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If y((3x-5)/2) =y and y#0, then x =
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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 =
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Re: If y((3x-5)/2) =y and y#0, then x =
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\(y(\frac{3x-5}{2}) =y\)
\(\frac{3x-5}{2}=1\)
\(3x-5=2\)
\(3x=7\)
\(x=7/3\)
Answer: 7/3
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Re: If y((3x-5)/2) =y and y#0, then x =
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Walkabout wrote:
If \(y(\frac{3x-5}{2}) =y\) and \(y\neq{0}\), then x =
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 =
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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 ? Posted from my mobile device
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Re: If y((3x-5)/2) =y and y#0, then x =
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Re: If y((3x-5)/2) =y and y#0, then x =
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y((3x-5)/2)=y
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Re: If y((3x-5)/2) =y and y#0, then x =
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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 =
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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 =
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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 =
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y((3x-5)/2) =y and y#0, then x = 7/3
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Re: If y((3x-5)/2) =y and y#0, then x =
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Walkabout wrote:
If \(y(\frac{3x-5}{2}) =y\) and \(y\neq{0}\), then x =
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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If y((3x-5)/2) =y and y#0, then x =
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Walkabout wrote:
If \(y(\frac{3x-5}{2}) =y\) and \(y\neq{0}\), then x =
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: C
For more on the Something Method, watch this video:
VIDEO For extra practice, try this question:
VIDEO Cheers,
Brent
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Originally posted by
GMATPrepNow on 16 Jul 2016, 12:01.
Last edited by
GMATPrepNow on 19 Mar 2018, 13:14, edited 2 times in total.
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Re: If y((3x-5)/2) =y and y#0, then x =
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( 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 =
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Bunuel wrote:
If \(y(\frac{3x-5}{2}) =y\) and \(y\neq{0}\), then x =
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 =
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cman2010 wrote:
Bunuel wrote:
If \(y(\frac{3x-5}{2}) =y\) and \(y\neq{0}\), then x =
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 =
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Bunuel wrote:
If \(y(\frac{3x-5}{2}) =y\) and \(y\neq{0}\), then x =
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 =
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Bunuel wrote:
If \(y(\frac{3x-5}{2}) =y\) and \(y\neq{0}\), then x =
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 =
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dave13 wrote:
Bunuel wrote:
If \(y(\frac{3x-5}{2}) =y\) and \(y\neq{0}\), then x =
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 = &nbs
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18 Mar 2018, 03:53
close
90% (00:32) correct 
please advice ) 
