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This problem states that \( y \) varies directly as \( x \), which means we can express this relationship as:

\[
y = kx
\]

where \( k \) is a constant of proportionality. Given that when \( x = 2 \), \( y = 11 \), we can substitute these values to find \( k \).

1. Substitute \( x = 2 \) and \( y = 11 \) into the equation:
   \[
   11 = k \cdot 2
   \]

2. Solve for \( k \):
   \[
   k = \frac{11}{2} = 5.5
   \]

Now that we know \( k = 5.5 \), we can use this to find \( y \) when \( x = 5 \):

\[
y = 5.5 \cdot 5 = 27.5
\]

**Answer:** When \( x = 5 \), \( y = 27.5 \).

Would you like further details or have any questions?

Here are some related questions to expand your understanding:
1. How would the solution change if \( y \) varies inversely as \( x \) instead?
2. If \( y = kx \) and \( k \) changes, what does this mean in real-world applications?
3. Can you find \( x \) if you know \( y \) and \( k \)?
4. What would happen if \( x = 0 \)? How does that affect \( y \) in direct variation?
5. How can we interpret direct variation graphically?

**Tip:** When two variables vary directly, as one increases, the other also increases proportionally. This is represented by a straight line through the origin on a graph.

y varies directly as x. If x = 2 then y = 11. Find y when x = 5.

Solve the direct variation problem where y varies directly as x. Given x = 2 and y = 11, determine the value of y when x = 5. This solution explores concepts of proportionality and includes step-by-step calculations. Suitable for students in Grades 7-9.

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