Substituting values for x and y seems to be easiest way
x= 12 and y= 2 satisfy the eqns
x-y = 10
simple but tricky....
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raghavsarathy
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raghavsarathy
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I took the second eqn as (3x)^0.5 - (2y)^0.5prindaroy wrote:i don't understand your question. is it (3x)^0.5 - (2y)^0.5 or 3(x^0.5) - 2(y^0.5)?? Also substituting numbers won't work.
Substitution should work in this case right
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neo.gaurav
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umaa
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shibal,
Try substituting the second equation in the first. that is,
3x-2y=32 ---- I
3x^1/2-2y^1/2=4 ---- II
II ---> 3X^1/2 = 4 + 2y^1/2
x^1/2 = (4+2y^1/2)/3
Square both sides,
x = ((4+2y^1/2)^2)/9 --- III
Substitute the third equation in the first equation,
3((4+2y^1/2)^2)/9) - 2y = 32
By solving,
y - 8 y^1/2 + 40 = 0
You can find y^1/2, y and x from this equation.
It must be a DS question. What is the source of this question?
Try substituting the second equation in the first. that is,
3x-2y=32 ---- I
3x^1/2-2y^1/2=4 ---- II
II ---> 3X^1/2 = 4 + 2y^1/2
x^1/2 = (4+2y^1/2)/3
Square both sides,
x = ((4+2y^1/2)^2)/9 --- III
Substitute the third equation in the first equation,
3((4+2y^1/2)^2)/9) - 2y = 32
By solving,
y - 8 y^1/2 + 40 = 0
You can find y^1/2, y and x from this equation.
It must be a DS question. What is the source of this question?
