Solve for b A=1/2*(h(B+b))

$A = \frac{1}{2} \cdot (h (B + b))$

Rewrite the equation as $\frac{1}{2} \cdot (h (B + b)) = A$ .

$\frac{1}{2} \cdot (h (B + b)) = A$

Multiply both sides of the equation by $2$ .

$2 \cdot \frac{1}{2} \cdot (h (B + b)) = 2 \cdot A$

Simplify $2 \cdot \frac{1}{2} \cdot (h (B + b))$ .

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Cancel the common factor of $2$ .

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Cancel the common factor.

$2 \cdot \frac{1}{2} \cdot (h (B + b)) = 2 \cdot A$

Rewrite the expression.

$1 \cdot (h (B + b)) = 2 \cdot A$

Multiply $h (B + b)$ by $1$ .

$h (B + b) = 2 \cdot A$

Apply the distributive property.

$h B + h b = 2 \cdot A$

$h B + h b = 2 A$

Subtract $h B$ from both sides of the equation.

$h b = 2 A - h B$

Divide each term by $h$ and simplify.

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Divide each term in $h b = 2 A - h B$ by $h$ .

$\frac{h b}{h} = \frac{2 A}{h} + \frac{- h B}{h}$

Cancel the common factor of $h$ .

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Cancel the common factor.

$\frac{h b}{h} = \frac{2 A}{h} + \frac{- h B}{h}$

Divide $b$ by $1$ .

$b = \frac{2 A}{h} + \frac{- h B}{h}$

Simplify each term.

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Cancel the common factor of $h$ .

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Cancel the common factor.

$b = \frac{2 A}{h} + \frac{- h B}{h}$

Divide $- 1 B$ by $1$ .

$b = \frac{2 A}{h} - 1 B$

Rewrite $- 1 B$ as $- B$ .

$b = \frac{2 A}{h} - B$

Solve for b A=1/2*(h(B+b))

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