Hi,

An athlete usually runs 80km at a steady speed of $\displaystyle v$km per hour. He decides to reduce his speed by 2.5km per hour resulting in his run taking an extra 2 hours and 40 minutes.

Derive $\displaystyle \frac{80}{v}+\frac{8}{3}=\frac{160}{2v-5}$

Here's what I did which unfortunately is wrong:

$\displaystyle \frac{80}{60(v-2.5)}-\frac{80}{60v}=160$ (Converted it to minutes)

$\displaystyle \frac{80}{60(v-2.5)}=160+\frac{80}{60v}$

$\displaystyle \frac{80}{60v-150}=160+\frac{80}{60v}$

$\displaystyle 80=160(60v-150)+\frac{80(60v-150)}{60v}$

$\displaystyle 80=9600v-24000+\frac{80(60v-150)}{60v}$

$\displaystyle 24080-9600v=\frac{80(60v-150)}{60v}$

$\displaystyle 60v(24080-9600v)=80(60v-150)$

$\displaystyle 1444800v-576000v^2=4800v-12000$

$\displaystyle 576000v^2-1440000v-12000=0$

$\displaystyle 48v^2-120v-1=0$

How on Earth did I get it so far from the answer?

Any help is appreciated