If $\dfrac{x-1}{3+i}+\dfrac{y-1}{3-i}=i$ then $(y, x)=$

Given the equation $\dfrac{x-1}{3+i} + \dfrac{y-1}{3-i} = i$.

Multiply the first term by $\dfrac{3-i}{3-i}$ and the second term by $\dfrac{3+i}{3+i}$ to rationalize the denominators:

$\dfrac{(x-1)(3-i)}{9+1} + \dfrac{(y-1)(3+i)}{9+1} = i$

$\dfrac{3x - xi - 3 + i}{10} + \dfrac{3y + yi - 3 - i}{10} = i$

Combine the numerators:

$(3x - 3 + 3y - 3) + i(-x + 1 + y - 1) = 10i$

$(3x + 3y - 6) + i(y - x) = 0 + 10i$

Equating the real and imaginary parts:

Real part: $3x + 3y - 6 = 0 \Rightarrow x + y = 2$

Imaginary part: $y - x = 10$

Adding the two equations: $(x + y) + (y - x) = 2 + 10 \Rightarrow 2y = 12 \Rightarrow y = 6$

Substituting $y = 6$ into $x + y = 2$: $x + 6 = 2 \Rightarrow x = -4$

Thus, $(y, x) = (6, -4)$.

Answer: $(6,-4)$