Hello,
I am trying to solve:
a0 = 2 = a · 3^{0} + b · 2^{0} = a + b
a1 = 6 = a · 3^{1} + b · 2^{1} = 3a + 2b
I am no so sure that is it "evident". yes, if a= 2 then 3a= 6 so a= 2, b= 0 satisfy the equation.
Or, as chiro suggested, from a+ b= 2, b= 2- a so the second equation becomes 3a+ 2(2- a)= a+ 4= 6 so a= 2 and then b= 2- 2= 0.
Or, as I suggested, multiplying the first equation by 2, 2a+ 2b= 4 and, subtracting that from 3a+ 2b= 6, a= 2. Multiplying the first equation by 3, 3a+ 3b= 6 and, subtracting that from 3a+ 2b= 6 from that b= 0.