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555-605 Level|   Algebra|                  
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Bunuel
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If k#0 and k - (3 -2k^2)/k = x/k, then x = 24 Feb 2014, 23:50
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If \(k\neq{0}\) and \(k - \frac{3 -2k^2}{k} = \frac{x}{k}\), then x =

(A) -3 - k^2
(B) k^2 -3
(C) 3k^2 - 3
(D) k - 3 - 2k^2
(E) k - 3 + 2k^2

PS58602.01
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C
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If k#0 and k - (3 -2k^2)/k = x/k, then x = 24 Feb 2014, 23:50
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SOLUTION

If \(k\neq{0}\) and \(k - \frac{3 -2k^2}{k} = \frac{x}{k}\), then x =

(A) -3 - k^2
(B) k^2 -3
(C) 3k^2 - 3
(D) k - 3 - 2k^2
(E) k - 3 + 2k^2

\(k - \frac{3 -2k^2}{k} = \frac{x}{k}\);

Multiply by k: \(k^2 - (3 -2k^2) = x\);

\(x=3k^2-3\).

Answer: C.
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If k#0 and k - (3 -2k^2)/k = x/k, then x = 25 Feb 2014, 23:26
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k - \frac{3 -2k^2}{k} = \frac{x}{k};
After simplifying and canceling k in the denominator, we get: k^2 - 3 + 2k^2 = x;
3k^2 - 3 = x;

Ans is (C).
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If k#0 and k - (3 -2k^2)/k = x/k, then x = 26 Feb 2014, 02:10
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Option C.
k-(3-2k^2)/k=x/k
Multiply both sides by k
k^2-3+2k^2=x
3k^2-3=x
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If k#0 and k - (3 -2k^2)/k = x/k, then x = 08 Jul 2014, 22:17
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Bunuel
SOLUTION

If \(k\neq{0}\) and \(k - \frac{3 -2k^2}{k} = \frac{x}{k}\), then x =

(A) -3 - k^2
(B) k^2 -3
(C) 3k^2 - 3
(D) k - 3 - 2k^2
(E) k - 3 + 2k^2

\(k - \frac{3 -2k^2}{k} = \frac{x}{k}\);

Multiply by k: \(k^2 - (3 -2k^2) = x\);

\(x=3k^2-3\).

Answer: C.

I just added a step after the highlighted one to remove -ve sign :) . Rest all did same

\(k - \frac{3 -2k^2}{k} = \frac{x}{k}\);

\(k + \frac{2k^2 - 3}{k} = \frac{x}{k}\)
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If k#0 and k - (3 -2k^2)/k = x/k, then x = 08 Jul 2014, 23:55
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Bunuel
If \(k\neq{0}\) and \(k - \frac{3 -2k^2}{k} = \frac{x}{k}\), then x =

(A) -3 - k^2
(B) k^2 -3
(C) 3k^2 - 3
(D) k - 3 - 2k^2
(E) k - 3 + 2k^2
You can also put in k = 1 in \(k - \frac{3 -2k^2}{k} = \frac{x}{k}\) to get x = 0.
When you put k = 1 in options, only (C) and (E) give x = 0.

Now put k = -1 in \(k - \frac{3 -2k^2}{k} = \frac{x}{k}\) to get x = 0.
When you put k = -1 in options, only (C) gives x = 0­
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If k#0 and k - (3 -2k^2)/k = x/k, then x = 10 Mar 2015, 20:07
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Why do we automatically assume there are parentheses around (3-2k^2)? I didn't assume that so then I didn't distribute the negative on the outside which of course then gave me answer A.
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If k#0 and k - (3 -2k^2)/k = x/k, then x = 10 Mar 2015, 20:28
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soniasawhney
Why do we automatically assume there are parentheses around (3-2k^2)? I didn't assume that so then I didn't distribute the negative on the outside which of course then gave me answer A.

hi soniasawhney..
it does not require any parenthesis as any sign in front of a fraction modifies the entire fraction irrespective of the term..
just an example..
let the fraction be - \(\frac{7-9}{2}\)..
two ways to do it ..
first change signs first and then find answer...
\(\frac{-7+9}{2}\)=\(\frac{2}{2}\)=1...

second simplify and then change the signs...
- \(\frac{7-9}{2}\)=- \(\frac{-2}{2}\)=-(-1)=1..

so both ways answer is same, which means parentheses is not required and a sign in front of a fraction automatically means that the entire fraction is inside bracket....
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If k#0 and k - (3 -2k^2)/k = x/k, then x = 10 Mar 2015, 21:24
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soniasawhney
Why do we automatically assume there are parentheses around (3-2k^2)? I didn't assume that so then I didn't distribute the negative on the outside which of course then gave me answer A.

