The following rational equation has denominators that contain variables. For this equation, a. Write the value or values of the variable that make a denominator zero. These are the restrictions on the variable. b. Keeping the restrictions in mind, solve the equation. \( \frac{x-7}{2 x}+1=\frac{x+3}{x} \) a. What is/are the value or values of the variable that make(s) the denominators zero? \( x=\square \) (Simplify your answer. Use a comma to separate answers as needed.) b. Solve the equation. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The solution set is \( \{\square\} \). (Type an integer or a simplified fraction. Use a comma to separate answers as needed.) B. The solution set is \( \{x \mid x \) is a real number\}. C. The solution set is \( \varnothing \).

a. The denominators in the equation are \(2x\) and \(x\). To find the restrictions, set each denominator equal to zero and solve for \(x\): \[ 2x = 0 \quad \Rightarrow \quad x = 0 \] \[ x = 0 \] Both denominators are zero when \(x = 0\). Therefore, the restriction is: \[ x = 0 \] b. To solve the equation \( \frac{x-7}{2x} + 1 = \frac{x+3}{x} \) while keeping the restriction \(x \neq 0\) in mind, follow these steps: 1. **Combine the terms on the left side:** \[ \frac{x-7}{2x} + 1 = \frac{x-7}{2x} + \frac{2x}{2x} = \frac{x-7 + 2x}{2x} = \frac{3x - 7}{2x} \] 2. **Set the two fractions equal to each other:** \[ \frac{3x - 7}{2x} = \frac{x + 3}{x} \] 3. **Cross-multiply to eliminate the denominators:** \[ (3x - 7) \cdot x = (x + 3) \cdot 2x \] \[ 3x^2 - 7x = 2x^2 + 6x \] 4. **Move all terms to one side to form a quadratic equation:** \[ 3x^2 - 7x - 2x^2 - 6x = 0 \] \[ x^2 - 13x = 0 \] 5. **Factor the quadratic equation:** \[ x(x - 13) = 0 \] 6. **Solve for \(x\):** \[ x = 0 \quad \text{or} \quad x = 13 \] 7. **Apply the restriction \(x \neq 0\):** \[ x = 0 \quad \text{is not valid} \] \[ x = 13 \quad \text{is valid} \] **Final Answer:** **a.** \( x = 0 \) **b.** A. The solution set is \( \{13\} \).