Let's denote:

• $d$ as the distance covered by the train (in kilometers).
• $s$ as the original speed of the train (in km/h).

Given conditions:

1. If the speed had been 10 km/h faster, the time taken would be 2 hours less.
2. If the speed had been 10 km/h slower, the time taken would be 3 hours more.

Using the relationship $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$:

First Condition:

• Original time taken at speed $s$ is $\frac{d}{s}$.
• New time taken at speed $s + 10$ is $\frac{d}{s + 10}$.
• According to the condition, $\frac{d}{s} - \frac{d}{s + 10} = 2$.

Second Condition:

• New time taken at speed $s - 10$ is $\frac{d}{s - 10}$.
• According to the condition, $\frac{d}{s - 10} - \frac{d}{s} = 3$.

Formulating Equations:

1. $\frac{d}{s} - \frac{d}{s + 10} = 2$ $\Rightarrow \frac{d(s + 10) - ds}{s(s + 10)} = 2 \quad \text{(Multiply both sides by } s(s + 10) \text{)}$ $\Rightarrow \frac{10d}{s(s + 10)} = 2$ $\Rightarrow 5d = s(s + 10)$ $\Rightarrow s^2 + 10s - 5d = 0 \quad \text{(Equation 1)}$

2. $\frac{d}{s - 10} - \frac{d}{s} = 3$ $\Rightarrow \frac{d(s - 10) - ds}{(s - 10)s} = 3 \quad \text{(Multiply both sides by } s(s - 10) \text{)}$ $\Rightarrow \frac{-10d}{s(s - 10)} = 3$ $\Rightarrow -10d = 3s(s - 10)$ $\Rightarrow 3s^2 - 30s + 10d = 0 \quad \text{(Equation 2)}$

Solving the System of Equations:

From Equation 1: $d = \frac{s^2 + 10s}{5}$

Substitute this into Equation 2: $3s^2 - 30s + 10\left(\frac{s^2 + 10s}{5}\right) = 0$ $\Rightarrow 3s^2 - 30s + 2(s^2 + 10s) = 0$ $\Rightarrow 3s^2 - 30s + 2s^2 + 20s = 0$ $\Rightarrow 5s^2 - 10s = 0$ $\Rightarrow 5s(s - 2) = 0$ $\Rightarrow s = 0 \quad \text{or} \quad s = 2$