If (x-iy)(3+5i) is the conjugate of -6-24i (w | vedclass

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(D) The conjugate of $-6-24i$ is $-6+24i$.Given that $(x-iy)(3+5i) = -6+24i$.Expanding the left side: $3x + 5xi - 3yi - 5yi^2 = -6+24i$.Since $i^2 = -

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$3, -3$

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(D) The conjugate of $-6-24i$ is $-6+24i$.Given that $(x-iy)(3+5i) = -6+24i$.Expanding the left side: $3x + 5xi - 3yi - 5yi^2 = -6+24i$.Since $i^2 = -1$,we have $(3x+5y) + (5x-3y)i = -6+24i$.Equating real and imaginary parts:$3x + 5y = -6$ (Equation $1$)$5x - 3y = 24$ (Equation $2$)Multiplying Equation $1$ by $3$ and Equation $2$ by $5$:$9x + 15y = -18$$25x - 15y = 120$Adding these equations: $34x = 102$,which gives $x = 3$.Substituting $x=3$ into Equation $1$: $3(3) + 5y = -6$ $\Rightarrow 9 + 5y = -6$ $\Rightarrow 5y = -15$ $\Rightarrow y = -3$.Thus,$x=3$ and $y=-3$.

If $(x-iy)(3+5i)$ is the conjugate of $-6-24i$ (where $x, y \in R$ and $i=\sqrt{-1}$),then the values of $x$ and $y$ are respectively:

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