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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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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

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

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03 Dec 2012, 04:05

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

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07 Dec 2012, 04:39

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

\(\frac{3x-5}{2}=1\)

\(3x-5=2\)

\(3x=7\)

\(x=7/3\)

Answer: 7/3

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

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11 Sep 2014, 03:57

Walkabout 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

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

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

since y <> 0

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

x= 7/3

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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 ? ThanksPosted from my mobile device

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

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16 Nov 2014, 13:36

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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
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y gets cancelled out in observation itself

\(\frac{3x-5}{2}=1\)

3x-5=2

3x=7

\(x=\frac{7}{3}\)

Answer = C

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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
Walkabout 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

The answer is A.

(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
Walkabout 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

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

Answer:

For more on the Something Method, watch this video:

VIDEO For extra practice, try this question:

VIDEO 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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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: \(\frac{3x-5}{2}=1\) --> \(3x-5=2\) --> \(x=\frac{7}{3}\). Answer: C.

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?

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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: \(\frac{3x-5}{2}=1\) --> \(3x-5=2\) --> \(x=\frac{7}{3}\). Answer: C.

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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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: \(\frac{3x-5}{2}=1\) --> \(3x-5=2\) --> \(x=\frac{7}{3}\). Answer: C.

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

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

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18 Mar 2018, 03:34

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: \(\frac{3x-5}{2} =1\) --> \(3x-5=2\) --> \(x=\frac{7}{3}\). Answer: C.

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: \(\frac{3x-5}{2} =1\) --> \(3x-5=2\) --> \(x=\frac{7}{3}\). Answer: C.

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 =
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18 Mar 2018, 03:53