Lesson: Basic Equation Solving

Comment on Basic Equation Solving

Hi Brent,
In your solution to this question

Given: (x − y)² = x² + y²
(x − y) (x − y) = x² -xy -xy +y² --> x² -2xy +y² = x² + y²
How you got it x² - xy + y² = x² + y²
where is the 2 in -2xy, although it will disappear in the next steps when we divide by -2, but I am just curious why it disappeared before, or maybe I am missing something..
Thank you.
gmat-admin's picture

Good catch - thanks!!
I have edited my solution here: https://gmatclub.com/forum/if-x-y-2-x-2-y-2-then-which-one-of-the-follow...


Hi Brent, I have a question about one of the practice problems related to this video, found here: http://www.beatthegmat.com/kaplan-is-x-0-t288237.html

Target question: is x>0
1. xy + y = y

I rearrange statement one to be x = (y-y)/y

Since any number minus itself is zero and since zero divided by any number is zero, the statement should never be greater than zero so I would conclude that statement one is sufficient. Why does this not work? I understand how you get two different answers by plugging in
(Case a: x = 1 and y = 0, x > 0
Case b: x = -1 and y = 0, x < 0 )

but I wouldn't have thought to plug in numbers for statement one if I would have first rearranged to get x = (y-y)/y

Thanks for your help,
gmat-admin's picture

Great question, Josh!!

Question link: http://www.beatthegmat.com/kaplan-is-x-0-t288237.html

It all comes down to your statement "zero divided by any number is zero"

This is not entirely true. 0/0 does not equal zero. 0/0 is undefined.

So, when we take xy + y = y and subtract y from both sides to get xy = 0, we must be very careful about our next step.

Our next step CANNOT be to divide both sides by y, because we may be inadvertently dividing by 0 (if it's the case that y = 0).

So, at this point (xy = 0), we must make the conclusion that either x = 0 or y = 0.

Does that help?


Thanks Brent, that clears it up!


why b
why not a?
gmat-admin's picture

Question link: https://gmatclub.com/forum/for-integers-a-b-and-c-if-ab-bc-then-which-of...

Given: ab = bc
Many students will want to divide both sides of the equation by b to get: a = c, but this is not true if b = 0.

For example, if a = 1, b = 0, and c = 2, the equation ab = bc holds true.
However, we cannot then conclude that a = c

Does that help?


Question link: https://gmatclub.com/forum/if-abc-b-3-which-of-the-following-must-be-true-109723.html

I think the trick here lies in the word "must". so the answer is none. If the question was "could"- the answer would be both 1 and 2 could be true.

Just checking the responses on the forum and no one seems to explain this key word.

Is my approach correct?
gmat-admin's picture

Question link: https://gmatclub.com/forum/if-abc-b-3-which-of-the-following-must-be-tru...

Yes, there's a huge difference between MUST and COULD.
If the question asks "What MUST be true," then that statement must be true for all possible values.
The great thing about "MUST be true" questions is that we can eliminate a statement if we can show an instance where that statement is NOT true.

If the question asks "What COULD be true" then a statement is true if we just need one instance where that statement is true.

So, if the question asked "What COULD be true," all three statements are valid.
Notice that, if a = b = c = 1, then statements I and III work.
Also, if a = b = c = 0, then statement II works.


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