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# If y((3x-5)/2) =y and y#0, then x =

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Manager
Joined: 02 Dec 2012
Posts: 178
If y((3x-5)/2) =y and y#0, then x = [#permalink]

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03 Dec 2012, 04:04
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Question Stats:

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If $$y(\frac{3x-5}{2}) =y$$ and $$y\neq{0}$$, then x =

(A) 2/3
(B) 5/3
(C) 7/3
(D) 1
(E) 4
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Sep 2009
Posts: 44298
Re: If y((3x-5)/2) =y and y#0, then x = [#permalink]

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03 Dec 2012, 04:05
Expert's post
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This post was
BOOKMARKED
If $$y(\frac{3x-5}{2}) =y$$ and $$y\neq{0}$$, then x =

(A) 2/3
(B) 5/3
(C) 7/3
(D) 1
(E) 4

Since $$y\neq{0}$$, then we can reduce the equation by it: $$\frac{3x-5}{2}=1$$ --> $$3x-5=2$$ --> $$x=\frac{7}{3}$$.

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Senior Manager
Joined: 13 Aug 2012
Posts: 453
Concentration: Marketing, Finance
GPA: 3.23
Re: If y((3x-5)/2) =y and y#0, then x = [#permalink]

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07 Dec 2012, 04:39
$$y(\frac{3x-5}{2}) =y$$
$$\frac{3x-5}{2}=1$$
$$3x-5=2$$
$$3x=7$$
$$x=7/3$$

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Manager
Joined: 07 Apr 2014
Posts: 130
Re: If y((3x-5)/2) =y and y#0, then x = [#permalink]

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11 Sep 2014, 03:57
If $$y(\frac{3x-5}{2}) =y$$ and $$y\neq{0}$$, then x =

(A) 2/3
(B) 5/3
(C) 7/3
(D) 1
(E) 4

y((3x-5)/2)=y

y[((3x-5)/2) - 1)] = 0

since y <> 0

(3x-5)/2 -1 = 0

x= 7/3
Manager
Joined: 15 Jan 2013
Posts: 57
Re: If y((3x-5)/2) =y and y#0, then x = [#permalink]

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16 Nov 2014, 13:32
Hey guys ... I'm just wondering .... Is this really a 600 level question ? It seems very easy to be in the 600. Can any one here make sure it is in 600 level ?

Thanks

Posted from my mobile device
Math Expert
Joined: 02 Sep 2009
Posts: 44298
Re: If y((3x-5)/2) =y and y#0, then x = [#permalink]

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16 Nov 2014, 13:36
shagalo wrote:
Hey guys ... I'm just wondering .... Is this really a 600 level question ? It seems very easy to be in the 600. Can any one here make sure it is in 600 level ?

Thanks

Posted from my mobile device

It's sub-600 level question, so below 600.
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Joined: 21 Jan 2014
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Re: If y((3x-5)/2) =y and y#0, then x = [#permalink]

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16 Nov 2014, 22:41
y((3x-5)/2)=y

since y not equal to zero

y((3x-5)/2)-y=0
y((3x-5)/2 - 1)=0
so (3x-5)/2 =1
x=7/3

Its C
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Re: If y((3x-5)/2) =y and y#0, then x = [#permalink]

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20 Nov 2014, 01:42
1
KUDOS
y gets cancelled out in observation itself

$$\frac{3x-5}{2}=1$$

3x-5=2

3x=7

$$x=\frac{7}{3}$$

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Intern
Joined: 17 Apr 2014
Posts: 1
GMAT Date: 01-07-2015
Re: If y((3x-5)/2) =y and y#0, then x = [#permalink]

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10 Dec 2014, 09:23
Why can you cancel out Y? What is the mathematic rule for that?
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Re: If y((3x-5)/2) =y and y#0, then x = [#permalink]

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10 Dec 2014, 20:27
JoanBarneveld wrote:
Why can you cancel out Y? What is the mathematic rule for that?

Y is common factor of LHS & RHS, so cancels out in the first instance
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Re: If y((3x-5)/2) =y and y#0, then x = [#permalink]

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05 Mar 2016, 09:07
y((3x-5)/2) =y and y#0, then x =

3xy/2-5y/2 =y

3xy/2= 5y/2+y

3xy/2 = (5y+2y)/2

6xy = 10y + 4y

6xy= 14y

x= 14/6 = 7/3
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Re: If y((3x-5)/2) =y and y#0, then x = [#permalink]

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27 May 2016, 05:39
Expert's post
2
This post was
BOOKMARKED
If $$y(\frac{3x-5}{2}) =y$$ and $$y\neq{0}$$, then x =

(A) 2/3
(B) 5/3
(C) 7/3
(D) 1
(E) 4

(Note: We can divide both sides by y because it’s given that y does not equal 0. Otherwise, we couldn't divide both sides by y.)
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Re: If y((3x-5)/2) =y and y#0, then x = [#permalink]

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16 Jul 2016, 12:01
Expert's post
Top Contributor
If $$y(\frac{3x-5}{2}) =y$$ and $$y\neq{0}$$, then x =

(A) 2/3
(B) 5/3
(C) 7/3
(D) 1
(E) 4

We can also apply a technique I call the Something Method

Given: y[(3x-5)/2] = y
We have y(something) = y, so that something must equal 1
That is, (3x-5)/2 = 1

If the fraction (3x-5)/2 =1, then: 3x - 5 = 2
Add 5 to both sides to get: 3x = 7
Solve: x = 7/3

