### Instruction: In trapezoid $ABCD$, sides $\overline{AB}$ and $\overline{CD}$ are parallel, $\angle A = 2\angle D$, and $\angle C = 3\angle B$.  Find $\angle A$.### Response: I know that the sum of the interior angles of a trapezoid is $360^\circ$, so I can write an equation using that fact.
Let $x$ be the measure of $\angle D$. Then $\angle A = 2x$, $\angle B = \frac{1}{3}\angle C$, and $\angle C = 3\angle B$.
Substituting these expressions into the equation $A + B + C + D = 360$, I get $2x + \frac{1}{3}\angle C + \angle C + x = 360$.
Simplifying, I get $3x + \frac{4}{3}\angle C = 360$.
I want to eliminate $\angle C$ and solve for $x$, so I use the fact that $\angle C = 3\angle B$ and substitute $\angle B = \frac{1}{3}\angle C$.
This gives me $3x + \frac{4}{3}(3\angle B) = 360$.
Expanding, I get $3x + 4\angle B = 360$.
Now I can use the fact that $\overline{AB}$ and $\overline{CD}$ are parallel, which means that $\angle A$ and $\angle B$ are supplementary.
This means that $\angle A + \angle B = 180$, or $\angle B = 180 - \angle A$.
Substituting $\angle A = 2x$, I get $\angle B = 180 - 2x$.
Plugging this into the equation $3x + 4\angle B = 360$, I get $3x + 4(180 - 2x) = 360$.
Simplifying, I get $-5x + 720 = 360$.
Subtracting $720$ from both sides, I get $-5x = -360$.
Dividing both sides by $-5$, I get $x = 72$.
This means that $\angle D = 72^\circ$, and $\angle A = 2\angle D = 2(72) = 144^\circ$.