To clarify further:
There are 3 ways in which you can make a fraction negative
\(\frac{-3}{5}\)

\(-\frac{3}{5}\) Take this to mean \(\frac{-3}{5}\)

and \(\frac{3}{-5}\)

Each one of these fractions is the same as \(\frac{-3}{5}\)

Now what happens in case you have multiple terms in numerator:

\(\frac{-x + 2}{4}\) The negative is only in front of x.

\(-\frac{x+2}{4}\) The negative is in effect, in front of the entire numerator. So this is the same as \(\frac{-(x+2)}{4}\)
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If k#0 and k - (3 -2k^2)/k = x/k, then x = 12 Mar 2015, 23:12
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Bunuel
If \(k\neq{0}\) and \(k - \frac{3 -2k^2}{k} = \frac{x}{k}\), then x =

(A) -3 - k^2
(B) k^2 -3
(C) 3k^2 - 3
(D) k - 3 - 2k^2
(E) k - 3 + 2k^2
Given that,
k - (3 - 2k^2)/k = x/k
Take LCM of denominators,
So, [k^2 - (3 - 2k^2)]/k = x/k
So, [k^2 - 3 + 2k^2]/k = x/k
Canceling out k from the denominators of both sides,
So, [k^2 - 3 + 2k^2] = x
So, 3k^2 - 3 = x
Hence option C.

--
Optimus Prep's GMAT On Demand course for only $299 covers all verbal and quant. concepts in detail. Visit the following link to get your 7 days free trial account: https://www.optimus-prep.com/gmat-on-demand-course­
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If k#0 and k - (3 -2k^2)/k = x/k, then x = 07 Apr 2016, 16:29
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I'm not sure whats wrong here:

1. k - [3-2k^2][/k] = [x][/k]

2. k - 3 + 2k^2 = x

I am assuming I can't move the denominator to the right side because of the first term (k), just not sure why.
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If k#0 and k - (3 -2k^2)/k = x/k, then x = 10 Apr 2016, 20:06
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Apollon
I'm not sure whats wrong here:

1. k - [3-2k^2][/k] = [x][/k]

2. k - 3 + 2k^2 = x

I am assuming I can't move the denominator to the right side because of the first term (k), just not sure why.

1. k - [3-2k^2][/k] = [x][/k]

2. k - 3 + 2k^2 = x

Hi Apollon,

If you see, only two of the three terms have the denominator k.
In order to cancel out the denominator from all the terms, you need to have the same denominator in each term.

In the given case,

1. k - (3-2k^2)/ k = x / k
We need to make sure that the denominator of the first term is also the same.

Hence
k^2/k - (3-2k^2)/ k = x / k
From here on, we can cancel the terms.
k^2 - 3 + 2k^2 = x
Or x = 3k^2 - 3
Option C

Does this help?
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If k#0 and k - (3 -2k^2)/k = x/k, then x = 10 Apr 2016, 23:10
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Apollon
I'm not sure whats wrong here:

1. k - [3-2k^2][/k] = [x][/k]

2. k - 3 + 2k^2 = x

I am assuming I can't move the denominator to the right side because of the first term (k), just not sure why.

Yes, you re right. To move the denominator to the other side, it must be the denominator of the entire expression.

Look at it from a very basic viewpoint:

1/2 = x
1 = 2x
Since x is 1/2, twice of x will be 1.

1 + 1/2 = x
1 + 1 = 2x ????
x is actually 3/2 or (1.5). If you double it, will you get 2? No.

