# Get results of solve

Hi,

Given

var("a b") solve([a+b==5,a-b==1],[a, b])

[[a == 3, b == 2]]

Is there a neat way to actually get a and b assigned to 3 and 2,

obviously this is a simplified equation.

Also sometimes one gets a == 3 and other times a = 3 why is that?

It would be good if the 'neat way' would work for both cases.

Thanks, Rob.

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You can use the solution_dict=True option:

sage: solve([a+b==5,a-b==1],[a, b], solution_dict=True)
[{b: 2, a: 3}]


Here, you get a list of solutions, each solution is a dictionary. You get the first solution (here the only one) with:

sage: solve([a+b==5,a-b==1],[a, b], solution_dict=True)
{b: 2, a: 3}


And it values as:

sage: solve([a+b==5,a-b==1],[a, b], solution_dict=True)[a]
3
sage: solve([a+b==5,a-b==1],[a, b], solution_dict=True)[b]
2


So, you just have to do:

sage: a = solve([a+b==5,a-b==1],[a, b], solution_dict=True)[a]


Also sometimes one gets a == 3 and other times a = 3 why is that?

Are you sure about that ? Could you please provide a way to reproduce, since this is not expected ?

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Sorry I am unfamiliar with this forum I posted as an answer by mistake.

The number of equals is just an artefact of print/show

var("a b")

show(solve([a+b-5,a-b-1],[a, b])) print(solve([a+b-5,a-b-1],[a, b]))

[[a=3,b=2] [ [a == 3, b == 2] ]

Thanks, I have now worked out how to do this and I will try the ictionary tip soon.

My session does not paste very well but at the bottom is the solution I want to parse into A, B etc. I hope you will see what I am trying to do, and you will also see the singke '='

# ratfr2ser rational fraction to power series á la Euler Archive June 2005

def ratfr2ser(ratfr, num_terms): var("A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z") lets=[A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z] rnt=range(num_terms) ser = [ lets[i] * x^i for i in rnt ] ser=sum(ser) # converts from list to symbolic show(ser) ratfr_num = ratfr.numerator() ratfr_den = ratfr.denominator() eqn = (serratfr_den - ratfr_num).expand().collect(x) show(eqn) cfs = [eqn.coefficient(x, n=p) for p in rnt] show(cfs) sol = [solve(ec, ec.variables()) for ec in cfs] show(sol) for i in rnt: sol = solve(cfs[i], cfs[i].variables()) #sage_eval(str(sol)) #show(A) ratfr=(1-x)/(1-x-2x^2) ratfr.show() ratfr2ser(ratfr, 4)

x−12x2+x−1 x 1 2 x 2 x 1 Dx3+Cx2+Bx+A D x 3 C x 2 B x A 2Dx5+(2C+D)x4+(2B+C−D)x3+(2A+B−C)x2+(A−B−1)x−A+1 2 D x 5 2 C D x 4 2 B C D x 3 2 A B C x 2 A B 1 x A 1 [−A+1,A−B−1,2A+B−C,2B+C−D] A 1 A B 1 2 A B C 2 B C D [[A=1],[[A=r1+1,B=r1]],[[A=12r2−12r3,B=r3,C=r2]],[[B=12r4−12r5,C=r5,D=r4]]]

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