System of equations

Question

Hal gives Tom the same amount of tickets as Tom first had and gave Gary the same amount of tickets as Gary first had. Tom then gives Hal and Gary as many tickets as each then has. Gary then gives Tom and Hal as many tickets as each then has. If each finally has 40 tickets, how many tickets did Tom first have? I already turned in the assignment, but this one is bugging me and I really want to know the answer!

Answer 1

In the beginning, Hal has H, Tom has T, and Gary has G

After the first transaction when Hal gives his tickets away:

Hal : H-T-G

Tom : 2T

Gary : 2G

After the 2nd transaction when Tom gives his tickets away:

Hal: 2H - 2T - 2G

Tom: 2T - H + T + G - 2G = 3T - G - H

Gary: 4G

After the 3rd transaction when Gary gives his tickets away:

Hal: 4H - 4T - 4G

Tom: 6T - 2G - 2h

Gary : 4G - 2h + 2t + 2g - 3t + g + h = 7g -h - t

These final expressions are equal to 40

SO.....

H-T-G = 10 after dividing both sides by 4 <--- original first equation

3T - G - H = 20 after dividing both sides by 2

7g - h - t = 40

Next, eliminates h by adding eq1 + eq2:

2T - 2g = 30

t - g = 15

THEN eliminates h by subtraction Eq 3 - Eq 2:

8g - 4t = 20

2g - t = 5

Finally eliminates t by adding these two new equations in bold:

t - g + 2g - t = 15 + 5

g = 20

Then per first bolded equation, t = 35.

this forces h = 65 per original first equation as noted above.

CHECK:

Hal: 65

Tom: 35

Gary : 20

Trans #1: Hal gives his tix

Hal: 10

Tom: 70

Gary 40

Trans #2: Tom gives his tix

Hal: 20

Tom: 20

Gary: 80

Trans #3: Gary gives his tix

which makes it 40 each