## Algebra question

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### Algebra question

by ornelaspedro » Wed Apr 29, 2009 7:02 pm
Can anybody tell me how to solve this;

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
e) x = 1 and z = 0

The answer is E, I solved for x and got one, however, I'm uncertain on how the answer is z = 0 and not y = 0

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by keizer Soze » Wed Apr 29, 2009 7:26 pm
The last equation is z=zx

so here x=1 and z must be 0

y could be 0 also, but it musn´t be 0, it could be any other number. I think that there is the key.

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by lilu » Wed Apr 29, 2009 7:47 pm
xy+z=xy+xz-->z=xz-->z-xz=0--->z(1-x)=0----->z=0, x=1------>E
The more you look, the more you see.

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### Re: Algebra question

by marcusking » Thu Apr 30, 2009 5:46 am
ornelaspedro wrote:Can anybody tell me how to solve this;

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
e) x = 1 and z = 0

The answer is E, I solved for x and got one, however, I'm uncertain on how the answer is z = 0 and not y = 0
So what I don't get is why c wouldn't be a possibility too.

x=-5
y=1
z=0
-5*1 + 0 = -5(1+0)
With c both sides would always be equal to the value of x

I understand E works but doesn't C also?

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by keizer Soze » Thu Apr 30, 2009 8:07 am
C is a posibility, but there Y could be any other number, and the question asked which MUST be true... you can´t be sure that y=1... it could be y=2 or y=3...

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