Question 100914
Let w be the walking speed, then w+2 must be the running speed (since we're told that his running speed is 2m/s more than his walking speed).

Let the walking time be t seconds and the running time be T seconds


Now using these facts: distance= average speed*time,

                       the distance is 15 metres in each case( walking and running),

                       the factors of 15 are 5 and 3 and

                       t+T= 8 (given)

                       
we can find w BY INSPECTION as follows:

Since 15=3*5=w*t
and
      15=5*3=(w+2)*T then it follows that w=3,t=5 and T=3.

So the walking speed,w,is 3m/sec.ANS 


You can of course solve the question using algebra but why do that with such a trivial question.
In an exam situation time is of the essence and a solution BY INSPECTION is totally valid provided you show what it is based upon.

Using algebra and the definitions given above a possible solution might be as follows:

15=w*t so t=15/w
Similarly,15=(w+2)*T so T=15/(w+2).

But t+T=8 and so (15/w)+(15/(w+2))=8
This leads to (1/w)+(1/(w+2))=8/15
Adding the two algebraic fractions together we get:

(w+2+w)/w(w+2) = 8/15

Hence 2w+2 = 8 (i) and w(w+2) = 15 (ii)
Considering 2w+2 = 8 it follows that w+1 = 4 (dividing both sides by 2)
And so w = 3
That is, the walking speed is 3m/sec.ANS