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This problem describes a direct variation between \( y \) and \( x \), meaning \( y = kx \), where \( k \) is the constant of variation.

We are given:
- \( y = 8 \) when \( x = 5 \).

Step 1: Find the constant \( k \).
\[
y = kx \implies 8 = k \cdot 5 \implies k = \frac{8}{5}.
\]

Step 2: Use \( k \) to find \( y \) when \( x = 15 \).
\[
y = kx = \frac{8}{5} \cdot 15 = \frac{8 \cdot 15}{5} = 24.
\]

Thus, \( y = 24 \) when \( x = 15 \).

Would you like more details or have any questions?

Here are some related questions:
1. What happens to \( y \) if \( x \) doubles in a direct variation?
2. How do you determine the constant of variation in general?
3. What is the formula for inverse variation and how does it differ from direct variation?
4. If \( y = 4x \), what is \( y \) when \( x = 20 \)?
5. How does changing \( x \) affect \( y \) when \( y \) varies directly with \( x \)?

**Tip**: Always check whether the variation is direct or inverse before setting up the equation.

If y varies directly as x, and y = 8 when x = 5, find y when x = 15.

This tutorial solves a direct variation problem where y varies directly as x. Given y = 8 when x = 5, find y when x = 15 using the formula y = kx. A step-by-step approach is provided to find the constant of variation and solve for y.

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