Let v = the velocity of the river
We know d = v *t where v is velocity d is distance and t is time
First, take the trip up river
78.2 = (v-3) *t1 Since the current is working against us
Now look at the trip down river
78.2 = ( v+3) *t2 since the current will help us
We know that t1+t2 is 8 hours
Solving each equation for the time
78.2 / (v-3) = t1
78.2 / ( v+3) = t2
78.2 / (v-3) + 78.2 / ( v+3) = t1+t2 = 8
78.2 / (v-3) + 78.2 / ( v+3) = 8
Now get a common denominator
I will multiply both sides by (v-3)(v+3) and clear the fractions
(v-3)(v+3) ( 78.2 / (v-3) + 78.2 / ( v+3) = 8)
78.2 * (v+3) + 78.2 ( v-3) = 8 * (v-3)(v+3)
78.2 v +234.6 + 78.2v - 234.6 = 8 ( v^2 +3v-3v -9)
156.4v = 8v^2 -72
Subtract 156.4v from each side
0 = 8v^2 -156.4v - 72
Using the quadratic formula
a= 8 b = 156.4 c = -72
I get v = 20 and v = -9/20
But the velocity cannot be negative so v= 20 kph