If a and B are fixed values, 2kx + A / 3 - x-6k / 6 = 2, no matter what the value of K is, his solution is always 1, and the value of a and B is obtained

If a and B are fixed values, 2kx + A / 3 - x-6k / 6 = 2, no matter what the value of K is, his solution is always 1, and the value of a and B is obtained

Substituting x = 1
(2k+a)/3-(1-bk)/6=2
Double six on both sides
4k+2a-1+bk=12
(b+4)k=13-2a
When B + 4 = 0 and 13-2a = 0, the constant holds
So a = 13 / 2, B = - 4