Daniel leaves school and runs 5 miles to his friend's house where he had left his bike the previous day. after picking up his bike, he then rides a distance of 2 miles to his home. his running speed is 3 mph slower than his biking speed. his total running and biking time is 2 hours. what is daniel's biking speed?

Let the running speed if Daniel be r mph, then his biking speed is (r+3) mph, because "His running speed is 3 mph slower than his biking speed. "

let the time it took Daniel to get to his friend's house be t hours, then the time it took him to go to his house, from his friend,s house is (2-t) hours , because "His total running and biking time is 2 hours."

the main formula we need to solve this problem is: 
                                           
                                      Distance=Speed*Time

Daniels biking speed, (r+3) , is Distance/Time=2/(2-t)

so (r+3)=2/(2-t)

we also have the equation  5=rt, because
"Daniel leaves school and runs 5 miles to his friend's house"

we substitute t, by 5/r in the previous equation:

(r+3)=2/(2-t)

(r+3)=2/(2-5/r)

(r+3)(2-5/r)=2

2r-5+6-15/r=2

2r-1-15/r=0

multiply by r:

2r^2-r-15=0

the expression in the left side can be factorized as (2r+5)(r-3), so the equation becomes:

(2r+5)(r-3)=0,

and the roots are r=-5/2, which is not possible in our problem,

and r=3

Daniel's biking speed is r+3=3+3=6 (mph)

6 mph

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