If xy + z = x(y + z), which of the following must be true?

\(xy+z=x(y+z)\);

\(xy+z=xy+xz\);

\(xy\) cancels out: \(xz-z=0\);

\(z(x-1)=0\).

Either \(z=0\) (in this case \(x\) can take ANY value) OR \(x=1\) (in this case \(z\) can take ANY value).

Answer: E.

To elaborate more:

As the expression with \(y\) cancels out, we can say that given expression \(xy+z=x(y+z)\) does not depend on value of \(y\). Which means that \(y\) can take ANY value. So all answer choices which specify the exact value of \(y\) are wrong.

Next: "AND" in answer choices means that BOTH values must be true, but "OR" in answer choices means that EITHER value must be true.

Hope it helps.

Thanks again Bunuel . I somehow always got stucked at option C Your explanation is awesome

Yeah the explanation was very helpful..

Hi there! I'm happy to help with this.

So, the question:
If xy + z = x(y + z), then
(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

So, as a very general rough approximation rule, when variables are multiplied, the mathematical consequences will far more often involve the word "or" than the word "and." That's just a rough-and-ready rule to use if all else is lost.

Let's look at this question. We have: xy + z = x(y + z)

Distribute on the right side: xy + z = xy + xz

Subtract xy from both sides: z = xz

So first of all, we get this funny equation with x & z in it, without y, so that means: there's no restriction on y, only restrictions on x & z.

In solving the equation z = xz, you may be tempted to divide by z, but WHOA THERE! Before you divide by a variable, you have to ask yourself the deeply heartfelt question: could this variable equal zero? If so, then dividing by it would break all mathematical laws, and Santa Claus would find out you've been bad. So, what do we do?

We'll look at two cases: (i) z = 0, and (ii) z =/ 0
(i) if z = 0, then both sides of the equation x = xz equal zero, and the equation is true. So, one possibility that makes the equation true is z = 0
(ii) if z is not equal to zero, then we can divide by z, and get 1 = x. That's the other solution that makes the equation true.

The solutions are x = 1 or z = 0, answer choice E.

Here's another practice question for you.

https://gmat.magoosh.com/questions/110

The question at that link should be followed by a video explanation.

Please let me know if you have any questions about what I've said. Good luck tomorrow!!

Mike

I agree with the answer i.e. X=1 OR Z=0
But can the answer be also X=1 AND Z=0 (provided there was an option) as both of them taken together also satisfy the equation???

Hi Bunuel,
As per your solution, after cancelling the xy terms, you subtracted z from xz i.e \(xz-z=0\) instead of cancelling the variable "z". Could you please explain that part. Is it because we run the risk of forgetting z=0 value?
Thanks
H

We cannot reduce (divide) \(z(x-1)=0\) by \(z\) since it can be zero and division by zero is not allowed. Also if we do that we exclude the possible solution of the equation: \(z=0\).

I am confused because answer choice (A) also satisfies the equation. X=0 AND Z=0 as it is one of many possible solutions. If Z=0 ==> (0)Y=(0)Y. X can take an an infinity amount of values, as well as Y. Very puzzled, it seems like we have two correct answer choices in the question.

We are asked "which of the following MUST be true".

Now, if \(z=0\) then \(x\) can take ANY value. So, \(x=0\) AND \(z=0\) is not necessarily true.

Hi Bunuel,

-Two questions: I took the approach of plugging in answer choices and multiple answers satisfied the "must be true part". Why is that wrong?

-When you came down to xz=z, why did you take the next step of factoring out the z and equating the whole thing to zero, meaning, why did you get to the point of z(x-1)=0?

Thanks

1.
"MUST BE TRUE" questions:
These questions ask which of the following MUST be true, or which of the following is ALWAYS true for ALL valid sets of numbers you choose. Generally for such kind of questions if you can prove that a statement is NOT true for one particular valid set of numbers, it will mean that this statement is not always true and hence not a correct answer.

So, for "MUST BE TRUE" questions plug-in method is good to discard an option but not 100% sure thing to prove that an option is ALWAYS true.


As for "COULD BE TRUE" questions:
The questions asking which of the following COULD be true are different: if you can prove that a statement is true for one particular set of numbers, it will mean that this statement could be true and hence is a correct answer.

So, for "COULD BE TRUE" questions plug-in method is fine to prove that an option could be true. But here, if for some set of numbers you'll see that an option is not true, it won't mean that there does not exist some other set which will make this option true.

2.
To get what x and z could be from xz = z, we should do the way shown in my solution.

Hope it helps.

After getting z=xz , I cancelled out the z and got x=1. So naturally, B, D and E were valid options. A also gave me the equation z=xz which is exactly the same thing so I went with A. What is wrong in my approach?

Once we know that z = xz, we cannot divide both sides by z, because it COULD be the case that z equals zero, and we should never divide by zero.

Here's why:
We know that (2)(0) = (1)(0)
However, if we try to divide by both sides by 0, we may believe that we get: 2 = 1, which of course is not true.

Cheers,
Brent

We are given the following equation, which we can simplify:

xy + z = x(y + z)

xy + z = xy + xz

z = xz

z – xz = 0

z(1 – x) = 0

z = 0

OR

1 – x = 0

x = 1

So x = 1 or z = 0.

Answer: E

Hi chetan2u, Bunuel, VeritasKarishma

Where exactly did I go wrong?

If xy + z = x(y + z), which of the following must be true?

\((y+z)= xy+z/x\)
\((y+z)=y+z/x\)

D. x=1 or y=0

0+z=0+z/1
z=z

Super confused :/

ueh55406

Note what is given to you and what is asked. The approach of plugging in values using the options is not correct in such "MUST BE TRUE" questions.

You are given that xy + z = x(y + z).
You are asked whether (D) is implied by this equation. Note that even if after plugging in values using (D) you find that the equation holds, it does not mean that (D) is correct.

Take a simpler example to understand this:

Given: x > 0
Which of the following must be true:
(A) x is not 0
(B) x > 2


(A) x is not 0
Put x = 2. Inequality x > 0 holds.
Put x = -1. Inequality x > 0 does not hold.
Does this mean option (A) must not be true? No.
If x is positive as given to us, it must be true that x is not 0.

(B) x > 2
Again, put x = 3. Inequality x > 0 holds.
Put x = 4. Inequality x > 0 holds.
Put x = 10. Inequality x > 0 holds.
For all values of x that are greater than 2. inequality will hold. Does this mean that option (B) must be true?
No.
If x is positive, it does not necessarily imply that x > 2. x could be 1 too.

What does this tell us? That we need to focus on what is given and what is implied from it.
Check this post: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2017/0 ... -question/

Hope this point is clear.

Now, note that in our original question, option (D) gives x = 0 or y = 1. This means that at least one of them must be true. So either x = 0 or y = 1. You are taking them both to be true simultaneously (which is possible but not necessary).
In any case, as we discussed above, even if (D) satisfied the equation, it DOES NOT mean that (D) must be true.

Then how do we solve such questions?
We start with the given inequality and solve that to see what must be true.
xy + z = x(y + z)
xy + z = xy + xz
x (z - 1) = 0
So either x = 0 or z = 1 (at least one of them must be true)

Its a MUST be true type of question.

x y +z = x(y +z)

=>x y +z = x y+ x z

=>x z−z=0

=>z (x−1)=0

Either z=0 (in this case x can take any value)
Or
x=1 (in this case z can take any value).
(option e)

Devmitra Sen
GMAT SME