balance a linear equation

Hello, Anyone knows a method for balancing a linear equation?

Thanks, Luis

Answer 1

[Walter Roberson]

What form is the linear equation in?

If you have the symbolic toolbox you could do it via solve()

[Walter Roberson]

Under the interpretation of

(sqrt(2)*x-b/lambda)^2*(2*x-1) = (G*sqrt(3/(2*Pi))/((2*Pi)*sqrt(x))-a/lambda)^2;

then that is not a linear equation: it is an octic (order 8) equation, and the values of x are the squares of the 8 roots of

roots( [-32*Pi^3*lambda^2, 0, 16*Pi^3*lambda^2+32*Pi^3*sqrt(2)*lambda*b, 0, -16*Pi^3*b^2-16*Pi^3*sqrt(2)*lambda*b, 0, 8*Pi^3*(a^2+b^2), -4*sqrt(3)*sqrt(2)*Pi^(3/2)*G*a*lambda, 3*G^2*lambda^2] )

[Walter Roberson]

Under the interpretation of

(sqrt(2)*x-b/lambda)^2*(2*x-1) = ((1/2)*G*Pi*sqrt((3/2)*Pi)/sqrt(x)-a/lambda)^2;

then that is not a linear equation: it is an octic (order 8) equation, and the values of x are the squares of the 8 roots of

roots( [32*lambda^2, 0, -8*sqrt(2)*(sqrt(2)*lambda+4*b)*lambda, 0, (16*(b+sqrt(2)*lambda))*b, 0, -8*(a^2+b^2), 4*sqrt(3)*sqrt(2)*Pi^(3/2)*G*a*lambda, -3*Pi^3*G^2*lambda^2] )

Answer 2

[Luis Manuel]

(√2x-b/λ)^2 (2x-1)=((G/2π) √(3/2π) 1/√x-a/λ)^2

where x is the variable to balance and other variables are constant

[Walter Roberson]

is that G/2 * pi, or G/(2*pi) ? And is it 3/2 * pi or 3/(2*pi) ?