Problem Solving

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
B. y = 1 and z = 0
D. x = 1 or y = 0
E. x = 1 or Z = 0

The key word here is must.
So, for example, consider answer choice A. While it's possible that x = 0 and z = 0, it need not be the case.
For example, x=1, y=1 and z=1 is a solution to the equation. So, this eliminates A.

The solution . . .
Given: xy+z = x(y+z)
Expand: xy+z = xy + xz
Subtract xy from both sides: z = xz
Rearrange: xz - z = 0
Factor: z(x-1) = 0

This tells us that z = 0 or x = 1
Answer: E

Cheers,
Brent

Hi GMATsid2016,

Notice how the answer choices to this question each involve two values and either an "and" or "or" qualifier. We can use those values against the prompt to determine what MUST be true.

To start, many of the answer choices include X=1, so let's see what happens when that 'condition' is met...

IF... X=1

XY + Z = X(Y+Z)
1(Y) + Z = 1(Y+Z)
Y + Z = Y + Z

X=1 certainly appears to be enough, on its own, to make the equation 'balance out' - so we wouldn't necessarily need any other restrictions on the values of Y or Z. There might be some other condition though that would also make the equation balance out, so we should take a good look at Answers D and E.

IF... Y=0

XY + Z = X(Y+Z)
X(0) + Z = X(0+Z)
X + Z = XZ

That isn't necessarily true (in fact, it's almost never true), so we can eliminate Answer D

IF... Z=0

XY + Z = X(Y+Z)
X(Y) + 0 = X(Y+0)
XY = XY

This is also balanced out, so this is another option.

Final Answer: E

GMAT assassins aren't born, they're made,
Rich

GMATsid2016 wrote: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 or y=0

E) x=1 or z=0

How come OA isE?

Please explain OR AND in choices?

I got x=1, but how come [spoiler]z=0[/spoiler]?

Thanks,

Sid

Hi Sid,

I think you have a challenge in dealing with z = xz.

Let us pick up from here.

z = xz

You can surely cancel out z from LHS and RHS, provided that you are sure that z is not equal to 0; but you are not. So the correct approach to deal is:

z = xz => z - xz = 0 => z(1 - x) = 0 => Either z = 0 or x - 1 = 0 => z = 0 or x = 1.

Hope this helps!

-Jay

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