Find the value of x and y which satisfy the following equation (x, y∈R).
`((x + "i"y))/(2 + 3"i") + (2 + "i")/(2 + 3"i") = 9/13(1 + "i")`
Solution
`((x + "i"y))/(2 + 3"i") + (2 + "i")/(2 + 3"i") = 9/13(1 + "i")`
∴ `((x + "i"y)(2 - 3"i") + (2 + "i")(2 + 3"i"))/((2 + 3"i")(2 - 3"i")) = 9/13(1 + "i")`
∴ `(2x - 3"i"x + 2"i"y - 3y"i"^2 + 4 + 6"i" + 2"i" + 3"i"^2)/(4 - 9"i"^2) = 9/13(1 + "i")`
∴ `(2x - 3"i"x + 2"i"y + 3y + 4 + 6"i" + 2"i" - 3)/(4 + 9) = 9/13(1 + "i")` ...[∵ i2 = – 1]
∴ `((2x + 3y + 1) + (-3"i"x + 2"i"y + 8"i"))/13 = 9/13(1 + "i")`
Equating the real and imaginary parts separately, we get,
2x + 3y + 1 = 9 and – 3x + 2y + 8 = 9
∴ 2x + 3y = 8 ...(1)
and – 3x + 2y = 1 ...(2)
Multiplying equation (1) by 3 and equation (2) by 2, we get,
6x + 9y = 24
and – 6x + 4y = 2
On adding, we get,
13y = 26
∴ y = 2
∴ from (1), 2x + 3(2) = 8
∴ 2x + 6 = 8
∴ 2x = 2
∴ x = 1
Hence, x = 1 and y = 2
