Two people leave from towns that are 195 miles apart at the same time and travel along the same road toward each other. the first person drives 5 miles slower than the second person. if they meet in three hours, what rate did they each drive at ?

The formula of distance-time-speed
d = v × t

From the question above, we know that
⇒ The total of distance they travel is 195 miles, or it could be written as
d₁ + d₂ = 195 (equation 1)
⇒ The first person is 5 miles slower than the second person, or it could be written as
v₁ = v₂ - 5 (equation 2)

Substitute formula to equation 1
d₁ + d₂ = 195
(v₁ × t) + (v₂ × t) = 195
t(v₁ + v₂) = 195
substitute equation 2
t(v₁ + v₂) = 195
t(v₂ - 5 + v₂) = 195
t(2v₂ - 5) = 195
remember, t = 3 hours
3(2v₂ - 5) = 195
6v₂ - 15 = 195
6v₂ = 195 + 15
6v₂ = 210
v₂ = 210/6
v₂ = 35

v₁ = v₂ - 5
v₁ = 35 - 5
v₁ = 30

The rate of first person is 30 miles/hour, the rate of second person is 35 miles/hour

The correct answer is $90