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this is translated to I'm sorry if the language is a bit unclear.

"the distance between city A to city B is 72km. a biker rode in a constant speed (which I'll refer to as x) from A to B to get to B at a set time. After 3 hours, the biker slowed down by 2km/h. As a result from that he will get there late and an hour after the time which he was supposed to get there (the set time), he was still 6km from city B, what is the speed at which he went at for the first 3 hours?"

This is my current solutionw ith all the data I've gathered

S = Speed, T = Time, D = Distance

Part A (the first 3 hours) S=x T=3 D=3x

Part B (after the 3 hours) S=x-2 T=$\left(\frac{72-3x}{x-2}\right)$ D=72-3x

I am now looking for an equation to solve for x but I fail to see how to use the 6km to form something meaningful. Can anyone explain this to me?

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  • $\begingroup$ You have to use the "set time" $T_S$, i.e. the planned time of arrival as distinct from the real time of arrival. The set time $T_S$ has the feature that at that time he has travelled $72-6=66$. $\endgroup$ Oct 2 '19 at 8:15
  • $\begingroup$ Ahh I see, I'll try this and try to form an equation from this thank you that makes more sense $\endgroup$
    – a new one
    Oct 2 '19 at 8:17
  • $\begingroup$ Hmm but now I come across the problem of how do i get the time he was supposed to go? I still can't seem to find an equtation for this :/ I know it's the two times combined +1 equals to something but I don't know what to $\endgroup$
    – a new one
    Oct 2 '19 at 8:19
  • $\begingroup$ @MauroALLEGRANZA I see thats the correct answer but how did you get that from $3x + (T-3)(x-2) = 66$? $\endgroup$
    – a new one
    Oct 2 '19 at 8:40
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We have to use the "set time" $T$, i.e. the planned time of arrival as distinct from the real time of arrival.

Let $T$ the expected time at arrival and $T+1$ the real time of arrival ("an hour after the time which he was supposed to get there").

This means that he traveleld for $3$ hours at speed $x$ and for $(T+1)−3=T-2$ hours at speed $x−2$.

Thus, we have the first equation :

$3x+(T−2)(x−2) = 72$.

But we know that at moment $T$ he has traveleld $72-6=66$.

This means that we have a second equation :

$3x+(T−3)(x−2) = 66$.

Subtracting the two we get :

$x-2=6$.

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  • $\begingroup$ How did you go from $3x+(T-3)(x-2)=66$ to $x-2=6$? where did the T go? $\endgroup$
    – a new one
    Oct 2 '19 at 8:52
  • $\begingroup$ @anewone - as said before, subtracting the two equations : $[3x+(T−2)(x−2)] - [3x+(T−3)(x−2)]=72-66=6$. From it : $[(T-2)-(T-3)](x-2)=6$. $\endgroup$ Oct 2 '19 at 9:17

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