Hi there! I'm happy to help with this.

So, the question:

If xy + z = x(y + z), then

(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

So, as a very general rough approximation rule, when variables are multiplied, the mathematical consequences will far more often involve the word "or" than the word "and." That's just a rough-and-ready rule to use if all else is lost.

Let's look at this question. We have: xy + z = x(y + z)

Distribute on the right side: xy + z = xy + xz

Subtract xy from both sides: z = xz

So first of all, we get this funny equation with x & z in it, without y, so that means: there's no restriction on y, only restrictions on x & z.

In solving the equation z = xz, you may be tempted to divide by z, but WHOA THERE! Before you divide by a variable, you have to ask yourself the deeply heartfelt question: could this variable equal zero? If so, then dividing by it would break all mathematical laws, and Santa Claus would find out you've been bad. So, what do we do?

We'll look at two cases: (i) z = 0, and (ii) z =/ 0

(i) if z = 0, then both sides of the equation x = xz equal zero, and the equation is true. So, one possibility that makes the equation true is z = 0

(ii) if z is not equal to zero, then we can divide by z, and get 1 = x. That's the other solution that makes the equation true.

The solutions are x = 1 or z = 0, answer choice E.

Here's another practice question for you.

http://gmat.magoosh.com/questions/110The question at that link should be followed by a video explanation.

Please let me know if you have any questions about what I've said. Good luck tomorrow!!

Mike

_________________

Mike McGarry

Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)