If frac{x-1}{3+i} + frac{y-1}{3-i} = i,then t | vedclass

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(A) Given: $\frac{x-1}{3+i} + \frac{y-1}{3-i} = i$ Multiply by the common denominator $(3+i)(3-i) = 3^2 + 1^2 = 10$: $(x-1)(3-i) + (y-1)(3+i) = i(10)$

Vedclass · Accepted Answer

$x = -4, y = 6$

Vedclass · Answer

(A) Given: $\frac{x-1}{3+i} + \frac{y-1}{3-i} = i$ Multiply by the common denominator $(3+i)(3-i) = 3^2 + 1^2 = 10$: $(x-1)(3-i) + (y-1)(3+i) = i(10)$ $3x - ix - 3 + i + 3y + iy - 3 - i = 10i$ $(3x + 3y - 6) + i(y - x) = 10i$ Comparing real and imaginary parts: $3x + 3y - 6 = 0 \Rightarrow x + y = 2$ $y - x = 10$ Adding the two equations: $2y = 12 \Rightarrow y = 6$ Substituting $y = 6$ into $x + y = 2$: $x + 6 = 2 \Rightarrow x = -4$ Thus,$x = -4$ and $y = 6$.

If $\frac{x-1}{3+i} + \frac{y-1}{3-i} = i$,then the true statement among the following is:

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