OG question

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

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??

x = 1 or z = 0

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.

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

All good until:

z - xz = 0

if you factor out z, you get:

z(1-x) = 0

So, z=0 or x=1: choose (5).

xy + z= x(y+z)
xy + z = xy + xz
z = xz
so, z=0 or x=1

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?

This is a great question to work on with the options as help.

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.