One student begins in Cleveland, Ohio, and walks towards Buffalo, New York, a distance of 200 miles. She walks at the rate of 2.
5 miles per hour. Another student begins riding his bicycle in Cleveland and rides towards Buffalo at the rate of 11 miles per hour. The walking student begins her journey at 8:00 AM. The riding student starts 4 hours later. Part A Write a system of two equations that could be used to find the distance each student is from Cleveland, Ohio, at any given time. Let x represent the distance that the walking student is from Cleveland, let y represent the distance that the riding student is from Cleveland, and let t be the time that has elapsed since the walking student left Cleveland.
Part B Solve the system of equations for x and y to determine how many hours elapse before the students meet.
Ok so student in cleveland walks 2.5 miles away from cleveland every hour from 8 so the distance, x is x=2.5t
the riding one rides 4 hours later, so t-4 11 mph y=11(t-4)
A. x=2.5t y=11(t-4)
when do they meet is when the distance when they are equal or when x=y 2.5t=x=y=11(t-4) solve for t 2.5t=11(t-4) 2.5t=11t-44 -8.5t=-44 divide both sides by -8.5 t=5.17647
First we combine the parenthesis to get 9/(-2-i) then we multiply both the numerator and the denomenator by the conjugate of -2-i (whic is -2+i) to get (-18+9i)/5
