Rational equation

Question

During a long-distance kayak race series, a competitor traveled for a total of 24 kilometers over the course of 8 hours on two rivers. 12 kilometers were traveled on the first river, and 12 kilometers were traveled on the second river. On the first river, the competitor traveled at an average speed 4 kilometers per hour greater than she traveled on the second river. What was the average speed of the competitor on the first river?

Answer 1

R is the slower speed ---> R+4 is the faster speed

T is the slower time ---> 8-T is the faster time

12 = RT and

12 = (R+4)(8-T) = 8R - RT + 32 - 4T

= 8R - 12 + 32 - 4T <--- substitutes RT=12 from first equation

= 8R + 20 - 4T

-8 = 8R - 4T

2 = -2R + T

2R + 2 = T

substitutes this into the first equation

RT = 12

R(2R+2) = 12

2R^2 + 2R = 12

R^2 + R = 6

R^2 + R - 6 = 0

( R + 3 )( R - 2 ) = 0

R+3 = 0 results in negative speeds

R-2=0 ---> R=2

So 2 km per hour for 6 hours and

6 km per hour for 2 hours

Answer 2

Well let's see - what do we know?

Let S1 be the average speed (km/hr) on River 1, S2 be the average speed (km/hr) on River 2. Then

S1 = S2 + 4km/hr

and

(12km / S1) + (12km/S2) = 8hr

Since we want S1, eliminate S2 by substitution:

S2 = S1 - 4km/hr

so that

(12/S1) + (12/(S1-4)) = 8

or

[12*(S1-4) + 12*S1] / [S1*(S1-4)] = 8

or

12*S1 - 48 + 12*S1 = S1*(S1-4)*8 = 8*S1² - 32*S1

Put is standard form (highest order of unknown first)

8*S1² - 56*S1 + 48 = 0

or

S1² -7*S1 + 6 = 0

This can be factored as

(S1 - 6 ) * ( S1 - 1) = 0

So we have two possible solutions:

S1 = 6 and S1 = 1

A speed of 1 km/hr would requre 12 hours for the first segment of the trip and so is not a solution, leaving:

S1 = 6

S2 = 2.

Checking:

12/6 + 12/2 = 2 + 6 = 8hr.