Flying

The airplane from Prague to Bratislava was flying at a speed of 60 km/h less and back by 70 km/h greater than the original speed. What was the original speed if the plane returned to Prague according to the timetable?

Correct result:

v =  840 km/h

Solution:

$$\begin{gathered} s\mathrm{/}(v-60)+s\mathrm{/}(v+70)=2s\mathrm{/}v \\ 1\mathrm{/}(v-60)+1\mathrm{/}(v+70)=2\mathrm{/}v \\ v(v+70)+v(v-60)=2\cdot (v+70)(v-60) \\ 10v=8400 \\ v=8400\mathrm{/}10=840 \\ =840\text{ km/h} \end{gathered}$$

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