# Find the value of x and y which satisfy the following equation (x, y∈R). (x+iy)2+3i+2+i2+3i=913(1+i) - Mathematics and Statistics

Sum

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

13y = 26

∴ y = 2

∴ from (1), 2x + 3(2) = 8

∴ 2x + 6 = 8

∴ 2x = 2

∴ x = 1

Hence, x = 1 and y = 2

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