Question

Please Help!!!
D= r*t???

For his morning workout, a boxer bicycles 8 miles and then jogs back to his house along the same route.
If he bicycles 6 miles per hour faster than he jogs and the entire workout was 2 hours, how fast does his jog?
a) Declare Variable
b) Equation
c) Solving
d) Answer
--------------------------------------------
So far I have
a) x: For how fast the boxer jogs
x+6: For how fast the boxer bicycles
Now this is where I get fuzzy is it
Distance= Rate * Time
Plane 16 x+6 2
Car 16 x 2

Answer 1

Hey Binky c:
Hmm what is this Plane/Car stuff about?

Answer 2

OH! This reminds me of something actually thank you! I knew it looked funny

Answer 3

Oh I didn't plug in the distance, hehe..

Answer 4

bahh i dunno >.< my brain so fried today..

Answer 5

who would jog for freaken 2 hrs?? it's suicidal!!!! D:

Answer 6

Well I think so

So the variables it would be
x: How fast the boxer jogs

Distance = Rate * Time
Bike 8 x+6 8/(x+6)
Jog 8 x 8/x

So the equation would be
8/x=8/(x+6) + 2

Answer 7

@zepdrix haha so is mine now... I thought this problem was all finished but my friend that helped me with it earlier this weekend texted me and said she had some doubts and maybe I should look at it again. SO that lead to a whole chain of problems and uncertainty.

Answer 8

@sourwing No kidding haha I am good for an hour, but two hours Im pretty dead.

Answer 9

@BinkyBoo , it should be
8/x + 8/(x+6) = 2
x = 6 mi/hr

Answer 10

so
\[\frac{ 8 }{ x } + \frac{ 8 }{ x+6 }=2\]
\[x(x+6) [\frac{ 8 }{ x}+\frac{ 8 }{ x+6 }]= [2] x(x+6)\]\[x(x+6)(\frac{ 8 }{ x }) + x(x+6)(\frac{ 8 }{ x+6})= 2 (x)(x+6)\]\[8(x+6) +8x=2x(x+6)\]
\[8x+48+8x=2x ^{2}+12x\]\[16x+48=2x ^{2}+12x\]
\[0=2x ^{2}-4x-48\]

Answer 11

\[0=2(x ^{2}-2x-24)\]
\[0=2(x+4)(x-6)\]
0=2 or x+4=0 or x-6=0
can't be x=-4 x=6
Cant be can be

Answer 12

so x= How fast the boxer jogs is 6 mph
???

Answer 13

yes

Answer 14

Thank you!