# Lesson: Eliminating Fractions

## Comment on Eliminating Fractions

### I don't understand how you

I don't understand how you rewrite The equation (m +n)/(4 + 5) = (m/4} + (n/5) can be rewritten as:

(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://greprepclub.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,

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

I'm confused about when to subtract for both sides. At 5:44, why do we subtract 7x from both sides and not 6x?

### Your approach will also yield

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://greprepclub.com/forum

https://greprepclub.com/forum/1-x-x-1-1-x-1837.html

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?

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

On the practice problems below the video:
(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?

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://greprepclub.com/forum

https://greprepclub.com/forum/qotd-8-when-the-decimal-point-of-a-certain-positive-decimal-2426.html

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

Thank You

### https://greprepclub.com/forum

https://greprepclub.com/forum/gre-math-challenge-111-n-4-r-8-s-8-t-717.html

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?

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,

Hi Brent,

For this Question.

https://greprepclub.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.

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.