What is the answer to (100)/(8+6i)x+(1-i)^3y=7 ?

The answer to (100)/(8+6i)x+(1-i)^3y=7 is x= 1/2 ,y=-3/2

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(100)/(8+6i)x+(1-i)^3y=7

\frac{100}{8 + 6 i} x + (1 - i)^{3} y = 7

x = \frac{1}{2}, y = - \frac{3}{2}

Solution steps

\frac{100}{8 + 6 i} x + (1 - i)^{3} y = 7

Expand

\frac{100}{8 + 6 i} x + (1 - i)^{3} y : (8 x - 2 y) + i (- 2 y - 6 x)

(8 x - 2 y) + i (- 2 y - 6 x) = 7

Rewrite

7

in standard complex form:

7 + 0 i

(8 x - 2 y) + i (- 2 y - 6 x) = 7 + 0 i

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

[8 x - 2 y = 7 - 2 y - 6 x = 0]

[8 x - 2 y = 7 - 2 y - 6 x = 0] : x = \frac{1}{2}, y = - \frac{3}{2}

x = \frac{1}{2}, y = - \frac{3}{2}

z^{2} + 16 i = 0

2 i = i x (2 - 3 i) - y (5 - 3 i)

(5 + 6 i) (x + y i) = 60 + 11 i

x, c u a d r a t i c a = x^{2} + 4

r, i = \frac{e}{r}

What is the answer to (100)/(8+6i)x+(1-i)^3y=7 ?
The answer to (100)/(8+6i)x+(1-i)^3y=7 is x= 1/2 ,y=-3/2