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by Abdulla » Sat Nov 21, 2009 2:57 pm
OA is 2
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Last edited by Abdulla on Sun Nov 22, 2009 10:21 pm, edited 1 time in total.
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by heshamelaziry » Sun Nov 22, 2009 8:19 pm
IMO D. (4x^2-9y^2)(4x^2+9y^2)/2x+3y = 3(4x^2+9y^2)

Cross multiply, we get:

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

3= (2x-3y)(2x+3y)/2x + 3y

(1) 2x - 3y = 3

(2) 4x + 3y = 9

subtrtact 1 from 2 we get 2x = 6 --------> x = 3
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by thephoenix » Sun Nov 22, 2009 8:39 pm
heshamelaziry wrote:IMO D. (4x^2-9y^2)(4x^2+9y^2)/2x+3y = 3(4x^2+9y^2)

Cross multiply, we get:

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

3= (2x-3y)(2x+3y)/2x + 3y

(1) 2x - 3y = 3

(2) 4x + 3y = 9

subtrtact 1 from 2 we get 2x = 6 --------> x = 3


solving above two eqn we get x=2...pls check...
IMO x is 2
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by thephoenix » Sun Nov 22, 2009 8:41 pm
Abdulla wrote:OA is 3
with OA is 3 do u mean its option three or value of x as 3

kindly check

as the value of x is coming as 2 which is option 3
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by heshamelaziry » Sun Nov 22, 2009 8:56 pm
thephoenix wrote:
heshamelaziry wrote:IMO D. (4x^2-9y^2)(4x^2+9y^2)/2x+3y = 3(4x^2+9y^2)

Cross multiply, we get:

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

3= (2x-3y)(2x+3y)/2x + 3y

(1) 2x - 3y = 3

(2) 4x + 3y = 9

subtrtact 1 from 2 we get 2x = 6 --------> x = 3

with 2x = 6, how can x = 2 ?

solving above two eqn we get x=2...pls check...
IMO x is 2
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by Abdulla » Sun Nov 22, 2009 10:20 pm
heshamelaziry wrote:
thephoenix wrote:
heshamelaziry wrote:IMO D. (4x^2-9y^2)(4x^2+9y^2)/2x+3y = 3(4x^2+9y^2)

Cross multiply, we get:

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

3= (2x-3y)(2x+3y)/2x + 3y

(1) 2x-3y = 3

(2) 4x + 3y = 9

subtrtact 1 from 2 we get 2x = 6 --------> x = 3

with 2x = 6, how can x = 2 ?

solving above two eqn we get x=2...pls check...
IMO x is 2

My bad the answer is 2..

Final step is ..

2x-3y = 3
4x + 3y = 9 Add 2 equations
6x = 12
x = 2
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Last edited by Abdulla on Sun Nov 22, 2009 10:30 pm, edited 6 times in total.
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by thephoenix » Sun Nov 22, 2009 10:25 pm
heshamelaziry wrote:
thephoenix wrote:
heshamelaziry wrote:IMO D. (4x^2-9y^2)(4x^2+9y^2)/2x+3y = 3(4x^2+9y^2)

Cross multiply, we get:

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

3= (2x-3y)(2x+3y)/2x + 3y

(1) 2x - 3y = 3

(2) 4x + 3y = 9

subtrtact 1 from 2 we get 2x = 6 --------> x = 3



with 2x = 6, how can x = 2 ?

solving above two eqn we get x=2...pls check...
IMO x is 2
subtracting 1 from 2 we get 2x+6y=6 [spoiler]not 2x=6[/spoiler]

whereas adding eqn1 and eqn2 we are getting 6x=12--------->x=2
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