If x, y in R and (x + iy)(3 + 2i) = 1 + i,the | vedclass

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(C) Given $(x + iy)(3 + 2i) = 1 + i$. $x + iy = \frac{1 + i}{3 + 2i}$. To simplify,multiply the numerator and denominator by the conjugate of the deno

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$\left( \frac{5}{13}, \frac{1}{13} \right)$

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(C) Given $(x + iy)(3 + 2i) = 1 + i$. $x + iy = \frac{1 + i}{3 + 2i}$. To simplify,multiply the numerator and denominator by the conjugate of the denominator,$(3 - 2i)$: $x + iy = \frac{(1 + i)(3 - 2i)}{(3 + 2i)(3 - 2i)} = \frac{3 - 2i + 3i - 2i^2}{3^2 + 2^2}$. Since $i^2 = -1$,we have: $x + iy = \frac{3 + i - 2(-1)}{9 + 4} = \frac{3 + i + 2}{13} = \frac{5 + i}{13} = \frac{5}{13} + i\frac{1}{13}$. Comparing the real and imaginary parts,we get $x = \frac{5}{13}$ and $y = \frac{1}{13}$. Thus,$(x, y) = \left( \frac{5}{13}, \frac{1}{13} \right)$.

If $x, y \in R$ and $(x + iy)(3 + 2i) = 1 + i$,then $(x, y)$ is

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