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Which MUST BE true? please help!!

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by gmatpup » Sat Dec 10, 2011 4:56 pm
If xy+z = x(y+z), which 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


Please provide an explanation, I am confused :( Thank you!!

Answer is E
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by GmatMathPro » Sat Dec 10, 2011 5:01 pm
gmatpup wrote:If xy+z = x(y+z), which 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


Please provide an explanation, I am confused :( Thank you!!
Answer is E
xy+z=x(y+z)

xy+z=xy+xz (distribute the x)

z=xz (subtract xy from both sides)

xz-z=0 (subtract z from both sides)

z(x-1)=0 factor out a z

z=0 or x-1=0 (set both factors equal to zero)

z=0 or x=1 solve
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by Anurag@Gurome » Sun Dec 11, 2011 4:08 am
gmatpup wrote:If xy+z = x(y+z), which MUST BE true?
--> (xy + z) = x(y + z)
--> (xy + z) = (xy + xz)
--> z = xz
--> (xz - z) = 0
--> z(x - 1) = 0

Hence, either z = 0 or x = 1.

The correct answer is E.
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by LalaB » Sun Dec 11, 2011 10:27 am
@Anurag@Gurome and GmatMathPro

I have got the same answer choice. But ...looking closely to answer choices, I have suddenly thought -why B must be wrong?
we have xy+z = x(y+z). if x=1 and y=1, then 1*1+z=1(1+z)

whats wrong in my logic?
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by kanwar86 » Sun Dec 11, 2011 11:01 am
LalaB wrote:@Anurag@Gurome and GmatMathPro

I have got the same answer choice. But ...looking closely to answer choices, I have suddenly thought -why B must be wrong?
we have xy+z = x(y+z). if x=1 and y=1, then 1*1+z=1(1+z)

whats wrong in my logic?
There is nothing wrong with your logic. It's just that option B is a subset of E.
Option E is if (x=1 or z=0), the given equation is valid.
That is, any value of y (as long as x=1) satisfies the given equation.
Option B, x=1 and y=1, is a particular case of E.
Regards

Kanwar

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