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# If sqrt(3x)-sqrt(4y)=3 and x-4/3y=5 then xy =

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Intern
Joined: 04 Jan 2017
Posts: 14
If sqrt(3x)-sqrt(4y)=3 and x-4/3y=5 then xy =  [#permalink]

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21 Feb 2017, 03:38
14
00:00

Difficulty:

75% (hard)

Question Stats:

58% (02:51) correct 42% (03:07) wrong based on 114 sessions

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If $$\sqrt{3x}-\sqrt{4y}=3$$ and $$x-\frac{4}{3}y=5$$ then xy =

a) 15
b) 9
c) 5
d) 1/3
e) 4/3

Source: NOVA Math
Intern
Joined: 04 Jan 2017
Posts: 14
Re: If sqrt(3x)-sqrt(4y)=3 and x-4/3y=5 then xy =  [#permalink]

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21 Feb 2017, 03:40
I'm stuck on this supposedly easy algebra problem... I guess there must be a fairly easy solution via special products but somehow I'm not seeing it right now... Plz help!
Manager
Joined: 17 May 2015
Posts: 247
If sqrt(3x)-sqrt(4y)=3 and x-4/3y=5 then xy =  [#permalink]

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Updated on: 21 Feb 2017, 05:31
2
Scyzo wrote:
If $$\sqrt{3x}-\sqrt{4y}=3$$ and $$x-\frac{4}{3}y=5$$ then xy =

a) 15
b) 9
c) 5
d) 1/3
e) 4/3

Source: NOVA Math

Hi Scyzo,

Given

$$\sqrt{3x}-\sqrt{4y}=3$$ --- (1)

$$x-\frac{4}{3}y=5 \Rightarrow 3x - 4y = 15 => (\sqrt{3x}+\sqrt{4y}) ((\sqrt{3x}-\sqrt{4y}) = 15$$

(Formula used $$a^{2} - b^{2} = (a+b)(a-b)$$)

=>$$(\sqrt{3x}+\sqrt{4y}) = 5$$ --- (2)

Now, by solving equation (1) and (2) we have x = 1/3 x = 16/3 and y = 4 y = 1/4. Hence, xy = 4/3.

Hope this helps.

Originally posted by ganand on 21 Feb 2017, 04:13.
Last edited by ganand on 21 Feb 2017, 05:31, edited 1 time in total.
Intern
Joined: 04 Jan 2017
Posts: 14
Re: If sqrt(3x)-sqrt(4y)=3 and x-4/3y=5 then xy =  [#permalink]

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21 Feb 2017, 04:55
Thanks! But could you elaborate on the last part a bit more? I do not quite get how you isolated x or y in the end...
Manager
Joined: 17 May 2015
Posts: 247
Re: If sqrt(3x)-sqrt(4y)=3 and x-4/3y=5 then xy =  [#permalink]

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21 Feb 2017, 05:29
1
Scyzo wrote:
Thanks! But could you elaborate on the last part a bit more? I do not quite get how you isolated x or y in the end...

Hi Scyzo",

Sure. There was an error in my last post. The value of x is 16/3 and y is 1/4. xy is still equal to 4/3. Sorry for that.

We have following equations:

$$\sqrt{3x}-\sqrt{4y}=3$$ --- (1)

and
$$\sqrt{3x}+\sqrt{4y} = 5$$ --- (2)

Add equation (1) and (2) we have following

$$2 \sqrt{3x} = 8 \Rightarrow \sqrt{3x} = 4$$. By squaring both side, we have $$3x = 16 \Rightarrow x = 16/3$$

Now substitute the value of x in equation (2).

$$\sqrt{3 \times \frac{16}{3}} + \sqrt{4y} = 5 \Rightarrow \sqrt{4y} = 5 - 4 = 1$$

By squaring both sides, we have $$4y = 1 \Rightarrow y = \frac{1}{4}$$.

$$xy = \frac{16}{3} \times \frac{1}{4} = \frac{4}{3}$$

Hope this helps. Please let me know, if something is not clear.
Intern
Joined: 04 Jan 2017
Posts: 14
Re: If sqrt(3x)-sqrt(4y)=3 and x-4/3y=5 then xy =  [#permalink]

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22 Feb 2017, 04:15
Thanks, yep I now get how this is supposed to be solved. Probably still wouldn't get it right on the actual test though. The question as such looks fairly straightforward but is actually quite a tough one as it involves a lot of steps. I'm still wondering if there's a simpler way of solving it but it doesn't really appear like it.
Intern
Joined: 21 Aug 2019
Posts: 1
Re: If sqrt(3x)-sqrt(4y)=3 and x-4/3y=5 then xy =  [#permalink]

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28 Sep 2019, 01:00
ganand wrote:
Scyzo wrote:
Thanks! But could you elaborate on the last part a bit more? I do not quite get how you isolated x or y in the end...

Hi Scyzo",

Sure. There was an error in my last post. The value of x is 16/3 and y is 1/4. xy is still equal to 4/3. Sorry for that.

We have following equations:

$$\sqrt{3x}-\sqrt{4y}=3$$ --- (1)

and
$$\sqrt{3x}+\sqrt{4y} = 5$$ --- (2)

Add equation (1) and (2) we have following

$$2 \sqrt{3x} = 8 \Rightarrow \sqrt{3x} = 4$$. By squaring both side, we have $$3x = 16 \Rightarrow x = 16/3$$

Now substitute the value of x in equation (2).

$$\sqrt{3 \times \frac{16}{3}} + \sqrt{4y} = 5 \Rightarrow \sqrt{4y} = 5 - 4 = 1$$

By squaring both sides, we have $$4y = 1 \Rightarrow y = \frac{1}{4}$$.

$$xy = \frac{16}{3} \times \frac{1}{4} = \frac{4}{3}$$

Hope this helps. Please let me know, if something is not clear.

How do you derive the term $$\sqrt{3x}+\sqrt{4y} = 5$$, how did you eliminate the term $$\sqrt{3x} - \sqrt{4y} = 5$$?
Manager
Joined: 09 May 2018
Posts: 76
Re: If sqrt(3x)-sqrt(4y)=3 and x-4/3y=5 then xy =  [#permalink]

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30 Sep 2019, 05:56
1
ganand wrote:
Scyzo wrote:
Thanks! But could you elaborate on the last part a bit more? I do not quite get how you isolated x or y in the end...

Hi Scyzo",

Sure. There was an error in my last post. The value of x is 16/3 and y is 1/4. xy is still equal to 4/3. Sorry for that.

We have following equations:

$$\sqrt{3x}-\sqrt{4y}=3$$ --- (1)

and
$$\sqrt{3x}+\sqrt{4y} = 5$$ --- (2)

Add equation (1) and (2) we have following

$$2 \sqrt{3x} = 8 \Rightarrow \sqrt{3x} = 4$$. By squaring both side, we have $$3x = 16 \Rightarrow x = 16/3$$

Now substitute the value of x in equation (2).

$$\sqrt{3 \times \frac{16}{3}} + \sqrt{4y} = 5 \Rightarrow \sqrt{4y} = 5 - 4 = 1$$

By squaring both sides, we have $$4y = 1 \Rightarrow y = \frac{1}{4}$$.

$$xy = \frac{16}{3} \times \frac{1}{4} = \frac{4}{3}$$

Hope this helps. Please let me know, if something is not clear.

In the last step, we can also, square both the equations and then subtracted the $$3x^1/2 +4y^1/2$$ to get multiplication directly.
Re: If sqrt(3x)-sqrt(4y)=3 and x-4/3y=5 then xy =   [#permalink] 30 Sep 2019, 05:56
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