Let's set up the problem with variables and equations:

• Let L be the price of a large candle.
• Let S be the price of a small candle.

From the problem, we know the following:

1. First customer:
   $3L + 4S = 64$ (Total payment by the first customer)

2. Second customer:
   $1L + 8S = 64 + 4 = 68$ (Total payment by the second customer)

Step 1: Solve one equation for L in terms of S

From the second equation:
$1L + 8S = 68 \quad \Rightarrow \quad L = 68 - 8S$

Step 2: Substitute into the first equation

Now substitute $L = 68 - 8S$ into the first equation $3L + 4S = 64$:

$3(68 - 8S) + 4S = 64$

Step 3: Simplify the equation

Distribute the 3:
$204 - 24S + 4S = 64$

Combine like terms:
$204 - 20S = 64$

Move 204 to the other side:
$-20S = 64 - 204 = -140$

Solve for $S$:
$S = \frac{-140}{-20} = 7$

Thus, the price of a small candle is: $S = 7$

Step 4: Find the price of a large candle

Now substitute $S = 7$ into the equation $L = 68 - 8S$:

$L = 68 - 8(7) = 68 - 56 = 12$

Thus, the price of a large candle is: $L = 12$

Final Answers:

• Small candle: $7
• Large candle: $12

Would you like to see these calculations visualized in another way or have any further questions?

---

Here are 5 questions related to this problem:

1. What happens if the second customer paid $10 more instead of $4 more?
2. How would the solution change if the two equations were inconsistent (no solution)?
3. Can you set up a system of equations for three variables with three customers?
4. What if the large candle’s price increased by 20%—how would the equations change?
5. How would you solve the problem if only the total price of 7 mixed candles was given?

Tip: When solving systems of equations, substitution works well with smaller numbers, while elimination might be faster for larger systems.