Question

An office manager booked 72 airline tickets. He booked 7 more tickets on Airlines A than on Airline B. On Airlines​ C, he booked 5 more than thrice as many tickets as on Airline B. How many tickets did he book on each​ airline?

He booked ___ tickets on Airlines​ A, ___ tickets on Airlines​ B, and ___ on Airlines C.

Answer 1

Hence, he booked 19 tickets for Airlines A, 12 tickets for Airline B, and 41 tickets for Airlines C.

Explanation:

Suppose the office manager booked a+b+c = 72 tickets.

a = b + 7

c = 3b + 5.

So, we have (b + 7) + b +(3b + 5) = 72, or

5b + 12 = 72, or

5b = 60, or

b = 12.

Answer 2

The manager booked 19 tickets on Airlines A, 12 tickets on Airlines B, and 41 tickets on Airlines C.

Let's denote the number of tickets booked on Airlines B as x.

The number of tickets booked on Airlines A is 7 more than on Airlines B, so the number on Airlines A is x + 7.

The number of tickets booked on Airlines C is 5 more than thrice the number on Airlines B, so the number on Airlines C is 3x + 5.

The total number of tickets booked is given as 72. Therefore, we can write the equation:

x + (x + 7) + (3x + 5) = 72

Combine like terms:

5x + 12 = 72

Subtract 12 from both sides:

5x = 60

Divide by 5:

x = 12

Now we can find the number of tickets for each airline:

Number of tickets on Airlines A: x + 7 = 12 + 7 = 19

Number of tickets on Airlines B: x = 12

Number of tickets on Airlines C: 3x + 5 = 3(12) + 5 = 36 + 5 = 41