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(16x^4 – 81y^4)/(2x + 3y) = 12x^2 + 27y^2 and 4x + 3y = 9, what is x?

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(16x^4 – 81y^4)/(2x + 3y) = 12x^2 + 27y^2 and 4x + 3y = 9, what is x?  [#permalink]

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New post 05 Nov 2014, 07:35
2
5
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

65% (02:53) correct 35% (02:57) wrong based on 138 sessions

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Tough and Tricky questions: Algebra.



(16x^4 – 81y^4)/(2x + 3y) = 12x^2 + 27y^2 and 4x + 3y = 9, what is x?

A. 1
B. 1.5
C. 2
D. 3
E. 4

Kudos for a correct solution.

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Re: (16x^4 – 81y^4)/(2x + 3y) = 12x^2 + 27y^2 and 4x + 3y = 9, what is x?  [#permalink]

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New post 10 Nov 2014, 00:58
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3
\(\frac{(16x^4 – 81y^4)}{(2x + 3y)} = 12x^2 + 27y^2\)

\(\frac{(4x^2 + 9y^2)(4x^2 - 9y^2)}{(2x + 3y}) = 12x^2 + 27y^2\)

\(\frac{(4x^2 + 9y^2)(2x + 3y)(2x - 3y)}{(2x + 3y)} = 3*(4x^2 + 9y^2)\)

2x - 3y = 3

4x + 3y = 9 .......... Info already given in problem

6x = 12; x = 2

Answer = C
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Re: (16x^4 – 81y^4)/(2x + 3y) = 12x^2 + 27y^2 and 4x + 3y = 9, what is x?  [#permalink]

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New post 05 Nov 2014, 17:07
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Factorizing the given: ((2x)^2+(3y)^2)(2x-3y)(2x+3y)/(2x+3y) = 3(4x^2+9y^2)

Further: (4x^2+9y^2)(2x-3y) = 3(4x^2+9y^2) ---> 2x-3y = 3 and 4x+3y = 9 (given) ----> Adding up, 6x = 12 or x = 2; Answer: C
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Re: (16x^4 – 81y^4)/(2x + 3y) = 12x^2 + 27y^2 and 4x + 3y = 9, what is x?  [#permalink]

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New post 29 May 2017, 16:46
Bunuel wrote:

Tough and Tricky questions: Algebra.



(16x^4 – 81y^4)/(2x + 3y) = 12x^2 + 27y^2 and 4x + 3y = 9, what is x?

A. 1
B. 1.5
C. 2
D. 3
E. 4

Kudos for a correct solution.


It's important to work neatly here.

(16x^4-81y^4)/(2x+3y) = 12x^2 + 27y^2

(4x^2+9y^2)(4x^2-9y^2)/(2x+3y) = 12x^2 + 27y^2

(2x+3y)(2x-3y)(4x^2+9y^2)/(2x+3y) = 12x^2 + 27y^2

(2x-3y)(4x^2+9y^2) = 12x^2 + 27y^2

(2x-3y)(4x^2+9y^2) = 3(4x^2 + 9y^2)

2x-3y = 3
+4x+3y = 9

6x = 12
x = 2
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Re: (16x^4 – 81y^4)/(2x + 3y) = 12x^2 + 27y^2 and 4x + 3y = 9, what is x?  [#permalink]

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New post 10 Sep 2018, 00:47
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Re: (16x^4 – 81y^4)/(2x + 3y) = 12x^2 + 27y^2 and 4x + 3y = 9, what is x? &nbs [#permalink] 10 Sep 2018, 00:47
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