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You’ve shared a set of proportion problems. Let's work through a few of them to show how they can be solved.

### Problem 19:  
\[
\frac{7}{3} = \frac{2x + 5}{x}
\]

**Step 1**: Cross-multiply the terms:
\[
7 \cdot x = 3 \cdot (2x + 5)
\]

**Step 2**: Expand the right-hand side:
\[
7x = 6x + 15
\]

**Step 3**: Move all terms involving \(x\) to one side:
\[
7x - 6x = 15 \quad \Rightarrow \quad x = 15
\]

So, the solution is \( x = 15 \).

---

### Problem 20:  
\[
\frac{a}{9a - 2} = \frac{1}{8}
\]

**Step 1**: Cross-multiply:
\[
a \cdot 8 = (9a - 2) \cdot 1
\]

**Step 2**: Simplify both sides:
\[
8a = 9a - 2
\]

**Step 3**: Move the terms involving \(a\) to one side:
\[
8a - 9a = -2 \quad \Rightarrow \quad -a = -2
\]

**Step 4**: Solve for \(a\):
\[
a = 2
\]

So, the solution is \( a = 2 \).

---

### Problem 23:
\[
\frac{k - 8}{7 + k} = -\frac{1}{5}
\]

**Step 1**: Cross-multiply:
\[
5(k - 8) = -(7 + k)
\]

**Step 2**: Expand both sides:
\[
5k - 40 = -7 - k
\]

**Step 3**: Move all terms involving \(k\) to one side:
\[
5k + k = -7 + 40 \quad \Rightarrow \quad 6k = 33
\]

**Step 4**: Solve for \(k\):
\[
k = \frac{33}{6} = 5.5
\]

So, the solution is \( k = 5.5 \).

---

Would you like further details for any other problems?

### Further Practice:
1. Solve problem 24. 
2. Solve problem 27.
3. Solve problem 29.
4. Explore the concept of cross-multiplication in proportions.
5. Check if cross-multiplication can be applied to all types of ratios.

**Tip**: Always simplify your expressions before cross-multiplying to reduce calculation complexity.

Learn how to solve the algebraic proportion 7/3 = (2x + 5)/x using cross-multiplication. This detailed guide covers key concepts in proportions and algebra, suitable for students in Grades 8-10.

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