The first step to translating word problems into mathematical equations is knowing how to phrase mathematical expressions. "Less than" indicates subtraction, "twice as many" is multiplication by 2 and "A rows of B" is A times B -- because in A rows of B there will be AB things. "In 10 years" would be Z + 10, and "four years ago" becomes "Y - 4." There are hundreds of these phrases; but after learning a few, the process of translation becomes easier. Look at some word problems and pay attention to how language phrases translate to mathematical ones.
There are common templates when it comes to how word problems are formulated. For example, "Two trains leave the station at..." should automatically trigger the formula "distance = rate x time." This formula plays a key role in any problems having to do with moving bodies. A problem that starts with "Paul can paint a room in two hours, and Egbert can paint..." should trigger the formula "1/together = 1/one + 1/other" which is central to problems involving people or processes working together.
If the word problem involves age, such as "A woman is five times as old as her son," a variable like W would represent the present, and the past would be represented by W - 5 for five years ago. The future would be represented by W + 2 for two years from now. Similarly, problems that contain phrases such as "The current of the river is 3 MPH" will have expressions like Z - 3 for movement against the current and Z + 3 for movement downstream.
Simultaneous equations involve a problem with two equations that are both true. Usually, these equations have two variables, and neither is solvable alone. The trick is to substitute one equation in the other so that one of the variables is eliminated. Then the problem is easy. For example, consider the problem: "A woman is five times as old as her son. In 10 years she will be three times as old as her son. How old is she now?" Let W be the woman's age and S be her son's age. Setting up the first sentence: W = 5S. Setting up the second sentence: W + 10 = 3 (S + 10). Neither equation is solvable by itself; but if you substitute the first equation in the second, you get: (5S) + 10 = 3 (S + 10), so 5S + 10 = 3S + 30 and 2S = 20. The son is 10, and the woman is 50.