Question 6 pause Zoom Mr. Beecher and Mrs. Carter are teachers at the same school. They leave their houses at the same time in the morning to get to school. Mr. Beecher frues 8 mikes away from school and rides his bicycle to work. Every minute, he gets \( \frac{1}{6} \) of a mile closer to schoof. Mrs. Carter lives 20 miles away from school and drives her car to work. Every minute, she gets \( \frac{1}{2} \) of a mile closer to school. After how many minutes will Mr. Beecher and Mrs. Carter first be the same distance away from school? \( \square \) minutes

To determine after how many minutes Mr. Beecher and Mrs. Carter will be the same distance away from school, let's set up equations based on their rates of approaching the school. **Given:** - **Mr. Beecher:** - Initial distance: 8 miles - Rate: \( \frac{1}{6} \) mile per minute - Distance after \( t \) minutes: \( 8 - \frac{1}{6}t \) - **Mrs. Carter:** - Initial distance: 20 miles - Rate: \( \frac{1}{2} \) mile per minute - Distance after \( t \) minutes: \( 20 - \frac{1}{2}t \) **Set the distances equal to find \( t \):** \[ 8 - \frac{1}{6}t = 20 - \frac{1}{2}t \] **Solve for \( t \):** 1. Multiply both sides by 6 to eliminate the fractions: \[ 6 \times 8 - t = 6 \times 20 - 3t \] \[ 48 - t = 120 - 3t \] 2. Rearrange the equation to gather like terms: \[ -t + 3t = 120 - 48 \] \[ 2t = 72 \] 3. Divide both sides by 2: \[ t = 36 \] **Conclusion:** Mr. Beecher and Mrs. Carter will first be the same distance away from school after **36 minutes**. **Answer:** 36 minutes