algebra help

Question

A man rode a bicycle for 12 miles and then hiked an additional 8 miles. The total time for the trip was 5 hours. If his rate when he was riding a bicycle was 10 miles per hour faster than his rate? walking, what was each? rate?

Answer 1

Use the formula

 

v = d/t

 

where:

v = rate (mph)

d = distance (miles)

t = time (hours)

 

Let the following variables:

 

d_Bike = 12

d_Hike = 8

t_Bike + t_Hike = 5

 

 

Now we can set up an equation using these variables and the formula above.

 

v_Bike - v_Hike = 10

 

(d_Bike / t_Bike) - (d_Hike / d_Hike) = 10

 

(12 / t_Bike) - [8 / (5 - t_Bike)] = 10

 

 

Now we solve the equation for t_Bike.  Let  x = t_Bike  to make the algebra easier.  Rewrite using the LCD.     LCD = x(5 - x)

 

 

[12(5 - x) - 8x] / [x(5 - x)] = 10x(5 - x) / [(x(5 - x)]

 

 

Equate numerators. 

 

60 - 20x = 50x - 10x²

 

10x² - 70x + 60 = 0

 

10(x² - 7x + 6) = 0

 

10(x - 6)(x - 1) = 0

 

x = 6      or      x = 1

 

 

Since the total time is 5 hours, we accept x=1.

 

t_Bike = 1 hour         ,         t_Hike = 4 hours

 

 

Now we find their rates.

 

v_Bike = 12 mph

v_Hike = 8 miles / 4 hours = 2 mph