The solution to \( \frac{15}{a^{2}-1}=\frac{5}{2 a-2} \) is \( a=\square \) The extraneous solution is \( a=\square \)

1. **Rewrite the equation:** \[ \frac{15}{a^2-1} = \frac{5}{2a-2} \] 2. **Factor the denominators:** \[ a^2-1 = (a-1)(a+1) \] \[ 2a-2 = 2(a-1) \] So the equation becomes: \[ \frac{15}{(a-1)(a+1)} = \frac{5}{2(a-1)} \] 3. **Identify the restrictions:** The denominators cannot be zero. - For \((a-1)(a+1)\): \(a \neq 1\) and \(a \neq -1\) - For \(2(a-1)\): \(a \neq 1\) 4. **Solve the equation (assuming \(a \neq 1\)):** Multiply both sides by the common denominator \(2(a-1)(a+1)\) to clear the fractions: \[ 15 \cdot 2(a-1) = 5 \cdot (a-1)(a+1) \] This simplifies to: \[ 30(a-1) = 5(a-1)(a+1) \] 5. **Cancel the common factor \((a-1)\):** Since \(a \neq 1\), we can cancel \((a-1)\) from both sides: \[ 30 = 5(a+1) \] 6. **Solve for \(a\):** Divide both sides by 5: \[ 6 = a+1 \] Subtract 1 from both sides: \[ a = 5 \] 7. **Determine the extraneous solution:** We earlier noted that \(a = 1\) is not allowed because it makes the denominators zero. \[ \boxed{a=5} \] \[ \text{Extraneous solution: } a=1 \]