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The problem you're dealing with is based on **direct variation**, which follows the formula:

\[
y = kx
\]

Where:
- \( y \) varies directly with \( x \),
- \( k \) is the constant of proportionality.

### Given:
- When \( y = 70 \), \( x = 14 \).

We can first solve for the constant \( k \) using the given values:

\[
70 = k \times 14
\]
\[
k = \frac{70}{14} = 5
\]

Now that we know \( k = 5 \), we can solve for \( y \) when \( x = 24 \).

\[
y = 5 \times 24 = 120
\]

### Final Answer:
\[
y = 120
\]

Would you like any further clarification or details?

Here are five related questions:
1. How would the formula change if \( y \) varied **inversely** with \( x \)?
2. Can you calculate \( y \) if \( x = 30 \) using the same \( k \)?
3. What happens to the value of \( y \) if \( x \) is doubled?
4. How would you express \( x \) in terms of \( y \) and \( k \)?
5. What is the graphical representation of a direct variation between \( x \) and \( y \)?

**Tip**: Always start by solving for the constant of proportionality \( k \) when working with direct variation problems!

The value of 'y' varies directly with 'x'. If y = 70, then x = 14. Solve for y when x = 24.

Learn how to solve a direct variation problem where y = 70 when x = 14, and find the value of y when x = 24. This algebra problem focuses on direct proportionality and is suitable for students in Grades 7-8.

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