On the other hand,
(1 + 3+ 5)/2 = x
Then (1 + 3 + 5) = 2x is correct.
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If k#0 and k - (3 -2k^2)/k = x/k, then x = 30 Apr 2017, 01:33
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VeritasPrepKarishma
soniasawhney
Why do we automatically assume there are parentheses around (3-2k^2)? I didn't assume that so then I didn't distribute the negative on the outside which of course then gave me answer A.

To clarify further:
There are 3 ways in which you can make a fraction negative
\(\frac{-3}{5}\)

\(-\frac{3}{5}\) Take this to mean \(\frac{-3}{5}\)

and \(\frac{3}{-5}\)

Each one of these fractions is the same as \(\frac{-3}{5}\)

Now what happens in case you have multiple terms in numerator:

\(\frac{-x + 2}{4}\) The negative is only in front of x.

\(-\frac{x+2}{4}\) The negative is in effect, in front of the entire numerator. So this is the same as \(\frac{-(x+2)}{4}\)

Thank you, VeritasPrepKarishma

I too fell for this rule.

So basically in this exercise.

\(-\cfrac { 3-2{ k }^{ 2 } }{ k } \neq \cfrac { -3-2{ k }^{ 2 } }{ k } \\ \\ -\cfrac { 3-2{ k }^{ 2 } }{ k } \rightarrow \cfrac { -\left( 3-2{ k }^{ 2 } \right) }{ k }\)
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If k#0 and k - (3 -2k^2)/k = x/k, then x = 14 Apr 2018, 03:41
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chetan2u
soniasawhney
Why do we automatically assume there are parentheses around (3-2k^2)? I didn't assume that so then I didn't distribute the negative on the outside which of course then gave me answer A.

hi soniasawhney..
it does not require any parenthesis as any sign in front of a fraction modifies the entire fraction irrespective of the term..
just an example..
let the fraction be - \(\frac{7-9}{2}\)..
two ways to do it ..
first change signs first and then find answer...
\(\frac{-7+9}{2}\)=\(\frac{2}{2}\)=1...

second simplify and then change the signs...
- \(\frac{7-9}{2}\)=- \(\frac{-2}{2}\)=-(-1)=1..

so both ways answer is same, which means parentheses is not required and a sign in front of a fraction automatically means that the entire fraction is inside bracket....

Hello chetan2u

if fraction is - \(\frac{7-9}{2}\)

then why didnt you rewrite is as \(\frac{-(7-9)}{2}\) ----> \(\frac{-7+9}{2}\) :?

have a great weekend :)
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If k#0 and k - (3 -2k^2)/k = x/k, then x = 15 Sep 2019, 18:57
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Bunuel
If \(k\neq{0}\) and \(k - \frac{3 -2k^2}{k} = \frac{x}{k}\), then x =

(A) -3 - k^2
(B) k^2 -3
(C) 3k^2 - 3
(D) k - 3 - 2k^2
(E) k - 3 + 2k^2
k^2-(3-2k^2) = x
x=3k^2-3

IMO C­
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If k#0 and k - (3 -2k^2)/k = x/k, then x = 14 Nov 2020, 14:04
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Bunuel
If \(k\neq{0}\) and \(k - \frac{3 -2k^2}{k} = \frac{x}{k}\), then x =

(A) -3 - k^2
(B) k^2 -3
(C) 3k^2 - 3
(D) k - 3 - 2k^2
(E) k - 3 + 2k^2

PS58602.01


\(k - \frac{3 -2k^2}{k} = \frac{x}{k}\)

\(k = \frac{x}{k }+ \frac{3 -2k^2}{k}\)

\(k = \frac{x+ 3 -2k^2}{k}\)

\(k^2 = x+ 3 -2k^2\)

\(3k^2 - 3 = x\)

Answer is C.
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If k#0 and k - (3 -2k^2)/k = x/k, then x = 28 Jan 2021, 03:35
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Plugging in k = 2 can get the answer quickly. It is better to plug in for the variable which appears most number of times. You will be surprised to get the answer faster and accurately with plugging in as compared to algebra. Some students may need more practice though, but it is worth it.

k = 2
x = 9

Target answer: x = 9 (circle it)

Now, plug the value of the variables in the answer choices. Only option C will match.

Thus, the answer is C.

Hope this helps. :)
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If k#0 and k - (3 -2k^2)/k = x/k, then x = 21 Jun 2025, 01:28
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