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Iet x hours be the time that he waled at 8km//h. Then, he walled at 10km//h for qquad kours qquad The diatance he walled = qquad * qquad Therefore the equation is qquad -61

Iet xx hours be the time that he waled at 8 km/h8\,\text{km/h}. Then, he walled at 10 km/h10\,\text{km/h} for \quad kours \quad The diatance he walled = ∗\quad * \quad Therefore the equation is −61\quad -61

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Full solution

Q. Iet xx hours be the time that he waled at 8 km/h8\,\text{km/h}. Then, he walled at 10 km/h10\,\text{km/h} for \quad kours \quad The diatance he walled = ∗\quad * \quad Therefore the equation is −61\quad -61
  1. Calculate Distance at 8 km/h8\,\text{km/h}: Let's calculate the distance walked at 8 km/h8\,\text{km/h}. Distance == speed ×\times time, so the distance walked at 8 km/h8\,\text{km/h} is 8×x km8 \times x\,\text{km}.
  2. Calculate Distance at 1010 km/h: Now, let's calculate the distance walked at 1010 km/h. The distance walked at 10 km/h10 \, \text{km/h} is 10×k km10 \times k \, \text{km}.
  3. Find Total Distance: To find the total distance walked, we add the distances from both speeds together. So, the total distance is 8×x+10×k8 \times x + 10 \times k km.
  4. Identify Mistake: The problem states that the total distance is −61 km-61\,\text{km}, which doesn't make sense because distance can't be negative. There must be a mistake in the problem statement.

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