Simultaneous equations—classify the solution set as finite, infinite, or none

Question

How do I find the values of $b$ and $c$ for the equations $x + 5 y = 4$ and $2 x + by = c$ with a unique solution, an infinite set of solutions and no solutions with Mathematica?

Answer 1

The quickest way to get there is to first Solve the first equation for x or y, then use Reduce on the second equation for the other variable:

Reduce[
 2*x + b*y == c /. Solve[x + 5*y == 4, y],
 x]

So either b=10, which makes c=8, or else c and b depend on the value of x.

It's interesting the different outputs you get depending on which variables you reduce by (thanks to Eric Towers for the tip on Backsubstitution)

DeleteDuplicates[
 LogicalExpand@
    Reduce[{x + 5 y == 4, 2 x + b y == c}, #, 
     Backsubstitution -> True] & /@ Permutations[{b, c, x, y}]
 ]

It's clear from these that b=10 and c=8 are the only solutions that are valid for all x and y. Is there a way to specify that you want the solution in this case?

Answer 2

I think this is somehow a problem from the perspective that Mathematica will NOT teach you math, it will help you computing stuff in several cases. You have to be able to interpret the output with your math understanding.

Your equations can be solved by

eqs = {
   x + 5*y == 4
   , 2*x + b*y == c
   };
vars = {x, y};
sol = Solve[eqs, vars]

Here Mathematica gave you the answer assuming that -10 + b will not be zero, although we never gave that information.

From your math class you should know how to differentiate between the cases. So if you solve first, e.g., in the first equation for x

solx = Solve[eqs[[1]], x][[1]]

and put this in the second equation

eqs[[2]] /. solx[[1]] // FullSimplify

you can clearly see, that if b equals 10 and c equals 8, then y can be anything, so you get an infinite number of solutions. But if b equals 10 and c is not specified or something like c==3, then this equations can not be fulfilled in general, so you dont get any solution (you can identify this cases in the solution given above in sol but Mathematica will not tell you that). You can check this as follows

Solve[eqs /. {b -> 10}, vars]
Solve[eqs /. {c -> 8, b -> 10}, vars]

Alternatively you can use

Reduce[eqs, vars]

but you have to be able to interpret this output by yourself, i.e., for the first case you get an infinite number of solutions for x and y (since they depend on each other) and for the second case you a specific x and then a specific y. If none of these cases are fulfilled, e.g., b==10 and c==3, then you dont have a solution.

My point is that Mathematica will NOT teach you math, in several cases you have to be able to interpret the output and know that may be Mathematica itself made some assumptions on its own (see the output of sol).