It is known that a and B are fixed values. No matter what the value of K is, the solution of the linear equation (3kx + A / 3) - (x-6k / 6) = 2 with respect to X is always x = 1 Try to find the value of a and B The equation is [(3kx + a) / 3] - [(x-bk) / 6] = 2

It is known that a and B are fixed values. No matter what the value of K is, the solution of the linear equation (3kx + A / 3) - (x-6k / 6) = 2 with respect to X is always x = 1 Try to find the value of a and B The equation is [(3kx + a) / 3] - [(x-bk) / 6] = 2

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