For the following equation, a. Write the value or values of the variable that make a denominator zel in mind, solve the equation. \( \frac{2}{5 x}+\frac{1}{4}=\frac{74}{10 x}-\frac{1}{3} \) a. Write the value or values of the variable that make a denominator zero. 0 (Use a comma to separate answers as needed.) b. What is the solution of the equation? Select the correct choice below and, if necessary, fill in the ans A. The solution set is \( \{ \) She solution set is \( \{x \mid x \) is a real number).

To solve the equation \[ \frac{2}{5x} + \frac{1}{4} = \frac{74}{10x} - \frac{1}{3}, \] we first need to identify the values of \( x \) that make any denominator zero. ### Part a: Values that make a denominator zero The denominators in the equation are \( 5x \), \( 4 \), \( 10x \), and \( 3 \). - The denominator \( 5x \) is zero when \( x = 0 \). - The denominator \( 4 \) is never zero. - The denominator \( 10x \) is zero when \( x = 0 \). - The denominator \( 3 \) is never zero. Thus, the only value that makes a denominator zero is: \[ \boxed{0} \] ### Part b: Solve the equation Now, we will solve the equation. First, we can eliminate the fractions by finding a common denominator. The least common multiple of the denominators \( 5x \), \( 4 \), \( 10x \), and \( 3 \) is \( 60x \). We multiply every term in the equation by \( 60x \): \[ 60x \left( \frac{2}{5x} \right) + 60x \left( \frac{1}{4} \right) = 60x \left( \frac{74}{10x} \right) - 60x \left( \frac{1}{3} \right). \] This simplifies to: \[ 12 \cdot 2 + 15x = 6 \cdot 74 - 20x. \] Calculating each term gives: \[ 24 + 15x = 444 - 20x. \] Now, we can combine like terms. First, add \( 20x \) to both sides: \[ 24 + 15x + 20x = 444. \] This simplifies to: \[ 24 + 35x = 444. \] Next, subtract \( 24 \) from both sides: \[ 35x = 444 - 24, \] which simplifies to: \[ 35x = 420. \] Now, divide both sides by \( 35 \): \[ x = \frac{420}{35} = 12. \] ### Conclusion The solution to the equation is: \[ \boxed{12}. \] ### Final Answer A. The value that makes a denominator zero is \( \{0\} \). B. The solution set is \( \{12\} \).