Antonio went on a bike ride of 72 miles. He realized that if he had gone 12 mph faster, he would have arrived 8 hours sooner. How fast did he actually ride?

Answer 1

Distance (d) = rate (r) times time (t)

Antonio went 72=(r)(t)

Divide both sides by R

T=72/r

He could have gone

12 mph Faster (r+12)

arrived 8 hrs sooner (t-8)

72=(r+12)(t-8)

Substitute T=72/r

72=(r+12)(72/r -8)

Distribute

72=72-8r+864/r -96

Subtract 72 both sides

0=-8r+864/r -96

Multiply through by r

0=-8r²+864-96r

Factor -8 and put in standard form

0=-8(r²+12r-108)

Factor

0=-8(r+18)(r-6)

Set (r+18)=0. r=-18 cannot have a negative rate so discard

Set r-6=0. r=6

So he actually rode 6mph

To check

T=72/r =72/6=12

If he went 12mph faster that would be 6+12=18 mph

If he arrived 8 hrs sooner that would be 12-8=4 hrs

Distance = rate times time

18 times 4 is 72 miles so it checks

Answer 2

Suppose Antonio traveled the 72 miles at a rate of r miles per hour in t hours.

If he went 12mph faster it would have taken 8 hours less to travel the 72 miles.

We need to know that rate*time = distance

So we have r*t=72

and (r+12)(t-8) = 72.

Can you solve from here?