Then find the value of (x, y) (4+5i)x+(3-2i)y+i²+6i=0
Answers
Answered by
1
Answer:
Yvonne I ea.geggghhhggg
Answered by
3
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
Similar questions