Question

Then find the value of (x, y) (4+5i)x+(3-2i)y+i²+6i=0​

Answer 1

Answer:

Yvonne I ea.geggghhhggg

Answer 2

Answer:

x = -16/23

y = 174/138

Step-by-step explanation:

(4 + 5i)x+ (3 - 2i)y + i² + 6i = 0

4x + 5ix + 3y - 2iy + i² + 6i = 0

(Value of i² = -1)

4x + 5ix + 3y - 2iy - 1 + 6i = 0

(4x + 3y - 1) + (5x - 2y + 6)i = 0

Real part = (4x + 3y - 1) × 2  -------(i)

Imaginary part = (5x - 2y + 6) × 3 -------(ii)

8x + 6y - 2 = 0 -------(iii)

15x - 6y + 18 = 0 -------(iv)

Add (iii) and (iv) ,

8x + 6y - 2 + 15x - 6y + 18 = 0

23x+ 16 = 0

23x = -16

x = -16/23

8(-16/23) + 6y - 2 = 0

-128/23 + 6y - 2 = 0

6y = 2 + 128/23

6y = (46 + 128)/23

y = 174/23(6)

y = 174/138