# The 4 Math Strategies Everyone Must Master: Testing Cases Redux

by , Apr 24, 2014

Today, Ive got some additional practice for you with regard to one of those strategies: Testing Cases.

Try this GMATPrep problem:

* If xy + z = x(y + z), which of the following must be true?

(A) x = 0 and z = 0

(B) x = 1 and y = 1

(C) y = 1 and z = 0

(D) x = 1 or y = 0

(E) x = 1 or z = 0

How did it go?

This question is called a theory question: there are just variables, no real numbers, and the answer depends on some characteristic of a category of numbers, not a specific number or set of numbers. Problem solving theory questions also usually ask what must or could be true (or what must not be true). When we have these kinds of questions, we can use theory to solvebut that can get very confusing very quickly. Testing real numbers to prove the theory to yourself will make the work easier.

The question stem contains a given equation:

xy + z = x(y + z)

Whenever the problem gives you a complicated equation, make your life easier: try to simplify the equation before you do any more work.

xy + z = x(y + z)

xy + z = xy + xz

z = xz

Very interesting! The y term subtracts completely out of the equation. What is the significance of that piece of info?

Nothing absolutely has to be true about the variable y. Glance at your answers. You can cross off (B), (C), and (D) right now!

Next, notice something. I stopped at z = xz. I didnt divide both sides by z. Why?

In general, never divide by a variable unless you know that the variable does not equal zero. Dividing by zero is an illegal move in algebraand it will cause you to lose a possible solution to the equation, increasing your chances of answering the problem incorrectly.

The best way to finish off this problem is to test possible cases. Notice a couple of things about the answers. First, they give you very specific possibilities to test; you dont even have to come up with your own numbers to try. Second, answer (A) says that both pieces must be true (and) while answer (E) says or. Keep that in mind while working through the rest of the problem.

z = xz

Lets see. z = 0 would make this equation true, so that is one possibility. This shows up in both remaining answers.

If x = 0, then the right-hand side would become 0. In that case, z would also have to be 0 in order for the equation to be true. That matches answer (A).

If x = 1, then it doesnt matter what z is; the equation will still be true. That matches answer (E).

Wait a secondwhats going on? Both answers cant be correct.

Be careful about how you test cases. The question asks what MUST be true. Go back to the starting point that worked for both answers: z = 0.

Its true that, for example, 0 = (3)(0).

Does z always have to equal 0? Can you come up with a case where z does not equal 0 but the equation is still true?

Try 2 = (1)(2). In this case, z = 2 and x = 1, and the equation is true. Heres the key to the and vs. or language. If z = 0, then the equation is always 0 = 0, but if not, then x must be 1; in that case, the equation is z = z. In other words, either x = 1 OR z = 0.