$ ext{If } \frac{x + 1}{x - 1} + \frac{x + 2}{x - 2} = \frac{22x + 30}{11x - 18} ext{, then the value of 'x' is :}$

Q: $\text{If } \frac{x + 1}{x - 1} + \frac{x + 2}{x - 2} = \frac{22x + 30}{11x - 18} \text{, then the value of 'x' is :}$

$\text{If } \frac{x + 1}{x - 1} + \frac{x + 2}{x - 2} = \frac{22x + 30}{11x - 18} \text{, then the value of 'x' is :}$

Question

$	ext{If } \frac{x + 1}{x - 1} + \frac{x + 2}{x - 2} = \frac{22x + 30}{11x - 18} 	ext{, then the value of 'x' is :}$

Accepted Answer

To solve the equation $\frac{x + 1}{x - 1} + \frac{x + 2}{x - 2} = \frac{22x + 30}{11x - 18}$ and verify why Option 2, which is $x = -6$, is the correct answer, we will follow these steps:

### Step-by-Step Solution:

1. **Combine the Left Side:**
   $$
   \frac{x + 1}{x - 1} + \frac{x + 2}{x - 2}
   $$
   To combine these fractions, find a common denominator, which is $(x - 1)(x - 2)$:
   $$
   \frac{(x + 1)(x - 2) + (x + 2)(x - 1)}{(x - 1)(x - 2)}
   $$

2. **Expand and Simplify the Numerator:**
   $$
   (x + 1)(x - 2) = x^2 - 2x + x - 2 = x^2 - x - 2
   $$
   $$
   (x + 2)(x - 1) = x^2 - x + 2x - 2 = x^2 + x - 2
   $$
   Adding these results:
   $$
   x^2 - x - 2 + x^2 + x - 2 = 2x^2 - 4
   $$

3. **Rewrite the Left Side:**
   $$
   \frac{2x^2 - 4}{(x - 1)(x - 2)}
   $$

4. **Equate to the Right Side:**
   $$
   \frac{2x^2 - 4}{(x - 1)(x - 2)} = \frac{22x + 30}{11x - 18}
   $$

5. **Cross-Multiply to Solve for x:**
   $$
   (2x^2 - 4)(11x - 18) = (22x + 30)(x - 1)(x - 2)
   $$

6. **Substitute $x = -6$ into the Original Equation:**
   $$
   \frac{-6 + 1}{-6 - 1} + \frac{-6 + 2}{-6 - 2} = \frac{22(-6) + 30}{11(-6) - 18}
   $$
   Simplifying each term:
   $$
   \frac{-5}{-7} + \frac{-4}{-8} = \frac{-132 + 30}{-66 - 18}
   $$
   $$
   \frac{5}{7} + \frac{1}{2} = \frac{-102}{-84}
   $$
   $$
   \frac{5}{7} + \frac{1}{2} = \frac{51}{42}
   $$
   $$
   \frac{10}{14} + \frac{7}{14} = \frac{51}{42}
   $$
   $$
   \frac{17}{14} = \frac{51}{42}
   $$
   Simplifying both sides confirms they are equal, verifying $x = -6$ is correct.

### Explanation for Other Options:

- **Option 1 (x = 6):** Substituting $x = 6$ into the equation does not satisfy the equality, as the left and right sides do not equate.
- **Option 3 (x = 5):** Substituting $x = 5$ into the equation also fails to satisfy the equality.
- **Option 4 (x = -5):** Similarly, substituting $x = -5$ does not result in equal sides.

Thus, the only value of $x$ that satisfies the equation is $-6$, confirming that Option 2 is indeed the correct answer.