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

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Question Stats: 71% (01:12) correct 29% (01:32) wrong based on 852 sessions

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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
Math Expert V
Joined: 02 Sep 2009
Posts: 65761
Re: if xy+z=x(y+z) which of the following must be true?  [#permalink]

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32
1
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rohitgoel15 wrote:
WishMasterUA 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=0
3)Y=1 AND Z=0
4)X=1 OR Y=0
5)X=1 OR Z=0

I agree with the answer i.e. X=1 OR Z=0
But can the answer be also X=1 AND Z=0 (provided there was an option) as both of them taken together also satisfy the equation???

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

Notice that we are asked "which of the following MUST be true"?

$$xy+z=x(y+z)$$ --> $$xy+z=xy+xz$$ --> $$xy$$ 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).

Addressing your questions: it's not necessary $$z=0$$ and $$x=1$$ both to be simultaneously true. "AND" in the answer choices means that BOTH values must be true, but "OR" in the answer choices means that EITHER value must be true.

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.

Hope it helps.
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Re: if xy+z=x(y+z) which of the following must be true?  [#permalink]

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26
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WishMasterUA 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=0
3)Y=1 AND Z=0
4)X=1 OR Y=0
5)X=1 OR Z=0

xy+z=xy+xz
z=xz
xz-z=0
z(x-1)=0

Means;
either z=0
OR
x-1=0 i.e. x=1

Ans: "E"
##### General Discussion
Manager  Joined: 07 Nov 2009
Posts: 198
Re: if xy+z=x(y+z) which of the following must be true?  [#permalink]

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WishMasterUA 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=0
3)Y=1 AND Z=0
4)X=1 OR Y=0
5)X=1 OR Z=0

I agree with the answer i.e. X=1 OR Z=0
But can the answer be also X=1 AND Z=0 (provided there was an option) as both of them taken together also satisfy the equation???
Manager  Joined: 15 Aug 2013
Posts: 223
Re: if xy+z=x(y+z) which of the following must be true?  [#permalink]

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Bunuel wrote:
rohitgoel15 wrote:
WishMasterUA 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=0
3)Y=1 AND Z=0
4)X=1 OR Y=0
5)X=1 OR Z=0

I agree with the answer i.e. X=1 OR Z=0
But can the answer be also X=1 AND Z=0 (provided there was an option) as both of them taken together also satisfy the equation???

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

Notice that we are asked "which of the following MUST be true"?

$$xy+z=x(y+z)$$ --> $$xy+z=xy+xz$$ --> $$xy$$ 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).

Addressing your questions: it's not necessary $$z=0$$ and $$x=1$$ both to be simultaneously true. "AND" in the answer choices means that BOTH values must be true, but "OR" in the answer choices means that EITHER value must be true.

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.

Hope it helps.

Hi Bunuel,

-Two questions: I took the approach of plugging in answer choices and multiple answers satisfied the "must be true part". Why is that wrong?

-When you came down to xz=z, why did you take the next step of factoring out the z and equating the whole thing to zero, meaning, why did you get to the point of z(x-1)=0?

Thanks
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Joined: 02 Sep 2009
Posts: 65761
Re: if xy+z=x(y+z) which of the following must be true?  [#permalink]

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russ9 wrote:
Bunuel wrote:
rohitgoel15 wrote:
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

Notice that we are asked "which of the following MUST be true"?

$$xy+z=x(y+z)$$ --> $$xy+z=xy+xz$$ --> $$xy$$ 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).

Addressing your questions: it's not necessary $$z=0$$ and $$x=1$$ both to be simultaneously true. "AND" in the answer choices means that BOTH values must be true, but "OR" in the answer choices means that EITHER value must be true.

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.

Hope it helps.

Hi Bunuel,

-Two questions: I took the approach of plugging in answer choices and multiple answers satisfied the "must be true part". Why is that wrong?

-When you came down to xz=z, why did you take the next step of factoring out the z and equating the whole thing to zero, meaning, why did you get to the point of z(x-1)=0?

Thanks

1.
"MUST BE TRUE" questions:
These questions ask which of the following MUST be true, or which of the following is ALWAYS true for ALL valid sets of numbers you choose. Generally for such kind of questions if you can prove that a statement is NOT true for one particular valid set of numbers, it will mean that this statement is not always true and hence not a correct answer.

So, for "MUST BE TRUE" questions plug-in method is good to discard an option but not 100% sure thing to prove that an option is ALWAYS true.

As for "COULD BE TRUE" questions:
The questions asking which of the following COULD be true are different: if you can prove that a statement is true for one particular set of numbers, it will mean that this statement could be true and hence is a correct answer.

So, for "COULD BE TRUE" questions plug-in method is fine to prove that an option could be true. But here, if for some set of numbers you'll see that an option is not true, it won't mean that there does not exist some other set which will make this option true.

2.
To get what x and z could be from xz = z, we should do the way shown in my solution.

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

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After getting z=xz , I cancelled out the z and got x=1. So naturally, B, D and E were valid options. A also gave me the equation z=xz which is exactly the same thing so I went with A. What is wrong in my approach?
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Re: If xy+z=x(y+z), which of the following must be true?  [#permalink]

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Top Contributor
hannahkagalwala wrote:
After getting z=xz , I cancelled out the z and got x=1. So naturally, B, D and E were valid options. A also gave me the equation z=xz which is exactly the same thing so I went with A. What is wrong in my approach?

Once we know that z = xz, we cannot divide both sides by z, because it COULD be the case that z equals zero, and we should never divide by zero.

Here's why:
We know that (2)(0) = (1)(0)
However, if we try to divide by both sides by 0, we may believe that we get: 2 = 1, which of course is not true.

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

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WishMasterUA wrote:
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

We are given the following equation, which we can simplify:

xy + z = x(y + z)

xy + z = xy + xz

z = xz

z – xz = 0

z(1 – x) = 0

z = 0

OR

1 – x = 0

x = 1

So x = 1 or z = 0.

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Re: If xy+z=x(y+z), which of the following must be true? A) x=0  [#permalink]

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_________________ Re: If xy+z=x(y+z), which of the following must be true? A) x=0   [#permalink] 08 Oct 2019, 03:26

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