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Let's solve the problem step by step. Here's the given mathematical expression:

\[
\frac{1}{7} + \frac{10}{3 - \frac{3}{5}}
\]

### Step 1: Simplify the denominator \(3 - \frac{3}{5}\)
The expression in the denominator is:
\[
3 - \frac{3}{5}.
\]

First, write \(3\) as a fraction with a denominator of \(5\):
\[
3 = \frac{15}{5}.
\]

Now subtract:
\[
\frac{15}{5} - \frac{3}{5} = \frac{12}{5}.
\]

So the denominator becomes \(\frac{12}{5}\).

### Step 2: Simplify \(\frac{10}{\frac{12}{5}}\)
Dividing by a fraction is equivalent to multiplying by its reciprocal. Thus:
\[
\frac{10}{\frac{12}{5}} = 10 \cdot \frac{5}{12}.
\]

Simplify the product:
\[
10 \cdot \frac{5}{12} = \frac{50}{12}.
\]

Now reduce \(\frac{50}{12}\) to its simplest form:
\[
\frac{50}{12} = \frac{25}{6}.
\]

### Step 3: Add \(\frac{1}{7} + \frac{25}{6}\)
To add the two fractions, find the least common denominator (LCD) of \(7\) and \(6\). The LCD is \(42\). Rewrite the fractions with the denominator \(42\):
\[
\frac{1}{7} = \frac{6}{42}, \quad \frac{25}{6} = \frac{175}{42}.
\]

Now add:
\[
\frac{6}{42} + \frac{175}{42} = \frac{181}{42}.
\]

### Final Answer:
The simplified result is:
\[
\frac{181}{42}.
\]

If needed, this can also be expressed as a mixed number:
\[
\frac{181}{42} = 4 \frac{13}{42}.
\]

Let me know if you'd like further details or have questions!

---

Here are 5 related questions to expand on this topic:
1. How do you find the least common denominator of two fractions?
2. How do you convert an improper fraction to a mixed number?
3. What are some strategies for simplifying fractions efficiently?
4. How can you verify your answer when adding or subtracting fractions?
5. What is the reciprocal of a fraction, and why is it used in division?

**Tip:** Always simplify fractions step by step to avoid calculation errors.

Simplify and express the result as a proper fraction or mixed number: \(\frac{1}{7} + \frac{10}{3 - \frac{3}{5}}\).

Learn how to simplify the expression \(\frac{1}{7} + \frac{10}{3 - \frac{3}{5}}\) step-by-step. This problem demonstrates fraction addition, reciprocals, and reducing fractions. Ideal for students in Grades 7-9.

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