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The problem states:

- John has 6 less than 8 times as many WOF tickets as Vince.
- Together, John and Vince have 147 tickets.
- We need to determine how many tickets John has.

Let \( V \) represent the number of tickets Vince has.

From the first statement, the number of tickets John has, \( J \), is:
\[
J = 8V - 6
\]

From the second statement, the total number of tickets is 147:
\[
J + V = 147
\]

Now we can substitute the expression for \( J \) into the second equation:
\[
(8V - 6) + V = 147
\]

Simplifying:
\[
9V - 6 = 147
\]

Adding 6 to both sides:
\[
9V = 153
\]

Dividing both sides by 9:
\[
V = 17
\]

Now that we know Vince has 17 tickets, we can find how many John has:
\[
J = 8(17) - 6 = 136 - 6 = 130
\]

So, John has 130 tickets. Therefore, the correct answer is **D. 130**.

Would you like more details or have any questions?

Here are some related questions:
1. What would happen if John had 10 less than 8 times Vince's tickets?
2. How many tickets would they have together if Vince had twice the tickets?
3. How can you set up a similar equation if John had \( 5 \) more than \( 4 \) times Vince's tickets?
4. What if they had 200 tickets together instead of 147?
5. How can you use a different method (e.g., substitution or graphing) to solve this system?

**Tip:** Always start by assigning variables for unknowns and translating word problems into equations.

John has 6 less than 8 times as many WOF tickets as Vince. Together, John and Vince have 147 tickets. How many tickets does John have?

Solve an algebraic problem where John has 6 less than 8 times as many WOF tickets as Vince, and together they have 147 tickets. This problem involves solving linear equations and is suitable for students in Grades 7-9.

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