Math problems from the *free* official practice tests and
problems from mba.com
CurtisVincentBorns
Forum Guests
 
Posts: 11
Joined: Wed Jun 20, 2012 11:39 am
 

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

by CurtisVincentBorns Tue Jul 24, 2012 1:01 pm

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

Correct Answer: E. x=1 or z = 0

This is what I did:

Step1: distribute the x

xy+z = xy+xz

Step2: Minus the xy from both sides

z= xz

Step 3: divide by z on both sides

x= 1

And this is where I got stuck!!

now I know x = 1 but how do i solve the rest of the equation?
tim
Course Students
 
Posts: 5665
Joined: Tue Sep 11, 2007 9:08 am
Location: Southwest Airlines, seat 21C
 

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

by tim Wed Jul 25, 2012 7:04 am

don't divide by z unless you know it can't be zero! what you should do instead is this:

zx = z
zx - z = 0
z(x-1) = 0

so either z=0 or x-1=0
Tim Sanders
Manhattan GMAT Instructor

Follow this link for some important tips to get the most out of your forum experience:
https://www.manhattanprep.com/gmat/forums/a-few-tips-t31405.html
CurtisVincentBorns
Forum Guests
 
Posts: 11
Joined: Wed Jun 20, 2012 11:39 am
 

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

by CurtisVincentBorns Wed Jul 25, 2012 12:53 pm

Thanks Tim!

I am just curious, what prompted you to attempt to solve the equation in that manner? As I naturally wouldn't think to go that route/

Was it just that you cant divide by a variable unless you know its not zero?
RonPurewal
Students
 
Posts: 19744
Joined: Tue Aug 14, 2007 8:23 am
 

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

by RonPurewal Fri Jul 27, 2012 1:47 am

please search the forum; there are already threads about this problem.

here:
if-xy-z-x-y-z-which-of-the-following-must-be-true-t6468.html

i'm going to lock this thread; if you still have questions, please post them on that thread, not on this one. thanks.