# How do I solve this rational equation word problem?

I’m having a very hard time figuring this out.

“Two joggers, one averaging 8mph and one averaging 6mph, start from a designated initial point. The slower jogger arrives at the end of the run a half hour after the other jogger. Find the distance of the run.

This is how I set it up.

8X=6X+180

Then, I cross multiply. I get the wrong answer though. The answer is 12 miles.

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## 19 Answers

I am no expert, but I believe you missed something. You do not tell how long anyone ran? The second jogger arrives an half hour after what? After 24 hours, 10 hours?

I am confused, sorry I can not help at this point.

I didn’t miss anything. This is the word problem the book gives. lol

Hmmm, I hope someone smarter than me answers then, sorry.

Where did you get the 180?

Response moderated (Off-Topic)

Pace = Distance / Time

P1 = 8mph

P2 = 6mph

D = D1 = D2

T2 = T1 + 0.5

P2 = D2 / T2

6 = D / (T1 + 0.5)

P1 = D1 / T1

8 = D / T1

Now we have two equations, two variables (D and T1)

Solve Equation 2 for D

D = 8 * T1

Plug D into Equation 1

6 = (8 * T1) / (T1 + 0.5)

6T1 + 3 = 8T1

3 = 2T1

T1 = 1.5

Plug T1=1.5 back into second equation

D = 8 * 1.5

D = 12

Therefore, the distance is 12 miles. The Faster runner completed it in 1.5 hours, and the slower runner completed it in (T2 = T1 + 0.5 = 2 hours

Response moderated (Off-Topic)

Response moderated (Off-Topic)

20/x=16/x-2

I got the first rate… cross multiply and it’s 10mph!

Now for the second one. I must work at it.

Response moderated (Writing Standards)

I solved a diff problem I’ve been having trouble with lol

Where did you get .5 from?

@chelle21689 Since the speed in in hours, you can’t use minutes on the other side of the equation. Half hour translates to .5

@chelle21689 0.5 = one half. The pace is in miles per **hour**. Therefore your time has to be in hours as well. One jogger was one half **hour** faster than the other.

Let “T” = the time run for the faster runner in hours.

Then his distance run is 8 mph x T, or 8T miles.

The time for the slower runner is T + .5; his distance is represented by 6mph x (T + .5) or 6 * (T + .5) or 6T + 3

You now have two terms that equate to each other, since the distances each runner ran are equal.

8T = 6T + 3

2T = 3

T = 1.5 hours

Distance for the first runner = 8mph x 1.5 hours = 12 miles

To check that, solve for the distance that the second runner ran, using those terms.

Let L= the lead that the faster runner has at the end of the race

Let t = time that the faster runner runs

Let d = total distance

L=distance slower runner runs in ½ hr = 6mph*½hr= 3 miles

L is also the lead that the faster runner builds up over the course of the race.

L=3mi = (8–6)mph* t, so t=1.5 hr

d=8mph *1.5 hr=12 mi

Or….

If X is the distance run

The faster runner will take X/8 hours to run this far

The slower runner will take X/6 hours

The difference is ½ hours so X/6 – X/8 = ½

Therefore X = 12

Thanks guys I really appreciate it.

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