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The image shows a pre-algebra problem asking to solve for \( x \) in the following equation:

\[
\frac{4}{5}x - \frac{1}{3}x = 7
\]

Let's break down the steps to solve for \( x \):

1. **Combine like terms**:  
   The terms \( \frac{4}{5}x \) and \( \frac{1}{3}x \) both have \( x \), so we can factor out \( x \).

   \[
   \left( \frac{4}{5} - \frac{1}{3} \right) x = 7
   \]

2. **Find the common denominator** between \( \frac{4}{5} \) and \( \frac{1}{3} \):  
   The least common denominator (LCD) of 5 and 3 is 15. We convert both fractions:

   \[
   \frac{4}{5} = \frac{12}{15}, \quad \frac{1}{3} = \frac{5}{15}
   \]

3. **Subtract the fractions**:  
   \[
   \frac{12}{15} - \frac{5}{15} = \frac{7}{15}
   \]

   So the equation becomes:
   \[
   \frac{7}{15}x = 7
   \]

4. **Solve for \( x \)** by multiplying both sides by the reciprocal of \( \frac{7}{15} \), which is \( \frac{15}{7} \):

   \[
   x = 7 \times \frac{15}{7}
   \]

   The 7 cancels out:
   \[
   x = 15
   \]

So, the value of \( x \) is **15**.

Let me know if you need further clarification or more details on the steps!

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Here are five questions that build on this problem:

1. How do you solve an equation with fractions if the denominators are different?
2. What is the least common denominator (LCD) and why is it important when working with fractions?
3. How do you check if your solution to a linear equation is correct?
4. What is the process of factoring like terms in algebra?
5. How can you rewrite fractions in equivalent forms to simplify an equation?

**Tip**: Always simplify fractions before performing operations to make solving equations easier!

Complete the steps to solve for x in the equation (4/5)x - (1/3)x = 7.

Learn how to solve a linear equation with fractions: (4/5)x - (1/3)x = 7. This solution involves combining like terms, finding the least common denominator, and solving for x. Suitable for students in Grades 7-9.

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