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## Comment on

Eliminating Fractions## I don't understand how you

(m +n)/9} = {(5m +4n)}{20}.

How do you get the (5m+4n)/20?

## You're referring to this

You're referring to this question: http://gre.myprepclub.com/forum/if-m-n-4-5-m-4-n-2181.html

Given: (m + n)/(4 + 5) = (m/4) + (n/5)

Simplify denominator on left side to get: (m + n)/(9) = (m/4) + (n/5)

Rewrite fractions on right side with SAME DENOMINATOR:(m + n)/9 = (5m/20) + (4n/20)

Combine fractions on right side: (m + n)/9 = (5m + 4n)/20

Simplify right side: (m + n)/9 = (9m)/20

## Hi there,

in the question from Barron's book (when C and D are positive integers)

When we reach this step: C = d/ (d+1)

we could have let it also = DC + C = D , then D = C/ (1-C) which makes quantity A bigger if i am not mistaken.

Even when we plug 2 for c we get D to be -2 ..

I hope that you answer my question, Thanx.

## It took me a while to

It took me a while to identify the issue with your solution :-)

You're right to say that, when c = 2, then d = -2

However, we're told that c and d are both POSITIVE.

So, it cannot be the case that d = -2

## I'm confused about when to

## Your approach will also yield

Your approach will also yield the correct answer.

Let's examine both approaches.

APPROACH #1 (mine)

Given: 7x + 4 = 6x

Subtract 7x from both sides: 4 = -1x

Divide both sides by -1 to get: -4 = x

APPROACH #2 (yours)

Given: 7x + 4 = 6x

Subtract 6x from both sides: x + 4 = 0

Subtract 4 from both sides: x = -4

As long as you follow the rule performing the same operation to both sides of the equation, you'll solve the equation.

Cheers,

Brent

## https://gre.myprepclub.com/forum

I'm a bit confused with this question. When i solved it i got the equation x^2 = 1 which can be solved to get +1 or -1. The explanation mentioned states that x cannot equal 1 or 0 as the equation would get undefined if we put in those values. But my question is that the problem did not mention that x cannot be 0 or 1. What stops us from selecting D as an answer?

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/1-x-x-1-1-x-1837.html

The key concept here is that, for any value of k, k/0 does not have a value.

We say that k/0 is undefined.

For example, 3/0, 19/0, and 0/0 are all undefined.

We're told that (1 - x)/(x - 1) = 1/x

This is a very subtle way of telling us that x CANNOT equal 0 or 1

To understand why this is the case, let's examine an easier equation first.

Let's solve: 15/x = 3

When we solve this, we get: x = 5

We can show that this is true by replacing the x in the original equation to get: 15/5 = 3, which turns out to be true.

Similarly, if x = 1 is a solution to the equation (1 - x)/(x - 1) = 1/x, then we should be able to replace x with 1 and get a resulting expression that's true. Let's see what happens.

We get: (1 - 1)/(1 - 1) = 1/1

Simplify: 0/0 = 1

This is NOT true.

So, x = 1 cannot be a solution.

In other words, based on the given equation, x CANNOT equal 1

Now let's see if x = 0 is a solution.

Take the equation (1 - x)/(x - 1) = 1/x, and replace x with 0.

We get: (1 - 0)/(0 - 1) = 1/0

Simplify: 1/-1 = 1/0

Simplify more: -1 = 1/0

This is NOT true.

So, x = 0 cannot be a solution.

In other words, based on the given equation, x CANNOT equal 0

Does that help?

Cheers,

Brent

## On the practice problems

(1-x)/(x-1)=1/x

x is not equal to 1

A: X

B: -1/2

I incorrectly solved by doing:

realizing that 1/1x = x so:

(1-x)/(x-1)=x

Multiply both sides by x-1 to get:

1-x=x^2-x

then

1=x^2

so x=√1

Therefore quantity A is greater. (incorrect) Where did I go wrong?

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/gre-math-challenge-142-1-x-x-1-1-x-800....

The problem occurs very early in your solution.

You say that 1/1x = x, but this is not true.

We can show it's not true, by testing a value of x.

For example, if x = 3, your equation becomes 1/(1)(3) = 3

Simplify to get: 1/3 = 3, which is not true.

Does that help?

Cheers,

Brent

## https://gre.myprepclub.com/forum

Could you explain the solution to this problem because I'm having a hard time understanding it.

Thank You

## Here's my full solution:

Here's my full solution: https://gre.myprepclub.com/forum/qotd-8-when-the-decimal-point-of-a-cert...

## https://gre.myprepclub.com/forum

2n+r ? 2s+t ->(divide by 2 both sides)

2n+r/2 ? 2s+t/2 ->(times 1/4)

2n+r/8 ? 2s+t/8 ->simplify

n/4 + r/8 ? 2* s/8 + t/8

At this point, since we do not know the value of t we cannot determine the result. So, D. is it correct that way?

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/gre-math-challenge-111-n-4-r-8-s-8-t-71...

I'm not sure I follow your logic.

You're correct to say that we don't know the value of t, but we also don't know the value of any of the four variables.

## Hi Brent,

For this Question.

https://gre.myprepclub.com/forum/x-x-1-or-x-1-x-28001.html

x/(x+1) - Can be simplified to 1 + x

(x+1)/x - Can be simplified to 1 + 1/x

We are left with x^2 and 1.

As we know x>0 So it could be 1 or anything over 1 Right. So answer needs to be D. Please tell me where i did go wrong.

Thanks.

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/x-x-1-or-x-1-x-28001.html

I'm terribly sorry, Nick. I don't know how this question slipped past me.

Be careful, one of your simplifications is incorrect.

I believe you're applying the fraction property that says: (a+b)/c = a/c + a/b

This means we can take (x+1)/x and rewrite it as x/x + 1/x, which simplifies to 1 + 1/x

So, your simplification of Quantity B is perfect.

However, there's no fraction property that says c/(a+b) = c/a + a/b

So we can't take x/(x+1) and rewrite it as x/x + x/1, which means you are simplification of Quantity A is incorrect.