Substituting values for x and y seems to be easiest way

x= 12 and y= 2 satisfy the eqns

x-y = 10

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.

There is no answer to this question. Where did you get it from? I just graphed the two equations and there are no real solutions. The two functions are asymptotic after x = 2.

prindaroy 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.

I took the second eqn as (3x)^0.5 - (2y)^0.5

Substitution should work in this case right

Multiply second equation by 3x^1/2 + 3y^1/2 and you will get the solution

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?

actually I don't have the OA... i saw that question on a CAT without answers...