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Sum

**Solve the following equation for x, y ∈ R: **

(4 – 5i) x + (2 + 3i) y = 10 – 7i

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#### Solution

(4 – 5i) x + (2 + 3i) y = 10 – 7i

∴ (4x + 2y) + (3y - 5x) i = 10 – 7i

Equating real and imaginary parts, we get

4x + 2y = 10

i.e., 2x + y = 5 ...(i)

and 3y – 5x = – 7 ...(ii)

Equation (i) x 3 – equation (ii) gives

11x = 22

∴ x = 2

Putting x = 2 in (i), we get

2(2) + y = 5

∴ y = 1

∴ x = 2 and y = 1.

Concept: Solution of a Quadratic Equation in Complex Number System

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