1. **Stating the problem:** We are given that the cost $C$ of registering for an examination is partly constant and partly varies with the number of subjects $N$. We know: - $C=70$ when $N=2$ - $C=85$ when $N=3$ We need to find an equation connecting $C$ and $N$. 2. **Formula and explanation:** Since the cost is partly constant and partly varies with $N$, we can write: $$C = a + bN$$ where $a$ is the constant part and $b$ is the cost per subject. 3. **Using the given data:** From $N=2$, $C=70$: $$70 = a + 2b$$ From $N=3$, $C=85$: $$85 = a + 3b$$ 4. **Solving the system:** Subtract the first equation from the second: $$85 - 70 = (a + 3b) - (a + 2b)$$ $$15 = b$$ 5. **Find $a$:** Substitute $b=15$ into the first equation: $$70 = a + 2 \times 15$$ $$70 = a + 30$$ $$a = 70 - 30 = 40$$ 6. **Final equation:** $$\boxed{C = 40 + 15N}$$ This means the registration cost has a fixed part of 40 and an additional 15 per subject.