Let's go through each of these proportions step by step. I'll start with problem 27 as highlighted, and then briefly outline solutions for the others.

Problem 27:

Given: $\frac{1}{c + 5} = \frac{3}{24}$ To solve for $c$, we cross-multiply: $1 \cdot 24 = 3 \cdot (c + 5)$ which simplifies to: $24 = 3c + 15$ Now, isolate $c$ by moving 15 to the other side: $24 - 15 = 3c$ $9 = 3c$ Then, divide by 3: $c = 3$

Solutions for Other Problems:

Here’s a quick approach for each of the remaining problems:

1. Problem 23: $\frac{6}{x} = \frac{3}{2}$
   Cross-multiply to find $x$.

2. Problem 24: $\frac{y}{20} = \frac{3}{10}$
   Solve for $y$ by cross-multiplying.

3. Problem 25: $\frac{2}{7} = \frac{12}{z}$
   Solve for $z$ by cross-multiplying.

4. Problem 26: $\frac{j + 1}{5} = \frac{4}{10}$
   Set up the equation and isolate $j$.

5. Problem 28: $\frac{4}{a - 3} = \frac{2}{5}$
   Cross-multiply and solve for $a$.

6. Problem 29: $1 + \frac{3b}{4} = \frac{5}{2}$
   Solve for $b$ by isolating terms.

7. Problem 30: $\frac{3}{2p + 5} = \frac{1}{9p}$
   Cross-multiply and isolate $p$.

Would you like more detailed solutions for each of these problems?

Relative Questions for Further Exploration:

1. How do you verify the solution of a proportion equation?
2. What happens if both sides of a proportion have variables in the denominators?
3. Can proportions be solved graphically? How would that look?
4. How does cross-multiplication help in solving proportions?
5. How can proportions be applied in real-world scenarios?

Tip:

When solving proportions, always cross-multiply to eliminate fractions; this simplifies solving for the unknown variable.