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

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

For more on the Something Method, watch this video:

For extra practice, try this question:

Cheers,
Brent

Originally posted by BrentGMATPrepNow on 16 Jul 2016, 12:01.
Last edited by BrentGMATPrepNow 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 = [#permalink]
( 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]
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?
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Re: If y((3x-5)/2) =y and y#0, then x = [#permalink]
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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If y((3x-5)/2) =y and y#0, then x = [#permalink]
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
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Re: If y((3x-5)/2) =y and y#0, then x = [#permalink]
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]
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
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If y((3x-5)/2) =y and y#0, then x = [#permalink]
Since y is not equal to 0 we can cancel out y
(3x-5)/(2)=1
solving gives us x=7/3 i.e C
If y((3x-5)/2) =y and y#0, then x = [#permalink]
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