Question

the formula for the area of a trapezoid is $a = \frac{1}{2}h(b_1 + b_2)$. the height, $h$, of the trapezoid may be expressed as
(1) $\frac{2a}{b_1 + b_2}$
(2) $\frac{1}{2}a(b_1 + b_2)$
(3) $\frac{b_1 + b_2}{2a}$
(4) $\frac{1}{2}a - (b_1 + b_2)$

the solution to the equation $\frac{2(3x - 1)}{3} = x + 2$ is
(1) $\frac{1}{3}$
(2) $\frac{2}{3}$
(3) $\frac{4}{3}$
(4) $\frac{8}{3}$

Explanation:

First Problem (Solve for $h$)

Step1: Multiply both sides by 2

$$2A = h(b_1 + b_2)$$

Step2: Divide by $(b_1 + b_2)$

$$h = \frac{2A}{b_1 + b_2}$$

Step1: Multiply both sides by 3

$$2(3x - 1) = 3(x + 2)$$

Step2: Expand both sides

$$6x - 2 = 3x + 6$$

Step3: Isolate $x$ terms

$$6x - 3x = 6 + 2$$

Step4: Simplify to solve for $x$

$$3x = 8 \implies x = \frac{8}{3}$$

Answer:

(1) $\boldsymbol{\frac{2A}{b_1 + b_2}}$

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Second Problem (Solve for $x$)