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Question
Find the real value of x and y, if
\[(3x - 2iy)(2 + i )^2 = 10(1 + i)\]
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Solution
\[ \left( 3x - 2iy \right) \left( 2 + i \right)^2 = 10 \left( 1 + i \right)\]
\[ \Rightarrow \left( 3x - 2iy \right)\left( 4 + i^2 + 4i \right) = 10\left( 1 + i \right)\]
\[ \Rightarrow \left( 3x - 2iy \right)\left( 3 + 4i \right) = 10\left( 1 + i \right)\]
\[ \Rightarrow 9x + 12xi - 6iy - 8 i^2 y = 10 + 10i\]
\[ \Rightarrow 9x + 8y + i\left( 12x - 6y \right) = 10 + 10i\]
\[\text{Comparing both the sides:} \]
\[9x + 8y = 10 . . . . (1)\]
\[12x - 6y = 10\]
\[or, 6x - 3y = 5 . . . (2)\]
\[\text { Multiplying equation (1) by 3 and equation (2) by 8 }, \]
\[27x + 24y = 30 . . . . (3) \]
\[48x - 24y = 40 . . . . (4)\]
\[\text {Adding equations (3) and (4):} \]
\[75x = 70\]
\[ \therefore x = \frac{14}{15}\]
\[\text { Substituting the value of x in equation (1): } \]
\[9 \times \frac{14}{15} + 8y = 10\]
\[ \Rightarrow \frac{126}{15} + 8y = 10\]
\[ \Rightarrow 8y = 10 - \frac{126}{15}\]
\[ \Rightarrow 8y = \frac{24}{15}\]
\[ \Rightarrow y = \frac{1}{5}\]
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