To solve this problem, we need to determine when the two candles will be the same height after burning for several hours.

Given:

• The small candle burns $\frac{1}{2}$ inch every hour and starts at 12 inches tall.
• The large candle burns $\frac{2}{3}$ inch every hour and starts at 15 inches tall.

Let $t$ represent the number of hours after which the two candles will be the same height.

Small candle:

The height of the small candle after $t$ hours is: $\text{Height of small candle} = 12 - \frac{1}{2}t$

Large candle:

The height of the large candle after $t$ hours is: $\text{Height of large candle} = 15 - \frac{2}{3}t$

To find when the heights are the same:

Set the heights of both candles equal to each other: $12 - \frac{1}{2}t = 15 - \frac{2}{3}t$

Solving for $t$:

1. First, subtract 12 from both sides: $-\frac{1}{2}t = 3 - \frac{2}{3}t$

2. Next, add $\frac{2}{3}t$ to both sides: $\frac{2}{3}t - \frac{1}{2}t = 3$

3. To simplify, find a common denominator for the fractions. The least common denominator of 2 and 3 is 6: $\left(\frac{4}{6}t - \frac{3}{6}t\right) = 3$ $\frac{1}{6}t = 3$

4. Multiply both sides by 6: $t = 18$

Conclusion:

The two candles will be the same height after 18 hours.

Thus, the correct answer is D. 18 hours.

Would you like further details on any step or clarification?

Here are 5 related questions to expand your understanding:

1. How would the problem change if the initial heights of the candles were different?
2. What would happen if the burning rates of the candles were reversed?
3. How can we generalize this solution for any initial height and burn rate?
4. Can we solve this problem graphically by plotting the height over time?
5. How would the equation change if one of the candles burned at a variable rate?

Tip:

When solving equations with fractions, it’s often helpful to first eliminate the fractions by multiplying through by the least common denominator.