You went on a 360km train ride but you are late! If the train had just been 5 km/hr faster, then the train ride would have been one hour shorter. How fast was the train going?
+1262 so we have (360/x)-(360/(x+5))=60 so we multiply all terms to get 360(x+5)+−360x=1x(x+5) and then −x^2−5x+1800=0 where x=-45 or x=40 but it can't be negative so x=40 and the speed is 40kph
+36915 x = k/hr
360 / x -1 = 360 / (x+5) solve for x
360/x - 360/(x+5) = 1
360(x+5) - 360x = x (x+5)
1800= x^2+5x
x^2+5x-1800 = 0 Quadratic formula shows x = 40 k/hr or -45(throw out)
