Maths Question, help please (1 Viewer)

Over97

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yo, im stuck on this question, n e assitance will be most appreciated.


if spliff walks at 1km/h faster than his usual speed, he takes 15 minutes less than usual for a distance of 5km/h.
Letting his usual rate of waling be xkm/h, use algebra to find his usual rate of walking.

the solution in my paper is

1) 5/x - 1/4 = 5/x+1

2) 20-x/4x = 5/x+1

a quadratic is formed, its solved
and we get 4.

i dont understand step 1, can n e 1 please explain this 2 me.

thanks
 

Lazarus

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note:
s<sub>1</sub>, x<sub>1</sub> = usual rate of walking
s<sub>2</sub> = faster rate of walking

t<sub>1</sub> = time for 5km at the usual rate of walking
t<sub>2</sub> = time for 5km at the faster rate of walking

15 minutes is 1/4 of an hour

step one:

distance = speed * time

s<sub>2</sub> = d/t<sub>2</sub>

x<sub>1</sub> + 1 = 5/(t<sub>1</sub> - 1/4)

but t<sub>1</sub> = d/s<sub>1</sub>

.'. t<sub>1</sub> = 5/x<sub>1</sub>

i.e.

x<sub>1</sub> + 1 = 5/((5/x<sub>1</sub>) - 1/4)

5/x<sub>1</sub> - 1/4 = 5/(x<sub>1</sub> + 1)
 

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