The problem from the image appears to describe a scenario where a train travels from station A to station B at a constant speed. Several conditions are given:

1. If the train travels 1.5 km/h faster than usual, it arrives 1 hour earlier.
2. If the train travels 1 km/h slower than usual, it arrives 1 hour later.

The goal is to determine the usual travel time in minutes from station A to station B.

Let's break this down mathematically:

Let:

• $v$ be the usual speed of the train in km/h.
• $t$ be the usual time it takes for the train to travel from A to B (in hours).
• The distance between station A and station B is $d = v \times t$.

Scenario 1: Train is 1.5 km/h faster

If the train travels 1.5 km/h faster than usual, it arrives 1 hour earlier. The new speed is $v + 1.5$, and the new time is $t - 1$.

Thus, the distance $d$ can be expressed as: $d = (v + 1.5)(t - 1)$ Since $d = v \times t$, we can set up the following equation: $v \times t = (v + 1.5)(t - 1)$ Expanding this: $v \times t = (v + 1.5)t - (v + 1.5)$ $v \times t = v \times t + 1.5t - v - 1.5$ Simplifying this equation: $0 = 1.5t - v - 1.5 \quad \text{(Equation 1)}$

Scenario 2: Train is 1 km/h slower

If the train travels 1 km/h slower than usual, it arrives 1 hour later. The new speed is $v - 1$, and the new time is $t + 1$.

Thus,