Tony went on an 8-mile run. For

the first 1/3 of an hour he ran at a

constant speed. For the last 3⁄4 of an

hour he ran at a constant speed that

was 2 miles per hour faster than his

original speed. What was his original

speed, and what was his ending scene on the run?

Original speed: 6 miles per hour

Ending: 8 miles per hour

Step-by-step explanation:

Speed = distance/time

Distance = speed * time

Let S be the original speed

(S * ⅓) + [ (S + 2) * ¾] = 8

S/3 + 3S/4 + 3/2 = 8

Lcm: 12

(S*4 + 3S*3) / 12 = 8 - 3/2

13S/12 = 13/2

S = 13/2 * 12/13

S = 6

The first rate is 6 miles per hour

The final rate is 6+2 or 8 miles per hour

Step-by-step explanation:

We know d = rt where distance is equal to rate times time

First part

d1 = r * 1/3

Second part

d2 = (r+2) (3/4)

We know the total distance is 8 miles and is

d1+d2 = 8

r * 1/3 + (r+2) (3/4) = 8

Multiply each side by 12 to get rid of the fractions

12 (r * 1/3 + (r+2) (3/4)) = 8*12

4r + (r+2) 9 = 96

Distribute

4r + 9r + 18 = 96

Combine like terms

13r + 18 = 96

Subtract 18 from each side

13r+18-18 = 96-18

13r = 78

Divide each side by 13

13r/13 = 78/13

r=6

The first rate is 6 miles per hour

The final rate is 6+2 or 8 miles per hour