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If xy+z=x(y+z), which of the following must be true? A. x=0 and y=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

\(xy+z=x(y+z)\) --> \(xy+z=xy+xz\) --> \(xy\) cancels out --> \(xz-z=0\) --> \(z(x-1)=0\) --> either \(z=0\) (in this case \(x\) can take ANY value) OR \(x=1\) (in this case \(z\) can take ANY value).

Answer: E.

To elaborate more: As the expression with \(y\) cancels out, we can say that given expression \(xy+z=x(y+z)\) does not depend on value of \(y\). Which means that \(y\) can take ANY value. So all answer choices which specify the exact value of \(y\) are wrong.

Next: "AND" in answer choices means that BOTH values must be true, but "OR" in answer choices means that EITHER value must be true.

Re: Which of the following must be true - Gmatprep [#permalink]

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10 Jan 2012, 13:05

1

This post received KUDOS

Expert's post

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.

xy+z=x(y+z). which of the following must be true? a) x=0 and y=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

\(xy+z=x(y+z)\) --> \(xy+z=xy+xz\) --> \(xy\) cancels out --> \(xz-z=0\) --> \(z(x-1)=0\) --> either \(z=0\) (in this case \(x\) can take ANY value) OR \(x=1\) (in this case \(z\) can take ANY value).

Answer: E.

To elaborate more: As the expression with \(y\) cancels out, we can say that given expression \(xy+z=x(y+z)\) does not depend on value of \(y\). Which means that \(y\) can take ANY value. So all answer choices which specify the exact value of \(y\) are wrong.

Next: "AND" in answer choices means that BOTH values must be true, but "OR" in answer choices means that EITHER value must be true.

Re: Which of the following must be true - Gmatprep [#permalink]

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10 Jan 2012, 21:01

mikemcgarry has given an exhaustive explanation!

Simply put, it's basic algebra. Simplify the expression and equate to zero. Watch out for the use of "and" and "or" in the answer choices. _________________

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 and y=0

A. X=1 and z=0

xy + z = x(y+z)-----------> xy + z = xy+xz ---------> z = xz --------> x=1 and Z= 0 or 1 Answer E. Note :- Here the value of y does not matter since xy = xy going to be zero for any value of y _________________

Re: If xy+z=x(y+z), which of the following must be true? [#permalink]

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09 Dec 2014, 10:16

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Re: If xy+z=x(y+z), which of the following must be true? [#permalink]

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14 Mar 2016, 01:43

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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