Question 956308
Mark walks 3 miles to the house of a friend and returns home on a bike. He averages 5 mph faster when cycling than walking, and the total time for both trips is 2 hours. Find his walking speed.

Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r

Let r=Mark's walking speed
Then r+5=Marks cycling speed
Time that Mark walked plus time that Mark cycled equals 2 hrs
Time Mark walked=3/r
Time Mark cycled=3/(r+5)
Soooo:
3/r + 3/(r+5)=2  multiply each term by r(r+5)
3(r+5)+3r=2r(r+5)  expand
3r+15+3r=2r^2+10r simplify
2r^2+10r-6r-15=0
2r^2+4r-15=0  Solve using the quadratic formula
r=(-b+-sqrt(b^2-4ac))/2a
r=(-4+-sqrt(16-4*2*(-15))/4
r=(-4+-sqrt(136))/4
r=(-4+-11.66)/4  negative answer is no good. The positive answer is
r=7.66/4=~~~1.92 mph---walking speed

Hope this helps----ptaylor