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.

Answer 1

Complete question is;

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:

Zac: y = 55a

Lynn: y = 60(a - ½)

6 hours after Zac started traveling, both Zac and Lynn would have covered 330 miles each.

Step-by-step explanation:

Zac has traveled 123.75 miles in 2.25 hours. Since he travels at constant speed, we can say;

zac's speed = 123.75/2.25 = 55 mi/hr

Similarly, Lynn traveled 105 miles in 1.75 hours. Thus, since she travels at a constant speed;

Lynn's speed = 105/1.75 = 60 mi/hr

Now, we are told that a represents the number of hours that have elapsed since Zac started traveling and y represents the number of miles traveled.

Thus;

a hours after Zac started travelling, his distance covered will be;

Zac: y = 55a

Now,for Lynn, since she started ½ an hour after Zac, it means a hours after Zac started, she had traveled (a - ½) hours.

Thus, Lynn's distance traveled after Zac started = 60(a - ½)

Lynn: y = 60(a - ½)

The solution will be when they have travelled equal distances a hours after Zac started. Thus;

55a = 60(a - ½)

55a = 60a - 30

60a - 55a = 30

5a = 30

a = 30/5

a = 6 hours

Putting 6 for a in y = 55a, we have;

y = 55 × 6

y = 330 miles