To solve the equation \( \frac{2}{5}(3x - 1) = \frac{1}{7}(5x + 4) \), follow these steps:

### Step 1: Eliminate the fractions by multiplying both sides by the least common denominator (LCD).
The denominators are 5 and 7. The LCD of 5 and 7 is 35. Multiply both sides of the equation by 35:

\[
35 \cdot \frac{2}{5}(3x - 1) = 35 \cdot \frac{1}{7}(5x + 4)
\]

This simplifies to:

\[
7 \cdot 2(3x - 1) = 5(5x + 4)
\]

### Step 2: Expand both sides.
Expand the terms on both sides:

\[
14(3x - 1) = 5(5x + 4)
\]

\[
42x - 14 = 25x + 20
\]

### Step 3: Collect like terms.
Move the terms involving \(x\) to one side and the constant terms to the other:

\[
42x - 25x = 20 + 14
\]

\[
17x = 34
\]

### Step 4: Solve for \(x\).
Divide both sides by 17:

\[
x = \frac{34}{17} = 2
\]

### Final Answer:
\[
x = 2
\]

Would you like further details or have any questions?

### Related Questions:
1. What happens if the equation had decimals instead of fractions?
2. Can this method be used if the denominators are larger or more complex?
3. What if \(x\) had powers in the equation, like \(x^2\)?
4. How can we check if \(x = 2\) is correct?
5. How do you handle equations with variables on both sides but no fractions?

### Tip:
Always check your solution by substituting the value of \(x\) back into the original equation.

Learn how to solve the linear equation 2/5(3x - 1) = 1/7(5x + 4) using step-by-step methods. This problem focuses on eliminating fractions by using the least common denominator, expanding, and solving for x. Suitable for students in Grades 8-10.

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