Use the basic formula that shows up in every motion problem: distance = rate X time. A good place to start is to write this basic formula for every moving object. Put in numbers, if they are available, and variables otherwise. Then combine the two equations in such a way that there is only one variable left, and solve for that variable. Often this will mean finding something -- distance, rate or time -- that is the same for both equations and combine the equations by eliminating this factor.
Let S be the rate of the slow car in the problem. The basic formula for the slow car is distance = S X 2.5 and the formula for the fast car is distance = (S + 10) X 2.5. The distances added together is 325, so the equation to solve is S X 2.5 + (S + 10) X 2.5 = 325. This means that S + S + 10 = 325/2.5 = 130; so 2S + 10 = 130 or 2S = 120, which means that the slow car is going 60 MPH and the fast car is going 70 MPH.
Factor in any constant force, if it is a part of the problem. If S represents the speed of a boat and C represents the speed of the current, the rate of speed going upstream is S - C and the rate of speed going downstream is S + C. If you know that your motorboat goes 3 MPH in still water, and it takes twice as long to go someplace upstream as it does to return from the journey, find the rate of the current in the river. The distance downstream = (3 + C)T and the distance upstream = 2(3 - C)T; so (3 + C)T = 2(3 - C)T or 3 + C = 6 - 2C or 3C = 3. The river current is 1 MPH.