[shahkb4]

Let x and y be real number such that $50\left(\frac{2 \mathrm{x}}{1+3 \mathrm{i}}-\frac{\mathrm{y}}{1-2 \mathrm{i}}\right)=31+17 \mathrm{i}, \mathrm{i}=\sqrt{-1}$. Then the value of $10(\mathrm{x}-3 \mathrm{y})$ is
[JEE MAIN 02 April 2026 S I]
(1) 20
(2) 31
(3) 35
(4) 75

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[shahkb4]

Solution

$\quad 50\left(\frac{2 \mathrm{x}(1-3 \mathrm{i})}{10}-\frac{\mathrm{y}(1+2 \mathrm{i})}{5}\right)=31+17 \mathrm{i} \Rightarrow 10 \mathrm{x}-(30 \mathrm{x}) \mathrm{i}-10 \mathrm{y}-20 \mathrm{yi}=31+17 \mathrm{i}$

\[ \begin{aligned} & 10(x-y)=31 \quad \ldots(1) \text { and }-30 x-20 y=17 \\ & \text { Eq.(1) } x 3+\text { Eq.(2) } \\ & \Rightarrow 30 x-30 y-30 x-20 y=93+17 \\ & \Rightarrow y=-\frac{110}{50}=-2.2, x=0.9 \\ & \therefore 10(x-3 y)=75 \end{aligned} \]

[CHAPTER M25 DETERMINANT TOPIC T2 ]

[shahkb4]

Answer

(4)

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