Question

Tina is training for a biathlon. To train for the running portion of the race, she runs 10 miles each day, over the same course. The first 4 miles of the course is on level ground, while the last 6 miles is downhill. She runs 4 miles per hour slower on level ground than she runs downhill. If the complete course takes 1 hour, how fast does she run on the downhill part of the course?

Answer 1

@Miracrown

Answer 2

Let the downhill speed = s
Then the level speed = s - 4
\[Time=\frac{distance}{speed}\]
\[\frac{4}{s-4}+\frac{6}{s}=1 ............(1)\]
Do you understand how I got equation (1)?

Answer 3

yes, i do

Answer 4

I multiply the left side by the right denominator & do the opposite the the right denominator, right?

Answer 5

Good. Next we need to put the left hand side of (1) into a single fraction:
\[\frac{4s+6(s-4)}{s(s-4)}=1\]

Answer 6

\[\frac{ 4s+6s-24 }{ s^2 }\]
right?

Answer 7

s^2-4s, i mean

Answer 8

Do I just add 4s+6s?

Answer 9

Now by cross multiplying we get:
\[4s+6s-24=s ^{2}-4s...........(2)\]
Now rearrange (2) to get a quadratic to gthen solve for s.

Answer 10

to then solve*

Answer 11

I know the s^2 is first & it'll end with =0?

Answer 12

Correct.

Answer 13

s^2-4s+6s-24=0, or do I have to change some signs?

Answer 14

\[s ^{2}-14s+24=0............(3)\]
Do you see where a sign change was needed?

Answer 15

yes. now I factor it right?

Answer 16

it's 12 & 2

Answer 17

(both negatives

Answer 18

(s-12)(s-4)

Answer 19

i mean 2

Answer 20

That is correct. Test each solution and discard the invalid one. Both possible solutions are positive,

Answer 21

12 & 2 ! Thank you, can you help with the last one like this? please?

Answer 22

If you plug the value s = 2 into equation (1) you get:
\[\frac{4}{2-4}+\frac{6}{2}=-2+3=1\]
This gives a negative value of time for the level part of the course. Therefore s = 2 is not a valid solution. Leaving 12 miles per hour as the solution for the downhill speed.

Answer 23

That's what I put on my work.