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 \( \varnothing \).

To identify the values of the variable that make the denominators zero, we inspect the denominators in the equation \( \frac{x-7}{2x} + 1 = \frac{x+3}{x} \). The denominators are \( 2x \) and \( x \). Setting these equal to zero: 1. \( 2x = 0 \) ⇒ \( x = 0 \) 2. \( x = 0 \) Hence, the only restriction on the variable is \( x = 0 \). Now, addressing the equation \( \frac{x-7}{2x} + 1 = \frac{x+3}{x} \), we first eliminate the denominators by multiplying through by \( 2x \) (remembering \( x \neq 0 \)): \[ (x - 7) + 2x = 2(x + 3) \] This simplifies to: \[ x - 7 + 2x = 2x + 6 \] Combining like terms gives us: \[ 3x - 7 = 2x + 6 \] Now subtract \( 2x \) from both sides: \[ x - 7 = 6 \] Adding 7 to both sides results in: \[ x = 13 \] Since \( x = 13 \) does not violate our earlier restriction of \( x \neq 0 \), we have our solution! A. The solution set is \( \{13\} \).