Math.NET Numerics can numerically solve any linear system , but I suggest that this is not what you are looking for.
Math.NET Symbolics can work with symbolic expressions, although this project is at an early stage and does not yet understand the concept of equations. However, we can still use it to solve simple systems like these with a bit of work - doing what we will do manually.
First, let's define a small function to solve a single linear equation of order up to 1:
using Expr = MathNet.Symbolics.Expression; Expr SolveSimpleRoot(Expr variable, Expr expr) { // try to bring expression into polynomial form Expr simple = Algebraic.Expand(Rational.Numerator(Rational.Simplify(variable,expr))); // extract coefficients, solve known forms of order up to 1 Expr[] coeff = Polynomial.Coefficients(variable,simple); switch(coeff.Length) { case 1: return Expr.Zero.Equals(coeff[0]) ? variable : Expr.Undefined; case 2: return Rational.Simplify(variable,Algebraic.Expand(-coeff[0]/coeff[1])); default: return Expr.Undefined; } }
Then we can use this to solve the system as follows:
// declare variables var x = Expr.Symbol("x"); var y = Expr.Symbol("y"); // Parse left and right side of both equations Expr aleft = Infix.ParseOrThrow("(4-x)*2"); Expr aright = Infix.ParseOrThrow("(y-1)*10+2"); Expr bleft = Infix.ParseOrThrow("x"); Expr bright = Infix.ParseOrThrow("y*2+1"); // Solve both equations to x Expr ax = SolveSimpleRoot(x,aleft-aright); // "8 - 5*y" Expr bx = SolveSimpleRoot(x,bleft-bright); // "1 + 2*y" // Equate both terms of x, solve to y Expr cy = SolveSimpleRoot(y,ax-bx); // "1" // Substitute term of y into one of the terms of x Expr cx = Algebraic.Expand(Structure.Substitute(y,cy,ax)); // "3" // Print expression in Infix notation Console.WriteLine(Infix.Print(cx)); // x=3 Console.WriteLine(Infix.Print(cy)); // y=1