Solve for b₂ A=1/2h(b_1+b_2)

$A = \frac{1}{2} h (b_{1} + b_{2})$

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

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

Multiply both sides of the equation by $2$ .

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

Simplify the left side.

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

Apply the distributive property.

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

Apply the distributive property.

$2 (\frac{1}{2} (h b_{1}) + \frac{1}{2} (h b_{2})) = 2 A$

Multiply $\frac{1}{2} (h b_{1})$ .

Combine $h$ and $\frac{1}{2}$ .

$2 (\frac{h}{2} b_{1} + \frac{1}{2} (h b_{2})) = 2 A$

Combine $\frac{h}{2}$ and $b_{1}$ .

$2 (\frac{h b_{1}}{2} + \frac{1}{2} (h b_{2})) = 2 A$

Multiply $\frac{1}{2} (h b_{2})$ .

Combine $h$ and $\frac{1}{2}$ .

$2 (\frac{h b_{1}}{2} + \frac{h}{2} b_{2}) = 2 A$

Combine $\frac{h}{2}$ and $b_{2}$ .

$2 (\frac{h b_{1}}{2} + \frac{h b_{2}}{2}) = 2 A$

Apply the distributive property.

$2 \frac{h b_{1}}{2} + 2 \frac{h b_{2}}{2} = 2 A$

Cancel the common factor of $2$ .

Cancel the common factor.

$2 \frac{h b_{1}}{2} + 2 \frac{h b_{2}}{2} = 2 A$

Rewrite the expression.

$h b_{1} + 2 \frac{h b_{2}}{2} = 2 A$

Cancel the common factor of $2$ .

Cancel the common factor.

$h b_{1} + 2 \frac{h b_{2}}{2} = 2 A$

Rewrite the expression.

$h b_{1} + h b_{2} = 2 A$

Subtract $h b_{1}$ from both sides of the equation.

$h b_{2} = 2 A - h b_{1}$

Divide each term in $h b_{2} = 2 A - h b_{1}$ by $h$ and simplify.

Divide each term in $h b_{2} = 2 A - h b_{1}$ by $h$ .

$\frac{h b_{2}}{h} = \frac{2 A}{h} + \frac{- h b_{1}}{h}$

Simplify the left side.

Cancel the common factor of $h$ .

Cancel the common factor.

$\frac{h b_{2}}{h} = \frac{2 A}{h} + \frac{- h b_{1}}{h}$

Divide $b_{2}$ by $1$ .

$b_{2} = \frac{2 A}{h} + \frac{- h b_{1}}{h}$

Simplify the right side.

Simplify each term.

Cancel the common factor of $h$ .

Cancel the common factor.

$b_{2} = \frac{2 A}{h} + \frac{- h b_{1}}{h}$

Divide $- 1 b_{1}$ by $1$ .

$b_{2} = \frac{2 A}{h} - 1 b_{1}$

Rewrite $- 1 b_{1}$ as $- b_{1}$ .

$b_{2} = \frac{2 A}{h} - b_{1}$

Solve for b₂ A=1/2h(b_1+b_2)

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