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Find the values of x and y which satisfy the following equations (x, y ∈ R),

\(\frac{x+iy}{2+3i}+\frac{2+i}{2-3i}=\frac9{13}(1+i)\)

(x + yi)/(2 + 3i) + (2 + i)/(2 - 3i) = 9/13 (1 + i)

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\(\frac{x+iy}{2+3i}+\frac{2+i}{2-3i}=\frac9{13}(1+i)\) 

(2x + 3y + 1) + (8 – 3x + 2y)i = 9 + 9i

Equating real and imaginary parts, we get

2x + 3y + 1 = 9 and 8 – 3x + 2y = 9

2x + 3y = 8 ……(i)

and 3x – 2y = – 1 ……(ii)

Equation (i) × 2 + equation (ii) × 3 gives

13x = 13

∴ x = 1

Substituting x = 1 in (i), we get

2(1) + 3y = 8

3y = 6

∴ y = 2

∴ x = 1 and y = 2

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