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If 3x - 2y = 32 and sqrt (3x) - sqrt (2y) = 4.

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MrWhite
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If 3x - 2y = 32 and sqrt (3x) - sqrt (2y) = 4. [#permalink] New post   04 Nov 2023, 02:57
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If \(3x - 2y = 32\) and \(\sqrt{3x} - \sqrt{2y} = 4\), what is the value of \(x - y\)?

(A) 4
(B) 8
(C) 10
(D) 12
(E) 14
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C
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gmatophobia
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If 3x - 2y = 32 and sqrt (3x) - sqrt (2y) = 4. [#permalink] New post   04 Nov 2023, 03:06
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MrWhite wrote:
If \(3x - 2y = 32\) and \(\sqrt{3x} - \sqrt{2y} = 4\), what is the value of \(x - y\)?

(A) 4
(B) 8
(C) 10
(D) 12
(E) 14


Given:

  • \(3x - 2y = 32\)
  • \(\sqrt{3x} - \sqrt{2y} = 4\) ------ (i)

\(3x - 2y = 32\) ⇒ Assume that the equation is the form of \(a^2 - b^2 \)

\((\sqrt{3x} - \sqrt{2y})(\sqrt{3x} + \sqrt{2y}) = 32\)

\(4*(\sqrt{3x} + \sqrt{2y}) = 32\)

\(\sqrt{3x} + \sqrt{2y} = 8\) ------ (ii)

Adding (i) and (ii)

\(2\sqrt{3x} = 12\)

\(\sqrt{3x} = 6\)

Squaring both sides

\(3x = 36\)

\(x = 12\)

We can replace the value of \(x\) in equation \(3x - 2y = 32\)

\(36 - 32 = 2y\)

\(y = 2\)

\(x - y = 10\)

Option C
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If 3x - 2y = 32 and sqrt (3x) - sqrt (2y) = 4. [#permalink] New post   Updated on: 04 Nov 2023, 03:10
My approach

\(3x - 2y = 32\)

\(1.5x - y = 16\) Dividing by 2 (to get something similar / close to desired x-y)

=> \((x-y) + \frac{x}{2} = 16\) Condition 1

\(\sqrt{3x} - \sqrt{2y} = 4\) Condition 2

Looking for values of \((x-y)\) from answer choices which yield x & y such that they satisfy Conditions 1 & 2

Cycling through options for \((x-y)\). Only when \((x-y) = 10\) solutions of x & y meet both conditions, since \(\sqrt{3x}\) becomes perfect square!

(C)

I am not too sure whether this is the intended / optimal way to approach the problem, so am open to other solution methods :)

Originally posted by MrWhite on 04 Nov 2023, 03:03.
Last edited by MrWhite on 04 Nov 2023, 03:10, edited 3 times in total.
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If 3x - 2y = 32 and sqrt (3x) - sqrt (2y) = 4. [#permalink] New post   04 Nov 2023, 17:13
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The problem can be solved pretty quickly by looking at square root 3x - square root 2y = 4

It becomes obvious that the two have to be perfect squares. If x is 3 and y is 2 answer will be 5 not acceptable.

But 36 and 4 are possible so x is 12 and y is 2. Plug in the first equation to see that it is also true. So 12-2 = 10. Option C answer.
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Re: If 3x - 2y = 32 and sqrt (3x) - sqrt (2y) = 4. [#permalink] New post   14 Jan 2024, 10:31
Bunuel chetan2u gmatphobia I'm not able to understand how you got 4 in 4∗(3x−−√+2y−−√)=32

Can you please explain?
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If 3x - 2y = 32 and sqrt (3x) - sqrt (2y) = 4. [#permalink] New post   14 Jan 2024, 11:30
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AnkurGMAT20 wrote:
Bunuel chetan2u gmatphobia I'm not able to understand how you got 4 in 4∗(3x−−√+2y−−√)=32

Can you please explain?


AnkurGMAT20 The value is taken from the question stem. Check if this alternate representation of the solution helps -
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Re: If 3x - 2y = 32 and sqrt (3x) - sqrt (2y) = 4. [#permalink] New post   13 Apr 2024, 04:41
Expert Reply
 
MrWhite wrote:
If \(3x - 2y = 32\) and \(\sqrt{3x} - \sqrt{2y} = 4\), what is the value of \(x - y\)?

(A) 4
(B) 8
(C) 10
(D) 12
(E) 14

­
Intuitive Method:

Since difference of x and y in an integer (from options) and
\(\sqrt{3x} - \sqrt{2y} = 4\) which is also an integer, most likely x is a multiple of 3 with another perfect square and y is a multiple of 2 with another perfect square. 
So y must be at least 2*1 = 2 and since 

\(\sqrt{3x} = 4 + \sqrt{2y}\),

x cannot be just 3. Let's try 3*4 = 12 (the next perfect square after 1 is 4)

It works! All numbers over here and simple and small hence it wouldn't require many iterations in any case. 

So x = 12 and y = 2. 
x - y = 10

Answer (C)

 ­
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Re: If 3x - 2y = 32 and sqrt (3x) - sqrt (2y) = 4. [#permalink] New post   13 Apr 2024, 04:57
Set 3X = v^2, 2Y=w^2

Substituting:

V^2-W^2=(V+W)(V-W)=32

V-W=4

Since V-W=4, V+W = 8

Adding the two:

2V=12, so V=6 and W=2

So X=V^2/3=36/3 =

12

and Y=W^2/2=4/2 =

2

So X-Y = 10

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Re: If 3x - 2y = 32 and sqrt (3x) - sqrt (2y) = 4. [#permalink]
13 Apr 2024, 04:57
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