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everyone! I am sorry, but I am an abcolute novice of Mathematica (to be more precise this is my first day of using it) and even after surfing the web and all documents I am not able to solve the following system:

Solve[{y*(((y*x)/(beta*b))^(1/(beta - 1)) - v) - c*alpha == 
   0, ((x/alpha))*(((y*x)/(alpha*beta*b))^(1/(beta - 1)) - 
       v) + (((y*x)/alpha) - 
       2*alpha*((yx)/(2*beta*b))^(1/(beta - 1)))*(1/(beta - 
         1))*(x/(alpha*beta*
         b))*((y*x)/(alpha*beta*b))^((2 - beta)/(1 - beta)) == 0}, {x,
   y}]

What I need is to solve following systems, getting x and y expressed through all these symbols. Is it even possible? Thank you in advance.

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Since you are evaluating x and y. So You can not put symbolic powers of x and y variables. See my answer... –  santosh Feb 14 at 16:21
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2 Answers

up vote 1 down vote accepted

Define

 \[Beta] = 2;

Solve[y*(((y*x)/(\[Beta]*b))^(1/(\[Beta] - 1)) - v) - c*\[Alpha] == 
0 && ((x/\[Alpha]))*(((y*x)/(\[Alpha]*\[Beta]*b))^(1/(\[Beta] - 
        1)) - v) + (((y*x)/\[Alpha]) - 
   2*\[Alpha]*((yx)/(2*\[Beta]*b))^(1/(\[Beta] - 
         1)))*(1/(\[Beta] - 1))*(x/(\[Alpha]*\[Beta]*
     b))*((y*x)/(\[Alpha]*\[Beta]*
      b))^((2 - \[Beta])/(1 - \[Beta])) == 0, {x, y}]

Answer:

  {{x -> 0, 
y -> -((c \[Alpha])/v)}, {x -> (-32 b^4 v^2 + 16 b^4 v^2 \[Alpha] - 
8 b^2 v yx \[Alpha] + 8 b^2 v yx \[Alpha]^2 + yx^2 \[Alpha]^3)/(
 32 b^3 c), 
y -> (8 b^2 c \[Alpha])/(-8 b^2 v + 4 b^2 v \[Alpha] + 
yx \[Alpha]^2)}}
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It seems to me that this is probably a transcendental set of equations, with no algebraic solution. You've got x and y raised to a bunch of incompatible powers (functions of beta). There's little hope of ever isolating x and y.

My advice is to defer to a numerical solution:

With[
{v = 0.1, b = 0.2, alpha = 0.3, beta = 0.4, c = 0.5},
NSolve[{y (-v + ((x y)/(b beta))^(1/(-1 + beta))) == 
alpha c, (x (-v + ((x y)/(alpha b beta))^(1/(-1 + beta))))/
alpha + (((x y)/(alpha b beta))^((-3 + 2 beta)/(-1 + 
beta)) ((x y)/alpha - 2^((-2 + beta)/(-1 + beta)) alpha 
((y x)/(b beta))^(1/(-1 + beta))))/((-1 + beta) y) == 0}, {x, y}]
]
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