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the answer is E (this is the answer that you get if you solve the equation), but I was just wondering why the answer couldn't also be (A) because if you plug the variables into the equation you find that the left side equals the right side.

Re: Algebra question on the GMAC practice test [#permalink]
16 Sep 2009, 11:55

2

This post received KUDOS

dhushan wrote:

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

1. x=0 and z=0 2. x=1 and y=1 3. y=1 and z=0 4. x=1 or y=0 5. x=1 or z=0

the answer is E (this is the answer that you get if you solve the equation), but I was just wondering why the answer couldn't also be (A) because if you plug the variables into the equation you find that the left side equals the right side.

Re: Algebra question on the GMAC practice test [#permalink]
16 Sep 2009, 12:15

1

This post received KUDOS

dhushan wrote:

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

1. x=0 and z=0 2. x=1 and y=1 3. y=1 and z=0 4. x=1 or y=0 5. x=1 or z=0

the answer is E (this is the answer that you get if you solve the equation), but I was just wondering why the answer couldn't also be (A) because if you plug the variables into the equation you find that the left side equals the right side.

i.e. xy + z = x(y+z) (0)y + 0 = 0(y+0) 0 = 0

what's wrong with my logic here.

I understand this, but what's wrong with plugging in 0, you still get the right hand equalling the left hand side, i.e. 0 = 0.

Re: Algebra question on the GMAC practice test [#permalink]
17 Sep 2009, 02:29

2

This post received KUDOS

dhushan wrote:

dhushan wrote:

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

1. x=0 and z=0 2. x=1 and y=1 3. y=1 and z=0 4. x=1 or y=0 5. x=1 or z=0

the answer is E (this is the answer that you get if you solve the equation), but I was just wondering why the answer couldn't also be (A) because if you plug the variables into the equation you find that the left side equals the right side.

i.e. xy + z = x(y+z) (0)y + 0 = 0(y+0) 0 = 0

what's wrong with my logic here.

I understand this, but what's wrong with plugging in 0, you still get the right hand equalling the left hand side, i.e. 0 = 0.

I did not find any reason except this: xy+z=x(y+z) can be rewritten as (xy+z)/x=(y+z) if x = 0 then the LHS will be infinite.

Re: Algebra question on the GMAC practice test [#permalink]
17 Sep 2009, 07:55

1

This post received KUDOS

maliyeci wrote:

dhushan wrote:

dhushan wrote:

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

1. x=0 and z=0 2. x=1 and y=1 3. y=1 and z=0 4. x=1 or y=0 5. x=1 or z=0

the answer is E (this is the answer that you get if you solve the equation), but I was just wondering why the answer couldn't also be (A) because if you plug the variables into the equation you find that the left side equals the right side.

i.e. xy + z = x(y+z) (0)y + 0 = 0(y+0) 0 = 0

what's wrong with my logic here.

I understand this, but what's wrong with plugging in 0, you still get the right hand equalling the left hand side, i.e. 0 = 0.

I did not find any reason except this: xy+z=x(y+z) can be rewritten as (xy+z)/x=(y+z) if x = 0 then the LHS will be infinite.

That makes sense. I guess plugging in answers is not always the best options in these situations. Thanks.

Re: Algebra question on the GMAC practice test [#permalink]
18 Sep 2009, 10:51

1

This post received KUDOS

this equation can be solved directly xy + z = x(y+z) xy + z = xy +xz z=xz or xz - z = 0 z(x-1) = 0 z= 0 or x = 1.

its all ways better to solve if the equations are simple like mentioned above....substituting values will be better if they r more complex invloving 3 or 4 variables .

Re: Algebra question on the GMAC practice test [#permalink]
18 Sep 2009, 11:17

1

This post received KUDOS

Expert's post

dhushan wrote:

dhushan wrote:

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

1. x=0 and z=0 2. x=1 and y=1 3. y=1 and z=0 4. x=1 or y=0 5. x=1 or z=0

the answer is E (this is the answer that you get if you solve the equation), but I was just wondering why the answer couldn't also be (A) because if you plug the variables into the equation you find that the left side equals the right side.

i.e. xy + z = x(y+z) (0)y + 0 = 0(y+0) 0 = 0

what's wrong with my logic here.

I understand this, but what's wrong with plugging in 0, you still get the right hand equalling the left hand side, i.e. 0 = 0.

The question asks what *must* be true. Sure, it *could* be true that x = 0 and z = 0, but that does not *need* to be true. We might instead have that x = 1 and z = 5, for example. If you plug in the numbers from the answer choices here, all you learn is what *may* be true, not what *must* be true. _________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

Re: Algebra question on the GMAC practice test [#permalink]
17 Sep 2013, 00:33

dhushan wrote:

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

1. x=0 and z=0 2. x=1 and y=1 3. y=1 and z=0 4. x=1 or y=0 5. x=1 or z=0

the answer is E (this is the answer that you get if you solve the equation), but I was just wondering why the answer couldn't also be (A) because if you plug the variables into the equation you find that the left side equals the right side.

i.e. xy + z = x(y+z) (0)y + 0 = 0(y+0) 0 = 0

what's wrong with my logic here.

The reason why the above logic is wrong is after simplifying the expression, we have, xz=z. For this to be true either x has to be 1 or z has to be 0 but choice A says that both x=0 and z=0 need to be true. We know that need not be the case.

For "which must be true" questions, if more than 1 choice seem to be correct, pick the one with the least restrictions. _________________

Re: If xy+z = x(y+z), which of the following must be true? [#permalink]
17 Sep 2013, 00:44

Expert's post

1

This post was BOOKMARKED

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

1. x=0 and z=0 2. x=1 and y=1 3. y=1 and z=0 4. x=1 or y=0 5. 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.

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