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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

α = 4β/3 = 24/7

αt = (24/7)21 = 72

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

α = 4β/3 = 24/7

αt = (24/7)21 = 72"

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,

α = 24/7

A travels at the rate of 24/7 miles per day.

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