J

I think this puzzle is a bit of a missed opportunity. They could have provided inputs with no solution or with a line of solutions, so that the cost optimization becomes meaningful. As it is, you just have to carry out Cramer's rule in extended precision rational arithmetic.

load 'regex'

data_file_name =: '13.data'
raw =: cutopen fread data_file_name
NB. a b sublist y gives elements [a..b) of y
sublist =: ({~(+i.)/)~"1 _
parse_button =: monad define
  match =. 'X\+([[:digit:]]+), Y\+([[:digit:]]+)' rxmatch y
  ". (}. match) sublist y
)
parse_prize =: monad define
  match =. 'X=([[:digit:]]+), Y=([[:digit:]]+)' rxmatch y
  ". (}. match) sublist y
)
parse_machine =: monad define
  3 2 $ (parse_button >0{y), (parse_button >1{y), (parse_prize >2{y)
)
NB. x: converts to extended precision, which gives us rational arithmetic
machines =: x: (parse_machine"1) _3 ]\ raw

NB. A machine is represented by an array 3 2 $ ax ay bx by tx ty, where button
NB. A moves the claw by ax ay, button B by bx by, and the target is at tx ty.
NB. We are looking for nonnegative integer solutions to ax*a + bx*b = tx,
NB. ay*a + by*b = ty; if there is more than one, we want the least by the cost
NB. function 3*a + b.

solution_rank =: monad define
  if. 0 ~: -/ . * }: y do. 0  NB. system is nonsingular
  elseif. */ (=/"1) 2 ]\ ({. % {:) |: y do. 1  NB. one equation is a multiple of the other
  else. _1 end.
)
NB. solve0 yields the cost of solving a machine of solution rank 0
solve0 =: monad define
  d =. -/ . * }: y
  a =. (-/ . * 2 1 { y) % d
  b =. (-/ . * 0 2 { y) % d
  if. (a >: 0) * (a = <. a) * (b >: 0) * (b = <. b) do. b + 3 * a else. 0 end.
)
NB. there are actually no machines of solution rank _1 or 1 in the test set
result1 =: +/ solve0"_1 machines

machines2 =: machines (+"2) 3 2 $ 0 0 0 0 10000000000000 10000000000000
NB. there are no machines of solution rank _1 or 1 in the modified set either
result2 =: +/ solve0"_1 machines2