Home work \( \begin{array}{ll}\text { a) } \frac{x-b}{x^{2}-4}-\frac{2}{x-2}=\frac{-1}{x+2} & \text { (b) } \frac{x^{2}-1}{x+1}=-2\end{array} \)

Sure, let's solve both parts of your homework step by step. --- ### **Problem a)** \[ \frac{x - b}{x^{2} - 4} - \frac{2}{x - 2} = \frac{-1}{x + 2} \] #### **Step 1: Factor the Denominator** Notice that \(x^{2} - 4\) is a difference of squares and can be factored: \[ x^{2} - 4 = (x - 2)(x + 2) \] So, the equation becomes: \[ \frac{x - b}{(x - 2)(x + 2)} - \frac{2}{x - 2} = \frac{-1}{x + 2} \] #### **Step 2: Find a Common Denominator** The common denominator for all terms is \((x - 2)(x + 2)\). Multiply each term by this denominator to eliminate the fractions: \[ (x - b) - 2(x + 2) = - (x - 2) \] #### **Step 3: Simplify the Equation** Expand and simplify: \[ x - b - 2x - 4 = -x + 2 \] Combine like terms: \[ (-x) - b - 4 = -x + 2 \] #### **Step 4: Solve for \(b\)** Add \(x\) to both sides to eliminate \(x\): \[ -b - 4 = 2 \] Now, solve for \(b\): \[ -b = 6 \\ b = -6 \] **Conclusion for Part a:** The equation holds true for any \(x\) except \(x = 2\) and \(x = -2\) if and only if \(b = -6\). --- ### **Problem b)** \[ \frac{x^{2} - 1}{x + 1} = -2 \] #### **Step 1: Factor the Numerator** The numerator \(x^{2} - 1\) is also a difference of squares: \[ x^{2} - 1 = (x - 1)(x + 1) \] So, the equation becomes: \[ \frac{(x - 1)(x + 1)}{x + 1} = -2 \] #### **Step 2: Simplify the Equation** Cancel out \((x + 1)\) from the numerator and the denominator, keeping in mind that \(x \neq -1\) to avoid division by zero: \[ x - 1 = -2 \] #### **Step 3: Solve for \(x\)** \[ x - 1 = -2 \\ x = -1 \] #### **Step 4: Check for Validity** We found \(x = -1\), but this value makes the original denominator zero (\(x + 1 = 0\)), which is undefined. **Conclusion for Part b:** There is **no solution** to the equation because the only potential solution \(x = -1\) is not valid. --- ### **Final Summary** - **a)** For the equation to hold, \(b = -6\). - **b)** The equation has no solution.