 2 c$

Factor $b$ out of $b \cdot 1 + b (- \frac{2 a}{3})$ .

$b (1 - \frac{2 a}{3}) = 2 c$

Divide each term in $b (1 - \frac{2 a}{3}) = 2 c$ by $1 - \frac{2 a}{3}$ and simplify.

Divide each term in $b (1 - \frac{2 a}{3}) = 2 c$ by $1 - \frac{2 a}{3}$ .

$\frac{b (1 - \frac{2 a}{3})}{1 - \frac{2 a}{3}} = \frac{2 c}{1 - \frac{2 a}{3}}$

Simplify the left side.

Cancel the common factor of $1 - \frac{2 a}{3}$ .

Cancel the common factor.

$\frac{b (1 - \frac{2 a}{3})}{1 - \frac{2 a}{3}} = \frac{2 c}{1 - \frac{2 a}{3}}$

Divide $b$ by $1$ .

$b = \frac{2 c}{1 - \frac{2 a}{3}}$

Simplify the right side.

Simplify the denominator.

Write $1$ as a fraction with a common denominator.

$b = \frac{2 c}{\frac{3}{3} - \frac{2 a}{3}}$

Combine the numerators over the common denominator.

$b = \frac{2 c}{\frac{3 - 2 a}{3}}$

Multiply the numerator by the reciprocal of the denominator.

$b = 2 c \frac{3}{3 - 2 a}$

Multiply $2 c \frac{3}{3 - 2 a}$ .

Combine $2$ and $\frac{3}{3 - 2 a}$ .

$b = c \frac{2 \cdot 3}{3 - 2 a}$

Multiply $2$ by $3$ .

$b = c \frac{6}{3 - 2 a}$

Combine $c$ and $\frac{6}{3 - 2 a}$ .

$b = \frac{c \cdot 6}{3 - 2 a}$

Move $6$ to the left of $c$ .

$b = \frac{6 c}{3 - 2 a}$

Solve for b b/2=c+(ab)/3

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