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05 Nov 2014, 08:35
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
73% (02:50) correct
27%
(02:48)
wrong
based on 439
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History
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\(\frac{(16x^4 – 81y^4)}{(2x + 3y) }= 12x^2 + 27y^2\) and \(4x + 3y = 9\), what is x?
A. 1
B. 1.5
C. 2
D. 3
E. 4
A. 1
B. 1.5
C. 2
D. 3
E. 4
10 Nov 2014, 01:58
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\(\frac{(16x^4 – 81y^4)}{(2x + 3y)} = 12x^2 + 27y^2\)
\(\frac{(4x^2 + 9y^2)(4x^2 - 9y^2)}{(2x + 3y}) = 12x^2 + 27y^2\)
\(\frac{(4x^2 + 9y^2)(2x + 3y)(2x - 3y)}{(2x + 3y)} = 3*(4x^2 + 9y^2)\)
2x - 3y = 3
4x + 3y = 9 .......... Info already given in problem
6x = 12; x = 2
Answer = C
\(\frac{(4x^2 + 9y^2)(4x^2 - 9y^2)}{(2x + 3y}) = 12x^2 + 27y^2\)
\(\frac{(4x^2 + 9y^2)(2x + 3y)(2x - 3y)}{(2x + 3y)} = 3*(4x^2 + 9y^2)\)
2x - 3y = 3
4x + 3y = 9 .......... Info already given in problem
6x = 12; x = 2
Answer = C
General Discussion
05 Nov 2014, 18:07
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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
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
