If (3x+2)-(5y-3)i and (6x+3)+(2y-4)i are conj | vedclass

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(B) Two complex numbers $z_1 = a+bi$ and $z_2 = c+di$ are conjugates if $a=c$ and $b=-d$.Given $z_1 = (3x+2) - (5y-3)i$ and $z_2 = (6x+3) + (2y-4)i$.E

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(B) Two complex numbers $z_1 = a+bi$ and $z_2 = c+di$ are conjugates if $a=c$ and $b=-d$.Given $z_1 = (3x+2) - (5y-3)i$ and $z_2 = (6x+3) + (2y-4)i$.Equating the real parts: $3x+2 = 6x+3$ $\Rightarrow 3x = -1$ $\Rightarrow x = -\frac{1}{3}$.Equating the imaginary parts: $-(5y-3) = 2y-4$ $\Rightarrow -5y+3 = 2y-4$ $\Rightarrow 7y = 7$ $\Rightarrow y = 1$.Wait,let's re-evaluate the conjugate condition: $z_1 = \overline{z_2}$.$(3x+2) - (5y-3)i = (6x+3) - (2y-4)i$.Real parts: $3x+2 = 6x+3$ $\Rightarrow 3x = -1$ $\Rightarrow x = -\frac{1}{3}$.Imaginary parts: $-(5y-3) = -(2y-4)$ $\Rightarrow 5y-3 = 2y-4$ $\Rightarrow 3y = -1$ $\Rightarrow y = -\frac{1}{3}$.Now,calculate $\frac{x-y}{x+y} = \frac{-\frac{1}{3} - (-\frac{1}{3})}{-\frac{1}{3} + (-\frac{1}{3})} = \frac{0}{-\frac{2}{3}} = 0$.

If $(3x+2)-(5y-3)i$ and $(6x+3)+(2y-4)i$ are conjugates of each other,then the value of $\frac{x-y}{x+y}$ is (where $i=\sqrt{-1}, x, y \in R$ ).

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