Before you can solve an equation, you first need an equation to solve. Word problems give you all the information you need to answer the question—but in English. On the SAT, you must be able to translate that information into math.

When you’re turning a word problem into an equation (or equations) there are four things you have to do:

Well, isn’t it obvious what the question is asking? It wants to know how many blue marbles are in the sack. It doesn’t get any more obvious than that. You’re right, but there are reasons we stress this point so seriously. “How many blue marbles are in the sack?” is not what the question is asking you. What it’s really asking is, “How many blue marbles are in a sack that contains 50 total marbles, out of which 20 more are red than blue?” This is the true question, because it identifies all the relationships and contains all the information that the question describes. SAT word problems can make even simple equations sound complicated and messy. Restating the question is a good way to clean things up.

The question tells you the total number of marbles, and it says that there are red and blue marbles. You don’t need a variable for total marbles, since you know the total is 50, but you do need variables for the red and blue marbles, since those are unknown quantities. Pick whatever ones you like. How about r and b?

Okay. You know from the question that the 50 total marbles are made up of red and blue marbles. So 50 = r + b. The question also tells you that the sack contains 20 more red marbles than blue marbles. So the number of red marbles is 20 more than the number of blue marbles: r = b + 20. Just like that, you’ve taken all the information and put it into mathematical form.

Once you’ve got the word problem translated, solving is a cinch. You’re looking for how many blue marbles are in a sack that contains 50 total marbles, out of which 20 more are red than blue. So, in the equation with total marbles, red marbles, and blue marbles, you’re looking for blue marbles. A little manipulating does the trick: b = 50 – r. And since you know that r = b + 20, you can substitute b + 20 for r to get b = 50 – (b + 20). Solve: r = 50 – r – 20; 2r = 30; r = 15.

Here’s a longer, more complicated problem. Conquer it with the same four-step technique:

What is the cost of the paint Gus needs to buy if he needs to cover a total of x square feet, when each can costs p dollars and covers y square feet?

The question assigns almost all of the variables for you: x is total square feet that need painting in the house, p is the price of a can of paint, and y is the square feet that each can of paint can cover. Nice! You just need a variable for that total cost. How about t?

Since p is the price of a can of paint, and you need to find the total cost of buying paint, you must find the number of cans of paint that Gus has to buy . . . look at that! You need another variable for numbers of cans of paint. How about n? (This happens sometimes: You’ll find out that there is a variable you need that is implied by the problem but not explicitly mentioned.) So t = np.

Now to find n. Use the information the problem gives you. It tells you the total square feet (x) and the square feet covered per can (y), which means that if you were to divide x by y, you’d get the number of cans:

You’ve got the relationships between the variables mapped out. Plug the equation for n into the t = np and you’ve got the total cost:

Unlike the rule of doing the same thing to both sides of the equation, there’s no universal rule stating that you must follow the three steps we’ve just explained. But we think following these steps will help you get a handle on every SAT world problem, no matter how long and gnarly it is.