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Question

Jeff bicycles 168 miles at the rate of r mph. The same trip would have taken 9 hours longer if he had decreased his speed by 6 mph. Find r.

Answer 1

d = rt means distance equals rate of speed multiply time
write these equations:
168 = r*t
168 = (r-6)(t+9) which becomes rt + 9r - 6t - 54
replace rt with 168 and t with 168/r in the second equation
168 = 168 + 9r - 6(168/r) - 54
subtract 168 both sides
0 = 9r - 1008/r - 54
multiply both sides by r
0 = 9r² - 1008 - 54r
put in standard form and factor out 9
0 = 9 (r² - 6r - 112)
factor this
0 = 9(r + 8)(r - 14)
solve and get r = -8 or r = 14
you cannot have rate be -8 so your answer is 14 mph

Answer 2

Time taken at the rate of r mph = 168/r hours

time taken at the rate of r-6 mph = 168/r-6 hours

so 168/(r-6) - 168/r = 9

1/r-6 - 1/r = 9/168

(r - r + 6)/r(r-6) = 9/168

6/(r^2 - 6r) = 9/168

1/(r^2 - 6r) = 9/(168)6 = 1/112

r^2 - 6r = 112

r^2 - 6r - 112 = 0

(r - 14)(r+8) = 0

r = 14 mph (ignoring negative value)

Answer 3

let r be the speed of Jeff
let t be the time it takes him to travel 168.

distance = speed * time
168 = rt

if he decrease his speed by 6mi/hr, r - 6. Then it will take him 9hrs longer to travel 168 miles. t + 9

168 = (r - 6) (t + 9)

solve for t
t = 168/r

substitube
168 = (r - 6) (168/r + 9)
168 = 168 + 9x - 1008/r - 54
0 = 9r - 54 - 1008/r
0 = 9r^2 - 54r - 1008

use quadratic formula and you'll get r = 14 mi/hr <== answer

Answer 4

distance = rate * time therefore 168 = rt but 168=(r-6)*9t

since you want r, eliminate t: t=168/r => 168=(r-6)9*168/r

simplify: 1/9=(r-6)/r => 1/9 = 1 - 6/r => 6/r =8/9 => r =6*9/8 or 6.75 mph.

I think. :-)

Answer 5

t = 168/r
t + 9 = 168/(r-6)

168/r = 168/(r-6) - 9

168(r-6) = 168r/(r-6) - 9r(r-6)
168r - 168*6 = 168r - 9r²+54r
9r² -54r - 168*6 = 0
r² -6r - 112 = 0
r₊ = 14
r- = -8 but he wasn't pedaling backwards, so r = 14 mi/hr

Answer 6

168 = r*t
168=(r-6)(t+9)
168 = rt +9r-6t-54
168=168+9r-6t-54 so 9r-6t=54 and t= 168/r
9r-1008/r-54=0
r^2-6r-112=0 so r=((6+sqrt(36+448))/2 =14 mph