Question

Find the real value of x and y, if `((1+i)x-2i)/(3+i) + ((2-3i)y+i)/(3-i) = i, xy ∈ R, i = sqrt-1`

Answer 1

\[ \frac{\left( 1 + i \right)x - 2i}{3 + i} + \frac{\left( 2 - 3i \right)y + i}{3 - i} = i\]

\[ \Rightarrow \frac{\left( 1 + i \right)\left( 3 - i \right)x - 2i\left( 3 - i \right) + \left( 2 - 3i \right)\left( 3 + i \right)y + i\left( 3 + i \right)}{\left( 3 + i \right)\left( 3 - i \right)} = i\]

\[ \Rightarrow \frac{3x - ix + 3ix - i^2 x - 6i + 2 i^2 + 6y + 2iy - 9iy - 3 i^2 y + 3i + i^2}{9 - i^2} = i\]

\[ \Rightarrow \frac{4x + 2ix - 3i + 9y - 7iy - 3}{10} = i\]

\[ \Rightarrow \left( 4x + 9y - 3 \right) + i\left( 2x - 3 - 7y \right) = 10i\]

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

\[4x + 9y - 3 = 0\]

\[ \Rightarrow 4x + 9y = 3 . . . (1) \]

\[2x - 3 - 7y = 10\]

\[ \Rightarrow 2x - 7y = 13 . . . (2)\]

\[\text{Multiplying equation (2) by 2:} \]

\[4x - 14y = 26 . . . (3) \]

\[\text { Subtracting equation (3) from (1): } \]

\[ 4x + 9y = 3 \]

\[ 4x - 14y = 26 \]

\[ - + - \]

\[ 23y = - 23\]

\[ \therefore y = - 1\]

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

\[4x - 9 = 3\]

\[ \Rightarrow 4x = 12\]

\[ \Rightarrow x = 3\]

\[ \therefore x = 3 \text { and y } = - 1\]