Given that XY + Z = X(Y + Z)
xy + z = xy + xz
xy + z - xy - xz = 0
z - xz = 0
z(x-z) = 0
z=0 OR x = z
OG question
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GMAT_crusher
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xy + z = xy +xz
i.e zx = z
i.e x = 1 for sure.
y does not matter.
so ..
x = 1; z =0 fits the most ... did I get that right ???
oa??
i.e zx = z
i.e x = 1 for sure.
y does not matter.
so ..
x = 1; z =0 fits the most ... did I get that right ???
oa??
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xy+z = x(y+z)
xy+z = xy+xz
z = xz ---> subtract xy from both sides
z - xz = 0 ---> move terms to one side
z(1-x) = 0 ---> factor left hand side of equation
So z = 0 or (1-x) = 0, which means x = 1.
So z=0, x=1.
xy+z = xy+xz
z = xz ---> subtract xy from both sides
z - xz = 0 ---> move terms to one side
z(1-x) = 0 ---> factor left hand side of equation
So z = 0 or (1-x) = 0, which means x = 1.
So z=0, x=1.
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All good until:mmukher wrote:Given that XY + Z = X(Y + Z)
xy + z = xy + xz
xy + z - xy - xz = 0
z - xz = 0
z(x-z) = 0
z=0 OR x = z
z - xz = 0
if you factor out z, you get:
z(1-x) = 0
So, z=0 or x=1: choose (5).

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richardwang6430
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Musiq
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This is a great question to work on with the options as help.Riggz 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
any explanation for this one?
XY + Z = XY + XZ.
Therefore, Z = XZ.
At this point eliminate, every option that has Y in it (That eliminates B / C and E).
Since the 2 remaining options have Z= 0, we know this must be true,so elminate Z from both sides of the equation to get X= 1.
You can do this since you have accounted for the Z=0 possibility.
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