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I want solution of x and y which should be positive real numbers

Solve[2*(1 - a)*((x)^-n) + 2*a*((y)^-n) - 1 == 0 && (1 - a)*((x)^(1 - n)) + a*((y)^(1 - n)) - 1 == 0 , {x, y}]

During evaluation of In[2]:= Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. >>

During evaluation of In[2]:= Solve::svars: Equations may not give solutions for all "solve" variables. >>

Out[2]= {{y -> 2^(1/n) ((a x^n)/(-2 + 2 a + x^n))^(1/n)}}

*Supppose now I plug in certain numeric values n=0.3 a=0.6, it still can't solve it .

NSolve[2*(1 - 0.6)*((x)^-0.3) + 2*0.6*((y)^-0.3) - 1 == 0 && (1 - 0.6)*((x)^(1 - 0.3)) + 0.6*((y)^(1 - 0.3)) - 1 == 0, {x, y}, Reals]

NSolve::ratnz: NSolve was unable to solve the system with inexact coefficients. The answer was obtained by solving a corresponding exact system and numericizing the result. >>

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  • $\begingroup$ Sorry can you please tell me how can you tell from the contour plot there is no real solutions? I ploted it and see two curves that are not intersecting $\endgroup$ Commented Mar 4, 2014 at 5:47

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Mathematica can solve it if you use explicit numerical values for the parameters, and it does return a result: {}. From the NSolve doc page:

NSolve gives {} if there are no solutions to the equations.

From a contour plot of your equations I believe this result is correct and there are no real valued solutions:

ContourPlot[{ -1 + 4/(5 x^(3/10)) + 6/(5 y^(3/10)) == 0, -1 + (2 x^(7/10))/5 + (3 y^(7/10))/5 == 0 }, {x, 0, 20}, {y, 0, 20}, MaxRecursion -> 5]

Generally, if Mathematica can't do a symbolic computation, it'll return the input command. If it returns something else, it has a specific meaning.

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  • $\begingroup$ I wonder what one should make of the 'special' solution y -> 2^(1/n) ((a x^n)/(-2 + 2 a + x^n))^(1/n). Any hint for which values of a and n this is a solution of the system? For a=0.6 and n=1/2 one can find the pairs {1,36} and {4,4} for {x,y} but these verify only the first equation and not the second. $\endgroup$ Commented Mar 3, 2014 at 15:06

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