Question

Zac and Lynn are each traveling on a trip. So far, Zac has traveled 123.75 miles in 2.25 hours. Lynn leaves half an hour after Zac. So far
she has traveled 105 miles in 1.75 hours. Assume Zac and Lynn travel at constant rates.
Let a represent the number of hours that have elapsed since Zac started traveling. Let y represent the number of miles traveled. Write a system of linear equations that represents the distance each of them has traveled since Zac left on his trip.
Assume Zac and Lynn continue to travel at the same constant rates and make no stops.

Determine the solution of the system of linear equations.​

Answer 1

Solution :-

Let

• x = Number of hours that have elapsed since Zac started traveling .
• y = Number of miles traveled .

we know that,

• Speed = Distance / Time.

so,

→ Speed of Zac = D / T = 123.75 / 2.25 = 55 miles / hour.

→ Speed of Lynn = D / T = 105 / 1.75 = 60 miles / hour.

then,

→ Distance travelled by Zac in x hours = S * T

→ y = 55 * x

→ y = 55x miles

and,

→ Distance travelled by Lynn in (x - 0.5) hours = S * T

→ y = 60 * (x - 0.5)

→ y = 60x - 30

therefore, a system of linear equations that represents the distance each of them has traveled since Zac left on his trip are :-

→ y = 55x ----------- Eqn.(1)

→ y = 60x - 30 ------------- Eqn.(2)

now, putting value of Eqn.(1) in Eqn.(2) , we get,

→ 55x = 60x - 30

→ 60x - 55x = 30

→ 5x = 30

→ x = 6 hours.

putting value of x in Eqn.(1),

→ y = 55 * 6

→ y = 330 miles.

hence, the solution of the system of linear equations is :-

• x = 6 hours .
• y = 330 miles.

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