Set up and solve a rational equation that models the following situation answer with a complete sentence

Question

Time taken down stream = t

Time taken upstream = t+4

Distance paddled each side = 42 Km

Speed in still water = X

Speed Upstream = X - 4 ( Speed relative to current)

Speed downstream = X +4

X=?

Speed = Distance /time

X- 4 = 42/(t -4)...... Eq 1

X+4 = 42/ ( t+4).... Eq 2

Solve these two equations for two variables X and t. Make a model, with distance(km) speed(km/h), time(h), upstream and downstream portions. Show step by step how to solve the equations. Thanks!

Answer 1

t(x+4)=42

(t+4)(x-4)=42

from first equation tx+4t=42 and so x=(42-4t)/t

substituting into second equation

(t+4)((42-4t)/t-4)=42

multiply by t

(t+4)(42-4t-4t)=42t

(t+4)(42-8t)=42t a little housekeeping gets

-8t²-32t+168=0 which you divide by -8 to get

t²+4t-21=0 factor (t+7)(t-3)=0 t=-7 or t=3

Assume t=3 so in first equation 3(x+4)=42 and so x+4=14 x=10

Answer 2

4 hours longer to travel the 42 km upstream than downstream...

current is 4 km/hr

x is speed of the boat

distance = rate * time

42 = (x+4)*t

= (x-4)*(t+4)

1st equation: 42 = xt + 4t

42 - 4t = xt

2nd equation: 42 = xt + 4x - 4t - 16

58 = xt + 4x - 4t

substitutes xt = 42-4t into 2nd equation:

58 = (42-4t) + 4x - 4t

= 42 + 4x - 8t

16 = 4x - 8t

4 = x - 2t

x = 2t+4

plugs into 1st equaiton:

42 - 4t = (2t+4)t

0 = (2t+4)t + 4t - 42

0 = 2t^2 + 4t + 4t - 42

0 = 2t^2 + 8t - 42

0 = t^2 + 4t - 21

0 = ( t + 7 )( t - 3 )

t+7=0 is disqualified because it results in negative measures

t-3=0 --> t = 3

Then x = 2t+4 = 2(3)+4 = 6+4=10

THe speed of the boat is 10

check:

upstream: (10-4)*(3+4) = 6*7=42

downstream (10+4)*3 = 14*3 = 42

Yes, the speed of the boat is 10