Question 416205
Solution: x mph the speed of train that travel east
          x+20 mph the speed of the train that travel north.
5(x+20) the distance traveled north
5x the distance traveled east. Write the equation: [5(x+20)]^2+(5x)^2=300^2
Solve this equation:x^2+20x-1600=0
Answer: The Eastbound train travel approximately 31 mph and the Northbound train travel 51 mph.(Hint: use the Pythagorean theorem )