Question

\( \frac{(1+i)x-2i}{3+i} + \frac{(2-3i)y+i}{3-i} = i \implies x+y = \)

• A. -1
• B. 1
• C. -2
• D. 2

Hint:

To simplify fractions with complex denominators like \(\frac{N}{a+ib}\), multiply numerator and denominator by the conjugate \(a-ib\).
After combining terms, equate the real parts on both sides of the equation and the imaginary parts on both sides.
This yields a system of linear equations in x and y, which can then be solved.

Answer 1

The Correct Option is D

The equation is \( \frac{(1+i)x-2i}{3+i} + \frac{(2-3i)y+i}{3-i} = i \). Multiply the first term by \(\frac{3-i}{3-i}\) and the second term by \(\frac{3+i}{3+i}\). Denominator common factor: \((3+i)(3-i) = 3^2 - i^2 = 9 - (-1) = 9+1 = 10\). Numerator of first term: \( ((1+i)x-2i)(3-i) = (3x-ix+3ix-i^2x -6i+2i^2) \) \( = (3x+2ix+x -6i-2) = (4x-2) + i(2x-6) \). Numerator of second term: \( ((2-3i)y+i)(3+i) = (2y-3iy+i)(3+i) \) \( = (6y+2iy -9iy-3i^2y +3i+i^2) = (6y-7iy+3y +3i-1) \) \( = (9y-1) + i(-7y+3) \). So the equation becomes: \( \frac{(4x-2) + i(2x-6)}{10} + \frac{(9y-1) + i(-7y+3)}{10} = i \) Multiply by 10: \( (4x-2) + i(2x-6) + (9y-1) + i(-7y+3) = 10i \) Group real and imaginary parts: Real part: \( (4x-2) + (9y-1) = 0 \) (since RHS real part is 0) \( 4x + 9y - 3 = 0 \Rightarrow 4x + 9y = 3 \) --- (1) Imaginary part: \( (2x-6) + (-7y+3) = 10 \) \( 2x - 7y - 3 = 10 \Rightarrow 2x - 7y = 13 \) --- (2) We have a system of two linear equations: 1) \(4x + 9y = 3\) 2) \(2x - 7y = 13\) Multiply equation (2) by 2: \(4x - 14y = 26\) --- (3) Subtract equation (3) from equation (1): \((4x + 9y) - (4x - 14y) = 3 - 26\) \(4x + 9y - 4x + 14y = -23\) \(23y = -23 \Rightarrow y = -1\). Substitute \(y=-1\) into equation (2): \(2x - 7(-1) = 13\) \(2x + 7 = 13\) \(2x = 13 - 7 = 6\) \(x = 3\). So, \(x=3\) and \(y=-1\). We need to find \(x+y\). \(x+y = 3 + (-1) = 3 - 1 = 2\). This matches option (d). \[ \boxed{2} \]