Find the real value of *x* and *y*, if

\[(3x - 2iy)(2 + i )^2 = 10(1 + i)\]

#### Solution

\[ \left( 3x - 2iy \right) \left( 2 + i \right)^2 = 10 \left( 1 + i \right)\]

\[ \Rightarrow \left( 3x - 2iy \right)\left( 4 + i^2 + 4i \right) = 10\left( 1 + i \right)\]

\[ \Rightarrow \left( 3x - 2iy \right)\left( 3 + 4i \right) = 10\left( 1 + i \right)\]

\[ \Rightarrow 9x + 12xi - 6iy - 8 i^2 y = 10 + 10i\]

\[ \Rightarrow 9x + 8y + i\left( 12x - 6y \right) = 10 + 10i\]

\[\text{Comparing both the sides:} \]

\[9x + 8y = 10 . . . . (1)\]

\[12x - 6y = 10\]

\[or, 6x - 3y = 5 . . . (2)\]

\[\text { Multiplying equation (1) by 3 and equation (2) by 8 }, \]

\[27x + 24y = 30 . . . . (3) \]

\[48x - 24y = 40 . . . . (4)\]

\[\text {Adding equations (3) and (4):} \]

\[75x = 70\]

\[ \therefore x = \frac{14}{15}\]

\[\text { Substituting the value of x in equation (1): } \]

\[9 \times \frac{14}{15} + 8y = 10\]

\[ \Rightarrow \frac{126}{15} + 8y = 10\]

\[ \Rightarrow 8y = 10 - \frac{126}{15}\]

\[ \Rightarrow 8y = \frac{24}{15}\]

\[ \Rightarrow y = \frac{1}{5}\]