What is the answer to (1+i)(x+2y)+(3-2i)(x-y)=8+3i3i ?

The answer to (1+i)(x+2y)+(3-2i)(x-y)=8+3i3i is x=-4/15 ,y=-1/15

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(1+i)(x+2y)+(3-2i)(x-y)=8+3i3i

(1 + i) (x + 2 y) + (3 - 2 i) (x - y) = 8 + 3 i 3 i

x = - \frac{4}{15}, y = - \frac{1}{15}

Solution steps

(1 + i) (x + 2 y) + (3 - 2 i) (x - y) = 8 + 3 i 3 i

Expand

(4 x - y) + i (- x + 4 y) = - 1

Rewrite

- 1

in standard complex form:

- 1 + 0 i

(4 x - y) + i (- x + 4 y) = - 1 + 0 i

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

[4 x - y = - 1 - x + 4 y = 0]

[4 x - y = - 1 - x + 4 y = 0] : x = - \frac{4}{15}, y = - \frac{1}{15}

x = - \frac{4}{15}, y = - \frac{1}{15}

3 x + (y - x) i = 6

z^{2} + 2 - 23 i = 0

4 x_{1} + 5 x_{2} = (1 + i) x_{1}

z^{2} - z - 1 + i = 0

+ i x^{2} = 1 - i, x^{1} = 1

What is the answer to (1+i)(x+2y)+(3-2i)(x-y)=8+3i3i ?
The answer to (1+i)(x+2y)+(3-2i)(x-y)=8+3i3i is x=-4/15 ,y=-1/15