6. \( \frac{a+3}{a-5} \div \frac{a+1}{a^{2}-2 a-15} \) a. \( \frac{(a+3)(a+3)}{a+1}, a \neq-3,-1 \) c. \( \frac{(a+3)(a+3)}{a+1}, a \neq 5,-1,-3 \) b. \( \frac{(a+3)(a+1)}{(a-5)^{2}(a+3)}, a \neq 5,-3 \) d. \( \frac{(a+3)(a+1)}{(a-5)^{2}(a+3)}, a \neq 5,-3,-1 \) 7. \( \frac{x^{2}-4}{x-8} \div(x-2) \) a. \( \frac{(x-2)(x+2)}{x-8} \) c. \( \frac{x-2}{x-8} \) b. \( \frac{x-8}{x+2} \) d. \( \frac{x+2}{x-8} \) 8. \( \frac{6}{x^{2}-9}-\frac{1}{x-3}=1 \) a. \( \quad-4 \) b. 3 or -4 c. \( \frac{-1 \pm \sqrt{73}}{2} \) d. 2 9. \( \frac{2}{5 c}+\frac{1}{4 c}=-6 \) a. \( -\frac{7}{60} \) b. \( \frac{13}{20} \) c. \( -\frac{13}{120} \) d. \( -\frac{1}{18} \) 10. \( \frac{1}{x-3}=\frac{-1}{x-1} \) a. 2 b. \( \frac{3}{2} \) c. 4 d. \( \frac{5}{2} \)

Let's solve each problem step by step. ### Problem 6: \[ \frac{a+3}{a-5} \div \frac{a+1}{a^{2}-2 a-15} \] **Step 1: Rewrite the division as multiplication by the reciprocal.** \[ \frac{a+3}{a-5} \times \frac{a^{2}-2a-15}{a+1} \] **Step 2: Factor the quadratic expression \(a^{2}-2a-15\).** To factor \(a^{2}-2a-15\), we look for two numbers that multiply to \(-15\) and add to \(-2\). These numbers are \(-5\) and \(3\). \[ a^{2}-2a-15 = (a-5)(a+3) \] **Step 3: Substitute the factored form back into the expression.** \[ \frac{a+3}{a-5} \times \frac{(a-5)(a+3)}{a+1} \] **Step 4: Cancel common factors.** The \((a-5)\) in the numerator and denominator cancels out: \[ \frac{(a+3)(a+3)}{a+1} \] **Step 5: State the restrictions.** The restrictions come from the original denominators: - \(a \neq 5\) (from \(a-5\)) - \(a \neq -3\) (from \(a+3\)) - \(a \neq -1\) (from \(a+1\)) Thus, the final answer is: \[ \frac{(a+3)(a+3)}{a+1}, \quad a \neq 5, -3, -1 \] ### Problem 7: \[ \frac{x^{2}-4}{x-8} \div (x-2) \] **Step 1: Rewrite the division as multiplication by the reciprocal.** \[ \frac{x^{2}-4}{x-8} \times \frac{1}{x-2} \] **Step 2: Factor \(x^{2}-4\) (difference of squares).** \[ x^{2}-4 = (x-2)(x+2) \] **Step 3: Substitute the factored form back into the expression.** \[ \frac{(x-2)(x+2)}{x-8} \times \frac{1}{x-2} \] **Step 4: Cancel the common factor \((x-2)\).** \[ \frac{x+2}{x-8} \] Thus, the final answer is: \[ \frac{x+2}{x-8} \] ### Problem 8: \[ \frac{6}{x^{2}-9}-\frac{1}{x-3}=1 \] **Step 1: Factor \(x^{2}-9\) (difference of squares).** \[ x^{2}-9 = (x-3)(x+3) \] **Step 2: Rewrite the equation.** \[ \frac{6}{(x-3)(x+3)} - \frac{1}{x-3} = 1 \] **Step 3: Find a common denominator.** The common denominator is \((x-3)(x+3)\): \[ \frac{6 - (x+3)}{(x-3)(x+3)} = 1 \] **Step 4: Simplify the numerator.** \[ \frac{6 - x - 3}{(x-3)(x+3)} = 1 \implies \frac{3 - x}{(x-3)(x+3)} = 1 \] **Step 5: Cross-multiply.** \[ 3 - x = (x-3)(x+3) \] **Step 6: Expand and rearrange.** \[ 3 - x = x^2 - 9 \implies x^2 + x - 12 = 0 \] **Step 7: Factor the quadratic.** \[ (x+4)(x-3) = 0 \] **Step 8: Solve for \(x\).** \[ x = -4 \quad \text{or} \quad x = 3 \] **Step 9: Check for extraneous solutions.** Since \(x = 3\) makes the denominator zero, we discard it. Thus, the solution is: \[ x = -4 \] ### Problem 9: \[ \frac{2}{5c}+\frac{1}{4c}=-6 \] **Step 1: Find a common denominator.** The common denominator is \(20c\): \[ \frac{8}{20c} + \frac{5}{20c} = -6 \] **Step 2: Combine the fractions.** \[ \frac{13}{20c} = -6 \] **Step 3: Cross-multiply.** \[ 13 = -120c \implies c = -\frac{13}{120} \] Thus, the final answer is: \[ c = -\frac{13}{120} \] ### Problem 10: \[ \frac{1}{x-3}=\frac{-1}{x-1} \] **Step 1: Cross-multiply.** \[ 1 \cdot (x-1) = -1 \cdot (x-3) \] **Step 2: Expand both sides.** \[ x - 1 = -x + 3 \] **Step 3: Rearrange the equation.** \[ 2x = 4 \implies x = 2 \] Thus, the final answer is: \[ x = 2 \] ### Summary of Answers: 6. \( \frac{(a+3)(a+3)}{a+1}, a \neq 5, -3, -1 \) 7. \( \frac{x+2}{x-8} \) 8. \( -4 \) 9. \( -\frac{13}{120} \) 10. \( 2 \)