position: \( 3 t-8 t+11=20 t-4 \) How often will they be in the same

To solve the equation \( 3t - 8t + 11 = 20t - 4 \), we first simplify the left side: \[ 3t - 8t + 11 = -5t + 11 \] Now, we can rewrite the equation: \[ -5t + 11 = 20t - 4 \] Next, we will move all terms involving \( t \) to one side and constant terms to the other side. Adding \( 5t \) to both sides gives: \[ 11 = 25t - 4 \] Now, add \( 4 \) to both sides: \[ 11 + 4 = 25t \] This simplifies to: \[ 15 = 25t \] Now, divide both sides by \( 25 \): \[ t = \frac{15}{25} = \frac{3}{5} \] Thus, the solution to the equation is \( t = \frac{3}{5} \). To answer the question of how often they will be in the same position, we need to clarify the context. If this equation represents the positions of two objects over time, they will be in the same position at \( t = \frac{3}{5} \) and potentially at other times depending on the nature of their motion (e.g., if they are moving in a periodic manner). If the motion is linear and continuous without periodicity, they will only be at the same position at \( t = \frac{3}{5} \). If there are periodic motions involved, we would need more information to determine how often they coincide.