Multiple Choice Identify the choice that best completes the statement or answers the question. - 1. \( \frac{-5}{x-1}=\frac{3}{x-3} \) a. \( \frac{18}{5} \) b. \( \frac{3}{8} \) c. \( \frac{9}{4} \) d. 0 2. \( \frac{-5}{x-3}=\frac{-3}{x+3} \) a. -12 b. -3 c. \( -\frac{9}{2} \) d. \( -\frac{24}{5} \) 3. \( \frac{a}{a^{2}-36}+\frac{2}{a-6}=\frac{1}{a+6} \) a. \( \quad-9 \) b. -9 and -6 c. 6 d. -6 4. \( \frac{6}{x^{2}-9}-\frac{1}{x-3}=1 \) a. \( \quad-4 \) b. 3 or -4 c. \( \frac{-1 \pm \sqrt{73}}{2} \) d. 2 5. \( \frac{5}{6 d}+\frac{5}{d}=2 \) a. \( \frac{35}{12} \) b. \( \frac{5}{7} \) c. \( \frac{7}{2} \) d. \( \frac{35}{6} \) 6. \( \frac{8}{5 d}+\frac{3}{5 d}=6 \) a. \( \frac{11}{48} \) b. \( \frac{11}{30} \) c. \( \frac{11}{5} \) d. \( \frac{11}{60} \) 7. The sum of the reciprocals of two consecutive even integers is \( \frac{11}{60} \). Write an equation that can be used to find the two integers. Find the two integers. a. \( \frac{1}{k}+\frac{1}{k+2}=\frac{11}{60} ; 8 \) and 10 c. \( \quad k+(k+2)=\frac{11}{60} ; 8 \) and 10 b. \( \frac{1}{k}+\frac{1}{k+2}=\frac{11}{60} ; 10 \) and 12 d. \( \quad k+(k+2)=\frac{11}{60} ; 10 \) and 12

Let's solve each of the multiple-choice questions step by step. --- **1. Solve \( \frac{-5}{x-1} = \frac{3}{x-3} \)** **Solution:** \[ -5(x - 3) = 3(x - 1) \\ -5x + 15 = 3x - 3 \\ -8x = -18 \\ x = \frac{18}{8} = \frac{9}{4} \] **Answer:** **c. \( \frac{9}{4} \)** --- **2. Solve \( \frac{-5}{x-3} = \frac{-3}{x+3} \)** **Solution:** \[ -5(x + 3) = -3(x - 3) \\ -5x - 15 = -3x + 9 \\ -2x = 24 \\ x = -12 \] **Answer:** **a. \(-12\)** --- **3. Solve \( \frac{a}{a^{2}-36} + \frac{2}{a-6} = \frac{1}{a+6} \)** **Solution:** \[ \frac{a}{(a-6)(a+6)} + \frac{2}{a-6} = \frac{1}{a+6} \\ \text{Multiply by } (a-6)(a+6): \\ a + 2(a + 6) = a - 6 \\ 3a + 12 = a - 6 \\ 2a = -18 \\ a = -9 \] (Note: \( a = -6 \) would make the denominator zero, so it's excluded.) **Answer:** **a. \(-9\)** --- **4. Solve \( \frac{6}{x^{2}-9} - \frac{1}{x-3} = 1 \)** **Solution:** \[ \frac{6}{(x-3)(x+3)} - \frac{1}{x-3} = 1 \\ \text{Multiply by } (x-3)(x+3): \\ 6 - (x + 3) = x^2 - 9 \\ 3 - x = x^2 - 9 \\ x^2 + x - 12 = 0 \\ x = \frac{-1 \pm \sqrt{1 + 48}}{2} = \frac{-1 \pm 7}{2} \\ x = 3 \text{ (invalid, causes division by zero)} \text{ or } x = -4 \] **Answer:** **a. \(-4\)** --- **5. Solve \( \frac{5}{6d} + \frac{5}{d} = 2 \)** **Solution:** \[ \frac{5}{6d} + \frac{5}{d} = \frac{5 + 30}{6d} = \frac{35}{6d} = 2 \\ 35 = 12d \\ d = \frac{35}{12} \] **Answer:** **a. \( \frac{35}{12} \)** --- **6. Solve \( \frac{8}{5d} + \frac{3}{5d} = 6 \)** **Solution:** \[ \frac{11}{5d} = 6 \\ 5d = \frac{11}{6} \\ d = \frac{11}{30} \] **Answer:** **b. \( \frac{11}{30} \)** --- **7. The sum of the reciprocals of two consecutive even integers is \( \frac{11}{60} \). Find the two integers.** **Solution:** Let the integers be \( k \) and \( k + 2 \). \[ \frac{1}{k} + \frac{1}{k+2} = \frac{11}{60} \] Testing the options: - **Option a:** \( k = 8 \) and \( k+2 = 10 \) \[ \frac{1}{8} + \frac{1}{10} = \frac{5}{40} + \frac{4}{40} = \frac{9}{40} \neq \frac{11}{60} \] - **Option b:** \( k = 10 \) and \( k+2 = 12 \) \[ \frac{1}{10} + \frac{1}{12} = \frac{6}{60} + \frac{5}{60} = \frac{11}{60} \] **Answer:** **b. \( \frac{1}{k} + \frac{1}{k+2} = \frac{11}{60}; \; 10 \) and \( 12 \)** ---