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The image contains two math problems written on a chalkboard. I will solve each of them step by step.

### Problem 1:
\[
\frac{6}{S + 5} + \frac{S - 5}{2} = 3
\]

#### Solution:
1. **Find the Least Common Denominator (LCD):**  
   The LCD for the fractions is \(2(S + 5)\).
   
2. **Multiply each term by the LCD to eliminate the denominators:**
   \[
   2(S + 5) \cdot \frac{6}{S + 5} + 2(S + 5) \cdot \frac{S - 5}{2} = 3 \cdot 2(S + 5)
   \]
   Simplifying each term:
   \[
   12 + (S - 5)(S + 5) = 6(S + 5)
   \]

3. **Expand and simplify the equation:**
   \[
   12 + S^2 - 25 = 6S + 30
   \]
   Combine like terms:
   \[
   S^2 - 6S - 43 = 0
   \]

4. **Solve the quadratic equation:**
   Use the quadratic formula:
   \[
   S = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-43)}}{2(1)}
   \]
   \[
   S = \frac{6 \pm \sqrt{36 + 172}}{2}
   \]
   \[
   S = \frac{6 \pm \sqrt{208}}{2}
   \]
   \[
   S = \frac{6 \pm 4\sqrt{13}}{2}
   \]
   \[
   S = 3 \pm 2\sqrt{13}
   \]

So the solutions are:
\[
S = 3 + 2\sqrt{13} \quad \text{or} \quad S = 3 - 2\sqrt{13}
\]

### Problem 2:
\[
\frac{3}{x} + \frac{4}{2x} = x - 1
\]

#### Solution:
1. **Find the LCD:**  
   The LCD here is \(2x\).

2. **Multiply each term by the LCD to eliminate the denominators:**
   \[
   2x \cdot \frac{3}{x} + 2x \cdot \frac{4}{2x} = 2x \cdot (x - 1)
   \]
   Simplifying:
   \[
   6 + 4 = 2x^2 - 2x
   \]
   Combine terms:
   \[
   10 = 2x^2 - 2x
   \]

3. **Rearrange to form a quadratic equation:**
   \[
   2x^2 - 2x - 10 = 0
   \]
   Divide the entire equation by 2:
   \[
   x^2 - x - 5 = 0
   \]

4. **Solve the quadratic equation:**
   Use the quadratic formula:
   \[
   x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-5)}}{2(1)}
   \]
   \[
   x = \frac{1 \pm \sqrt{1 + 20}}{2}
   \]
   \[
   x = \frac{1 \pm \sqrt{21}}{2}
   \]

So the solutions are:
\[
x = \frac{1 + \sqrt{21}}{2} \quad \text{or} \quad x = \frac{1 - \sqrt{21}}{2}
\]

---

Would you like more details on any part of the solutions or have any questions? Here are some related questions for further exploration:

1. How does the quadratic formula work, and why is it effective for solving quadratic equations?
2. What is the significance of finding the least common denominator in rational equations?
3. Can you explain how to simplify expressions involving square roots?
4. How do you determine whether a quadratic equation has real or complex solutions?
5. What are the steps to factor quadratic equations instead of using the quadratic formula?

**Tip:** Always double-check your work when solving quadratic equations by substituting the solution back into the original equation.

Solve the following algebraic equations: (1) 6/(S+5) + (S-5)/2 = 3 and (2) 3/x + 4/(2x) = x - 1

This page provides a detailed step-by-step solution to two algebraic equations involving rational and quadratic expressions. Learn how to solve equations like 6/(S+5) + (S-5)/2 = 3 and 3/x + 4/(2x) = x - 1 using the quadratic formula and the least common denominator (LCD). Ideal for students in Grades 9-11.

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