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The given equation to solve for \( X \) is:

\[
\frac{x}{3} + \frac{4}{15} - x = \frac{1}{6} \cdot \frac{7x}{10}
\]

Let's solve this step-by-step.

### Step 1: Simplify the right-hand side

First, compute the product on the right-hand side:

\[
\frac{1}{6} \cdot \frac{7x}{10} = \frac{7x}{60}
\]

So the equation becomes:

\[
\frac{x}{3} + \frac{4}{15} - x = \frac{7x}{60}
\]

### Step 2: Find a common denominator

To simplify the fractions on the left side, we should find a common denominator. The least common multiple of 3, 15, and 60 is 60. So, we rewrite each fraction with 60 as the denominator:

\[
\frac{x}{3} = \frac{20x}{60}, \quad \frac{4}{15} = \frac{16}{60}, \quad -x = -\frac{60x}{60}
\]

Substitute these into the equation:

\[
\frac{20x}{60} + \frac{16}{60} - \frac{60x}{60} = \frac{7x}{60}
\]

### Step 3: Combine the fractions

Combine all fractions into a single fraction on the left side:

\[
\frac{20x + 16 - 60x}{60} = \frac{7x}{60}
\]

Simplify the numerator:

\[
\frac{-40x + 16}{60} = \frac{7x}{60}
\]

### Step 4: Eliminate the denominators

Since the denominators are the same, we can equate the numerators:

\[
-40x + 16 = 7x
\]

Combine like terms:

\[
16 = 47x
\]

### Step 5: Solve for \( x \)

Divide both sides by 47:

\[
x = \frac{16}{47}
\]

### Conclusion

The value of \( x \) is:

\[
\boxed{x = \frac{16}{47}}
\]

Would you like more details or have any questions about the steps? Here are some related questions to consider:

1. What would happen if there were different denominators on both sides?
2. How can you check if the solution \( x = \frac{16}{47} \) is correct?
3. How do you handle equations with variables in both the numerator and denominator?
4. What are some common mistakes to avoid when solving fractions?
5. Can this method be used for more complex algebraic equations?

**Tip:** Always remember to find a common denominator when dealing with fractions in equations to simplify the problem effectively.

Calcule el valor de X para la siguiente ecuación: (x/3) + (4/15) - x = (1/6) * (7x/10)

Learn how to solve an algebraic equation involving fractions. The problem demonstrates step-by-step simplification, finding a common denominator, and solving for X. Suitable for Grades 9-11 students.

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