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

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

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03 Dec 2012, 03:04
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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
[Reveal] Spoiler: OA
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Re: If y((3x-5)/2) =y and y#0, then x = [#permalink]

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03 Dec 2012, 03:05
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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

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

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07 Dec 2012, 03:39
$$y(\frac{3x-5}{2}) =y$$
$$\frac{3x-5}{2}=1$$
$$3x-5=2$$
$$3x=7$$
$$x=7/3$$

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

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11 Sep 2014, 02:57
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
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Re: If y((3x-5)/2) =y and y#0, then x = [#permalink]

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16 Nov 2014, 12:32
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

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

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16 Nov 2014, 12:36
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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Re: If y((3x-5)/2) =y and y#0, then x = [#permalink]

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16 Nov 2014, 21:41
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, 00:42
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]

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10 Dec 2014, 08:23
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, 19:27
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, 08:07
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, 04:39
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, 11:01
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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

[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] 16 Jul 2016, 11:01
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