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. -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

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Let's solve each of the multiple-choice questions step by step.

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### **1. Solve $$ \frac{-5}{x-1} = \frac{3}{x-3} $$**

**Solution:**
$$
\begin{align*}
\frac{-5}{x-1} &= \frac{3}{x-3} \
-5(x - 3) &= 3(x - 1) \quad \text{(Cross-multiplying)} \
-5x + 15 &= 3x - 3 \
-5x - 3x &= -3 - 15 \
-8x &= -18 \
x &= \frac{-18}{-8} = \frac{9}{4}
\end{align*}
$$

**Answer:**  
**c. $$ \frac{9}{4} $$**

---

### **2. Solve $$ \frac{-5}{x-3} = \frac{-3}{x+3} $$**

**Solution:**
$$
\begin{align*}
\frac{-5}{x-3} &= \frac{-3}{x+3} \
-5(x + 3) &= -3(x - 3) \quad \text{(Cross-multiplying)} \
-5x - 15 &= -3x + 9 \
-5x + 3x &= 9 + 15 \
-2x &= 24 \
x &= -12
\end{align*}
$$

**Answer:**  
**a. -12**

---

### **3. Solve $$ \frac{a}{a^{2}-36} + \frac{2}{a-6} = \frac{1}{a+6} $$**

**Solution:**
$$
\begin{align*}
a^{2} - 36 &= (a - 6)(a + 6) \
\frac{a}{(a - 6)(a + 6)} + \frac{2}{a - 6} &= \frac{1}{a + 6} \
\text{Multiply all terms by } (a - 6)(a + 6): \
a + 2(a + 6) &= a - 6 \
a + 2a + 12 &= a - 6 \
3a + 12 &= a - 6 \
2a &= -18 \
a &= -9
\end{align*}
$$
*Note: $$a = -6$$ would make the denominator zero, which is invalid.*

**Answer:**  
**a. -9**

---

### **4. Solve $$ \frac{6}{x^{2}-9} - \frac{1}{x-3} = 1 $$**

**Solution:**
$$
\begin{align*}
x^{2} - 9 &= (x - 3)(x + 3) \
\frac{6}{(x - 3)(x + 3)} - \frac{1}{x - 3} &= 1 \
\text{Multiply all terms by } (x - 3)(x + 3): \
6 - (x + 3) &= (x - 3)(x + 3) \
6 - x - 3 &= x^{2} - 9 \
3 - x &= x^{2} - 9 \
0 &= x^{2} + x - 12 \
x &= \frac{-1 \pm \sqrt{1 + 48}}{2} = \frac{-1 \pm 7}{2}
\end{align*}
$$
Possible solutions: $$x = 3$$ or $$x = -4$$.  
However, $$x = 3$$ makes the denominator zero, so it's invalid.

**Answer:**  
**a. $$-4$$**

---

### **5. Solve $$ \frac{5}{6d} + \frac{5}{d} = 2 $$**

**Solution:**
$$
\begin{align*}
\frac{5}{6d} + \frac{5}{d} &= 2 \
\frac{5}{6d} + \frac{30}{6d} &= 2 \quad \text{(Common denominator)} \
\frac{35}{6d} &= 2 \
35 &= 12d \
d &= \frac{35}{12}
\end{align*}
$$

**Answer:**  
**a. $$ \frac{35}{12} $$**

---

### **6. Solve $$ \frac{8}{5d} + \frac{3}{5d} = 6 $$**

**Solution:**
$$
\begin{align*}
\frac{8 + 3}{5d} &= 6 \
\frac{11}{5d} &= 6 \
11 &= 30d \
d &= \frac{11}{30}
\end{align*}
$$

**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 two consecutive even integers be $$ k $$ and $$ k + 2 $$.

$$
\frac{1}{k} + \frac{1}{k + 2} = \frac{11}{60}
$$

**Solving the equation:**

$$
\begin{align*}
\frac{2k + 2}{k(k + 2)} &= \frac{11}{60} \
2(k + 1) &= \frac{11}{60} k(k + 2) \
120(k + 1) = 11k(k + 2) \quad \text{(Multiplying both sides by 60k(k+2))} \
120k + 120 &= 11k^2 + 22k \
11k^2 - 98k - 120 &= 0 \
k &= \frac{98 \pm \sqrt{(-98)^2 - 4 \times 11 \times (-120)}}{2 \times 11} \
k &= \frac{98 \pm \sqrt{9604 + 5280}}{22} \
k &= \frac{98 \pm 122}{22}
\end{align*}
$$

Possible values:
$$
k = \frac{220}{22} = 10 \quad \text{or} \quad k = \frac{-24}{22} = -\frac{12}{11} \quad (\text{Discard non-integer value})
$$

Thus, the integers are **10** and **12**.

**Answer:**  
**b. $$ \frac{1}{k}+\frac{1}{k+2}=\frac{11}{60} ; 10 \text{ and } 12 $$**

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1. **a. $$ \frac{9}{4} $$** 2. **a. -12** 3. **a. -9** 4. **a. -4** 5. **a. $$ \frac{35}{12} $$** 6. **b. $$ \frac{11}{30} $$** 7. **b. $$ \frac{1}{k}+\frac{1}{k+2}=\frac{11}{60} ; 10 \text{ and } 12 $$**

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