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QUESTION: The statement should be:

A can travel the distance B traveled in 15 3/4 days.  B can  travel the distance A traveled in 28 days.

Maybe that will help.  Thanks again.

Sorry for the typo.

ANSWER: They must be walking, and slowly!

A and B travel α miles per day and β miles per day, respectively.
They meet after t days, when A has traveled αt miles and B has traveled βt miles.

A can travel βt miles in 15¾ days:
βt miles × 1 day/(α miles) = 15.75 days

B can travel αt miles in 28 days:
αt miles × 1 day/(β miles) = 28 days
α = 28β/t

Substitution:
βt/(28β/t) = 15.75
t² = 15.75×28 = 441
t = 21 days

When A and B meet, A has traveled 18 miles more than B.
αt = βt+18
21α = 21β + 18
α = β + 6/7

α = 28β/t = 4β/3

β +6/7 = 4β/3
β = 18/7
βt = (18/7)21 = 54

Distance between starting points = 54+72 = 126 miles.

---------- FOLLOW-UP ----------

QUESTION: "α = β + 6/7

α = 28β/t = 4β/3

β +6/7 = 4β/3
β = 18/7
βt = (18/7)21 = 54

I don't understand this part of your solution.  Thanks again.

ANSWER: distance = rate × time
We know time (21 days), so now we need rates α and β.

Eq. I:  α = β+(6/7)
Eq. II:  α = 28β/t = 28β/21 = 4β/3

Combine Eqs. I & II to determine rate β:
β+(6/7) = 4β/3
Eq. III:  β = 18/7

Distance traveled by B:
βt = (18/7)21 = 54 miles

Use Eqs. II and III to determine rate α:
α = 4β/3 = 4(18/7)/3 = 24/7

Distance traveled by A:
αt = (24/7)21 = 72 miles

---------- FOLLOW-UP ----------

QUESTION: "Combine Eqs. I & II to determine rate β:
β+(6/7) = 4β/3
Eq. III:  β = 18/7"

"Use Eqs. II and III to determine rate α:
α = 4β/3 = 4(18/7)/3 = 24/7"

I still don't understand these steps.  Thanks.

You know that α = 4β/3 and α = β+(6/7).
By transitivity,
4β/3 = β+(6/7)
4β/3 - β = 6/7
(4/3 - 1)β = 6/7
(⅓)β = 6/7
β = 3·6/7 = 18/7
B travels at the rate of 18/7 miles per day.

You know that α = 4β/3 and β = 18/7.
By substitution,
A travels at the rate of 24/7 miles per day.
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