# Lesson: Basic Equation Solving

## Comment on Basic Equation Solving

### Hi Brent,

Hi Brent,
In your solution to this question
https://gmatclub.com/forum/if-x-y-2-x-2-y-2-then-which-one-of-the-following-statements-241756.html

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.

### Good catch - thanks!!

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...

Cheers,
Brent

### Hi Brent, I have a question

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

Josh

### Great question, Josh!!

Great question, Josh!!

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?

Cheers,
Brent

### Thanks Brent, that clears it

Thanks Brent, that clears it up!

### https://gmatclub.com/forum

https://gmatclub.com/forum/for-integers-a-b-and-c-if-ab-bc-then-which-of-the-foll-147063.html

why b
why not a?

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?

Cheers,
Brent

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?

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.

Cheers,
Brent

Hi Brent, why is option A incorrect? One of the answers said that if b is 0 then a may not be equal to c. However, if we apply the same rule then if b=0 then a^2 may not be equal to c^2. Then why is option B correct?

The solution doesn't say "if b is 0 then a may not be equal to c"
It says either b = 0 or a = c. This includes the possibility that b = 0 AND a = c

Does that help?

Cheers,
Brent

### Hi Brent,

Hi Brent,

Could you please explain this question: https://gmatclub.com/forum/if-1-y-1-y-1-1-y-4-what-is-the-sum-of-all-solutions-for-y-242757.html

I don't understand the multiplying to get to y(y+1)=(y+4).

In order to eliminate all of the fractions (and to make the equation easier to solve), we must multiply both sides of the equation by the product of all three denominators: (y)(y+1)(y+4)

Take: 1/y + 1/(y+1) = 1/(y+4)
Multiply both sides of the equation by (y)(y+1)(y+4) to get: (y)(y+1)(y+4)/y + (y)(y+1)(y+4)/(y+1) = (y)(y+1)(y+4)/(y+4)
Now simplify to get: (y+1)(y+4) + (y)(y+4) = (y)(y+1)

Now that we have eliminated all of the fractions, we can solve the remaining equation for y.

### Hi BRent, I have a doubt

Hi Brent, I have a doubt regarding this question. Shall I place it here?

In the IT department, there are four people who are experienced and earn more than \$4000, two who are inexperienced and earn more than \$4000. There are twice as many people who are just from the IT department as are in the IT department and experienced but earning less than \$4000. How many inexperienced junior staff (junior staff earns lesser than \$4000) are in IT?
A) 6
B) 14
C) 9
D) 7
E) Cannot be determined from the information given

### ASIDE: This is a poorly

ASIDE: This is a poorly-worded question. You've posted some other low-quality questions here. What's the source?
There are thousands of official GMAT practice questions out there (as well as tons of high-quality unofficial practice questions). I think you'd be better served by focusing on those questions rather than the questions from the resource you've been using.

Having said all of that, here's my response.....

This is an Overlapping Sets question, which means we can solve it using the Double Matrix method.

Here's the video on the Double Matrix method: https://www.gmatprepnow.com/module/gmat-word-problems/video/919

I suggest that you watch the video, and then see if you can answer the question.
To help you get started, notice that we have a population of IT employees, and the two characteristics are:
- experienced or inexperienced
- earns more than \$4000 or does not earn more than \$4,000

Hint: If we let x = the number of employees who are experienced AND earning less than \$4000, then 2x = the total number of IT employees

### Hi Brent!

Hi Brent!
I got stuck in this one: https://gmatclub.com/forum/what-is-the-number-of-solutions-of-the-equation-2x-y-40-where-x-an-240112.html

I got 20 solutions, not 19. Values for x=1,2,3,4,5,6,7,8,9,10 ; corresponding y values = 38,36,34,32,30,28,26,24,22,20

The question stem asks for total number of solutions. Since the nearest is 19 I selected it.

As per the solutions in the thread, x=10 is incorrect. But isn't x=10 and y=20 a possible solution? It gives 2x + y = 40. I'm unable to see how that's not one of the solutions.

If you're referring to my solution in the thread, I meant to write that the solutions range from x=1 & y=38 all the way to x=19 & y=2
So, the solutions are:
x=1 & y=38
x=2 & y=36
x=3 & y=34
.
.
.
x=19 & y=2

I've edited my solution accordingly.