Answer: A. Zero

Explanation:1. Understand the problem: We are given a system of three linear equations: 1) $x + 4y = -16$ 2) $5x + 20y = 0$ 3) $3x + 12y = 34$ We need to find the number of solutions $(x, y)$ for this system.

2. Apply method (analyze slopes and intercepts): Convert each equation to the slope-intercept form $y = mx + c$, where $m$ is the slope and $c$ is the y-intercept.

3. Perform calculation: Equation 1: $x + 4y = -16$ $4y = -x - 16$ $y = -\dfrac{1}{4}x - 4$ Slope $m\_1 = -\dfrac{1}{4}$, y-intercept $c\_1 = -4$.

Equation 2: $5x + 20y = 0$ $20y = -5x$ $y = -\dfrac{5}{20}x$ $y = -\dfrac{1}{4}x$ Slope $m\_2 = -\dfrac{1}{4}$, y-intercept $c\_2 = 0$.

Equation 3: $3x + 12y = 34$ $12y = -3x + 34$ $y = -\dfrac{3}{12}x + \dfrac{34}{12}$ $y = -\dfrac{1}{4}x + \dfrac{17}{6}$ Slope $m\_3 = -\dfrac{1}{4}$, y-intercept $c\_3 = \dfrac{17}{6}$.

All three lines have the same slope $m = -\dfrac{1}{4}$. Their y-intercepts are $-4$, $0$, and $\dfrac{17}{6}$, which are all different. This means the three lines are parallel and distinct.

4. Determine the number of solutions: Since the three lines are parallel and distinct, they never intersect at a common point. Therefore, there is no point $(x, y)$ that satisfies all three equations simultaneously.

5. Conclusion: The system of three equations has zero solutions.