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# If 3x/2 = y, and 2 - 3y = y + 2, then x =

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Math Expert
Joined: 02 Sep 2009
Posts: 52230
If 3x/2 = y, and 2 - 3y = y + 2, then x =  [#permalink]

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23 Oct 2018, 23:17
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00:00

Difficulty:

5% (low)

Question Stats:

94% (00:52) correct 6% (00:16) wrong based on 47 sessions

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If $$\frac{3x}{2} = y$$, and $$2 - 3y = y + 2$$, then x =

A. -3
B. -2
C. 0
D. 2
E. 3

_________________
Director
Joined: 19 Oct 2013
Posts: 509
Location: Kuwait
GPA: 3.2
WE: Engineering (Real Estate)
Re: If 3x/2 = y, and 2 - 3y = y + 2, then x =  [#permalink]

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23 Oct 2018, 23:32
2
Bunuel wrote:
If $$\frac{3x}{2} = y$$, and $$2 - 3y = y + 2$$, then x =

A. -3
B. -2
C. 0
D. 2
E. 3

2 - 3y = y + 2

2 - 2 = 4y

y = 0 x = 0

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e-GMAT Representative
Joined: 04 Jan 2015
Posts: 2447
Re: If 3x/2 = y, and 2 - 3y = y + 2, then x =  [#permalink]

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24 Oct 2018, 02:32

Solution

Given:
• 3x/2 = y ……... (1)
• 2 – 3y = y + 2 ……… (2)

To find:
• The value of x

Approach and Working:
• From equation (2), we get, 4y = 0
o Implies, y = 0
• Substituting y = 0 in equation (1), we get, x = 0

Hence, the correct answer is Option C

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Re: If 3x/2 = y, and 2 - 3y = y + 2, then x =  [#permalink]

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24 Oct 2018, 03:52
Bunuel wrote:
If $$\frac{3x}{2} = y$$, and $$2 - 3y = y + 2$$, then x =

A. -3
B. -2
C. 0
D. 2
E. 3

2 - 3y = y + 2

4y = 0

y = 0.

Given

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

$$\frac{3x}{2}$$= 0

3x = 0

x = 0.

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Re: If 3x/2 = y, and 2 - 3y = y + 2, then x =  [#permalink]

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24 Oct 2018, 07:57
Bunuel wrote:
If $$\frac{3x}{2} = y$$, and $$2 - 3y = y + 2$$, then x =

A. -3
B. -2
C. 0
D. 2
E. 3

$$2 - 3y = y + 2$$

Or, $$4y = 0$$

So, $$y = 0$$

Thus, $$\frac{3x}{2} = 0$$

Hence, x = 0 ; Answer must be (C)
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Joined: 14 Oct 2015
Posts: 4527
Location: United States (CA)
Re: If 3x/2 = y, and 2 - 3y = y + 2, then x =  [#permalink]

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28 Oct 2018, 17:19
Bunuel wrote:
If $$\frac{3x}{2} = y$$, and $$2 - 3y = y + 2$$, then x =

A. -3
B. -2
C. 0
D. 2
E. 3

Simplifying the second equation we have:

2 = 4y + 2

0 = 4y

0 = y

Thus,

3x/2 = 0

3x = 0

x = 0

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Re: If 3x/2 = y, and 2 - 3y = y + 2, then x = &nbs [#permalink] 28 Oct 2018, 17:19
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