The image you uploaded contains three rational equations that need to be solved, along with the Least Common Denominator (LCD) identified for each equation. Let’s go through the problems step by step:

1. $\frac{5}{6} - \frac{x}{4} = \frac{1}{12}$

The LCD here is 12 (as noted in the image).

• Rewrite each fraction with a denominator of 12: $\frac{5}{6} = \frac{10}{12}, \quad \frac{x}{4} = \frac{3x}{12}, \quad \frac{1}{12} = \frac{1}{12}$
• Now rewrite the equation: $\frac{10}{12} - \frac{3x}{12} = \frac{1}{12}$
• Multiply the entire equation by 12 to eliminate the denominators: $10 - 3x = 1$
• Solve for $x$: $10 - 1 = 3x \implies 9 = 3x \implies x = 3$

2. $\frac{x + 2}{x - 5} = \frac{7}{2}$

The LCD is $2(x - 5)$ (as noted in the image).

• Cross-multiply: $2(x + 2) = 7(x - 5)$
• Expand both sides: $2x + 4 = 7x - 35$
• Solve for $x$: $2x - 7x = -35 - 4 \implies -5x = -39 \implies x = \frac{39}{5}$

3. $\frac{3x}{x + 1} = \frac{12}{x^2 - 1} + 2$

The LCD is $(x + 1)(x - 1)$ (since $x^2 - 1 = (x + 1)(x - 1)$).

• First, rewrite the equation: $\frac{3x}{x + 1} = \frac{12}{(x + 1)(x - 1)} + 2$
• Multiply everything by the LCD $(x + 1)(x - 1)$ to eliminate denominators: $3x(x - 1) = 12 + 2(x + 1)(x - 1)$
• Expand both sides: $3x^2 - 3x = 12 + 2(x^2 - 1)$ $3x^2 - 3x = 12 + 2x^2 - 2$
• Simplify: $3x^2 - 3x = 2x^2 + 10$
• Bring all terms to one side: $3x^2 - 2x^2 - 3x - 10 = 0 \implies x^2 - 3x - 10 = 0$
• Solve using the quadratic formula: $x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-10)}}{2(1)}$ $x = \frac{3 \pm \sqrt{9 + 40}}{2} \implies x = \frac{3 \pm \sqrt{49}}{2} \implies x = \frac{3 \pm 7}{2}$
• So, $x = 5$ or $x = -2$.

Would you like more details on any of these steps? Let me know if you have any questions or if you need further clarification.

Relative questions:

1. What is the significance of the Least Common Denominator in rational equations?
2. How do you determine when to use cross-multiplication in rational equations?
3. What are the possible issues when solving rational equations involving quadratic terms?
4. How does the quadratic formula help in solving non-linear equations like in problem 3?
5. How can you check for extraneous solutions in rational equations?

Tip:

Always check your solutions in the original equation to ensure they do not result in undefined expressions (like division by zero).