Let $ n$ be the number of number-pairs $ (x,y)$ which satisfy $ 5y \minus{} 3x \equal{} 15$ and $ x^2 \plus{} y^2 \le 16$.  Then $ n$ is:

$ \textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \text{more than two, but finite} \qquad \textbf{(E)}\ \text{greater than any finite number}$
1. **Rewrite the linear equation in slope-intercept form:**
   \[
   5y - 3x = 15 \implies 5y = 3x + 15 \implies y = \frac{3}{5}x + 3
   \]
   This is the equation of a line with slope \(\frac{3}{5}\) and y-intercept 3.

2. **Interpret the inequality \(x^2 + y^2 \leq 16\):**
   This inequality represents a circle centered at the origin with radius 4.

3. **Determine the intersection points of the line and the circle:**
   Substitute \(y = \frac{3}{5}x + 3\) into the circle's equation \(x^2 + y^2 = 16\):
   \[
   x^2 + \left(\frac{3}{5}x + 3\right)^2 = 16
   \]
   Simplify the equation:
   \[
   x^2 + \left(\frac{3}{5}x + 3\right)^2 = x^2 + \left(\frac{9}{25}x^2 + \frac{18}{5}x + 9\right) = 16
   \]
   \[
   x^2 + \frac{9}{25}x^2 + \frac{18}{5}x + 9 = 16
   \]
   Combine like terms:
   \[
   \left(1 + \frac{9}{25}\right)x^2 + \frac{18}{5}x + 9 = 16
   \]
   \[
   \frac{34}{25}x^2 + \frac{18}{5}x + 9 = 16
   \]
   Multiply through by 25 to clear the fractions:
   \[
   34x^2 + 90x + 225 = 400
   \]
   Simplify:
   \[
   34x^2 + 90x - 175 = 0
   \]

4. **Solve the quadratic equation:**
   Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 34\), \(b = 90\), and \(c = -175\):
   \[
   x = \frac{-90 \pm \sqrt{90^2 - 4 \cdot 34 \cdot (-175)}}{2 \cdot 34}
   \]
   \[
   x = \frac{-90 \pm \sqrt{8100 + 23800}}{68}
   \]
   \[
   x = \frac{-90 \pm \sqrt{31900}}{68}
   \]
   \[
   x = \frac{-90 \pm 178.52}{68}
   \]
   Calculate the two solutions:
   \[
   x_1 = \frac{-90 + 178.52}{68} \approx 1.30
   \]
   \[
   x_2 = \frac{-90 - 178.52}{68} \approx -3.94
   \]

5. **Find the corresponding \(y\) values:**
   For \(x_1 \approx 1.30\):
   \[
   y_1 = \frac{3}{5}(1.30) + 3 \approx 3.78
   \]
   For \(x_2 \approx -3.94\):
   \[
   y_2 = \frac{3}{5}(-3.94) + 3 \approx 0.64
   \]

6. **Verify the points \((x_1, y_1)\) and \((x_2, y_2)\) are within the circle:**
   \[
   (1.30)^2 + (3.78)^2 \approx 1.69 + 14.29 = 15.98 \leq 16
   \]
   \[
   (-3.94)^2 + (0.64)^2 \approx 15.52 + 0.41 = 15.93 \leq 16
   \]

Since both points lie within the circle, there are exactly two points of intersection.

The final answer is \(\boxed{2}\)