[Reveal] Spoiler:
C

For more on the Something Method, watch this video:

For extra practice, try this question:

Cheers,
Brent
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Re: If y((3x-5)/2) =y and y#0, then x = [#permalink]

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18 Sep 2017, 22:36
( 3x-5)/2 = y/y
3x - 5 = 2
3x = 7
x = 7/3
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Joined: 10 Sep 2017
Posts: 5
Re: If y((3x-5)/2) =y and y#0, then x = [#permalink]

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18 Dec 2017, 22:09
Bunuel wrote:
If $$y(\frac{3x-5}{2}) =y$$ and $$y\neq{0}$$, then x =

(A) 2/3
(B) 5/3
(C) 7/3
(D) 1
(E) 4

Since $$y\neq{0}$$, then we can reduce the equation by it: $$\frac{3x-5}{2}=1$$ --> $$3x-5=2$$ --> $$x=\frac{7}{3}$$.

Hi Brunel - I am grateful for your participation and explanation of PS questions.

Question: does y =/= 0 mean that y is 1? or that y is simply not equal to zero?
Math Expert
Joined: 02 Sep 2009
Posts: 44298
Re: If y((3x-5)/2) =y and y#0, then x = [#permalink]

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18 Dec 2017, 22:52
cman2010 wrote:
Bunuel wrote:
If $$y(\frac{3x-5}{2}) =y$$ and $$y\neq{0}$$, then x =

(A) 2/3
(B) 5/3
(C) 7/3
(D) 1
(E) 4

Since $$y\neq{0}$$, then we can reduce the equation by it: $$\frac{3x-5}{2}=1$$ --> $$3x-5=2$$ --> $$x=\frac{7}{3}$$.

Hi Brunel - I am grateful for your participation and explanation of PS questions.

Question: does y =/= 0 mean that y is 1? or that y is simply not equal to zero?

$$y\neq{0}$$ means that y is not 0. It can be any other value. It's given so that we could reduce by y. If y were 0, then we could not reduce because we cannot divide by 0. y being 0, would also imply that $$y(\frac{3x-5}{2}) =y$$ would hold true for any value of x, because $$y(\frac{3x-5}{2})=0 =y$$ and we would not be able to solve for x.

Hope it's clear.
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Posts: 327
If y((3x-5)/2) =y and y#0, then x = [#permalink]

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24 Feb 2018, 12:43
Bunuel wrote:
If $$y(\frac{3x-5}{2}) =y$$ and $$y\neq{0}$$, then x =

(A) 2/3
(B) 5/3
(C) 7/3
(D) 1
(E) 4

Since $$y\neq{0}$$, then we can reduce the equation by it: $$\frac{3x-5}{2}=1$$ --> $$3x-5=2$$ --> $$x=\frac{7}{3}$$.

Hi Bunuel i did this, i got correct answer but i doubt of my approach ... can you please point out where am i wrong

$$y(\frac{3x-5}{2}) =y$$

$$y(3x-5)=2y$$

$$3xy-5y=2y$$

$$3xy-5y-2y$$

$$3xy-7y$$

$$3xy =7y$$

$$x=7y/3y$$ (i got confused here, should i divide by 3y, or 3xy, or 3x please advice )

many thanks
Senior Manager
Joined: 09 Mar 2016
Posts: 327
Re: If y((3x-5)/2) =y and y#0, then x = [#permalink]

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18 Mar 2018, 03:34
Bunuel wrote:
If $$y(\frac{3x-5}{2}) =y$$ and $$y\neq{0}$$, then x =

(A) 2/3
(B) 5/3
(C) 7/3
(D) 1
(E) 4

Since $$y\neq{0}$$, then we can reduce the equation by it: $$\frac{3x-5}{2} =1$$ --> $$3x-5=2$$ --> $$x=\frac{7}{3}$$.

Hi niks18, can you advice if my solution is correct, and where from did Bunuel get 1 ?

$$y(\frac{3x-5}{2}) =y$$

$$\frac{y(3x-5)}{2}=y$$

$$y3x-5y = 2y$$

$$3xy= 2y +5y$$ now lets divide by 3y since we need to isolate x

$$x= \frac{2y +5y}{3y}$$

$$x= 7y/3y$$

hence OA = C
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Re: If y((3x-5)/2) =y and y#0, then x = [#permalink]

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18 Mar 2018, 03:53
dave13 wrote:
Bunuel wrote:
If $$y(\frac{3x-5}{2}) =y$$ and $$y\neq{0}$$, then x =

(A) 2/3
(B) 5/3
(C) 7/3
(D) 1
(E) 4

Since $$y\neq{0}$$, then we can reduce the equation by it: $$\frac{3x-5}{2} =1$$ --> $$3x-5=2$$ --> $$x=\frac{7}{3}$$.

Hi niks18, can you advice if my solution is correct, and where from did Bunuel get 1 ?

$$y(\frac{3x-5}{2}) =y$$

$$\frac{y(3x-5)}{2}=y$$

$$y3x-5y = 2y$$

$$3xy= 2y +5y$$ now lets divide by 3y since we need to isolate x

$$x= \frac{2y +5y}{3y}$$

$$x= 7y/3y$$

hence OA = C

Hi dave13

you solution is correct. you can also simply divide both sides of the equation y. So in the RHS y/y=1
Re: If y((3x-5)/2) =y and y#0, then x =   [#permalink] 18 Mar 2018, 03:53
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