Simultaneous equation problem...?

Multiply the first equation by "a", and you get...

a^2 + ab + ac = 16a

Now, subtract the second equation to the new one, and you get

a^2 - bc = 16a - 43................(*)

From the third eq. we get

- bc = 60 /a, then, substituting this onto (*) we have

a^2 + 60/a = 16a - 43 or

a^3 - 16a^2 + 43a + 60 = 0...........(**)

We can find analogous expressions for b and c.

Obviously a = -1 is a solution to that equation, then (**) can be rewritten as:

(a+1) (a^2 -17a +60) = 0

Solving a^2 - 17a +60=0 for a, we get that the other solutions are a=12 and a =5... Then, (**) must be

(a+1) (a-5) (a-12) = 0.

Since you have the same kind of expressions for b and c, you get

(b+1) (b-5) (b-12) = 0.

(c+1) (c-5) (c-12) = 0.

So, a, b and c, must be different numbers from the set {-1, 5, 12}. Since the equations you provided are symmetric to a, b and c, this was to be expected.

(a,b,c) = (-1,5,12) , (-1, 12, 5) , (5, -1, 12) , (5, 12, -1), (12, -1, 5) and (12, 5, -1) are all the possible solutions.