Solve for b 2/3b+5=20-b

$\frac{2}{3} b + 5 = 20 - b$

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

$\frac{2 b}{3} + 5 = 20 - b$

Move all terms containing $b$ to the left side of the equation.

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Add $- b$ to both sides of the equation.

$\frac{2 b}{3} + 5 + b = 20$

Simplify the left side of the equation.

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To write $\frac{b}{1}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{2 b}{3} + \frac{b}{1} \cdot \frac{3}{3} + 5 = 20$

Write each expression with a common denominator of $3$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$\frac{2 b}{3} + \frac{b \cdot 3}{1 \cdot 3} + 5 = 20$

Multiply $3$ by $1$ .

$\frac{2 b}{3} + \frac{b \cdot 3}{3} + 5 = 20$

Combine the numerators over the common denominator.

$\frac{2 b + b \cdot 3}{3} + 5 = 20$

Add $2 b$ and $b \cdot 3$ .

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Reorder $b$ and $3$ .

$\frac{2 b + 3 \cdot b}{3} + 5 = 20$

Add $2 b$ and $3 \cdot b$ .

$\frac{5 b}{3} + 5 = 20$

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

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Subtract $5$ from both sides of the equation.

$\frac{5 b}{3} = 20 - 5$

Subtract $5$ from $20$ .

$\frac{5 b}{3} = 15$

Multiply both sides of the equation by $\frac{3}{5}$ .

$\frac{3}{5} \cdot \frac{5 b}{3} = \frac{3}{5} \cdot 15$

Simplify both sides of the equation.

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Simplify the left side.

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Cancel the common factor of $3$ .

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Cancel the common factor.

$\frac{3}{5} \cdot \frac{5 b}{3} = \frac{3}{5} \cdot 15$

Rewrite the expression.

$\frac{1}{5} \cdot (5 b) = \frac{3}{5} \cdot 15$

Cancel the common factor of $5$ .

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Factor $5$ out of $5 b$ .

$\frac{1}{5} \cdot (5 (b)) = \frac{3}{5} \cdot 15$

Cancel the common factor.

$\frac{1}{5} \cdot (5 b) = \frac{3}{5} \cdot 15$

Rewrite the expression.

$b = \frac{3}{5} \cdot 15$

Simplify $\frac{3}{5} \cdot 15$ .

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Cancel the common factor of $5$ .

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Factor $5$ out of $15$ .

$b = \frac{3}{5} \cdot (5 (3))$

Cancel the common factor.

$b = \frac{3}{5} \cdot (5 \cdot 3)$

Rewrite the expression.

$b = 3 \cdot 3$

Multiply $3$ by $3$ .

$b = 9$

Solve for b 2/3b+5=20-b

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