What is the answer to (7+3i)x+(3-2i)y=1+7i ?

The answer to (7+3i)x+(3-2i)y=1+7i is x=1,y=-2

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(7+3i)x+(3-2i)y=1+7i

(7 + 3 i) x + (3 - 2 i) y = 1 + 7 i

x = 1, y = - 2

Solution steps

(7 + 3 i) x + (3 - 2 i) y = 1 + 7 i

Rewrite

(7 + 3 i) x + (3 - 2 i) y

in standard complex form:

(7 x + 3 y) + (3 x - 2 y) i

(7 x + 3 y) + (3 x - 2 y) i = 1 + 7 i

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

[7 x + 3 y = 1 3 x - 2 y = 7]

[7 x + 3 y = 1 3 x - 2 y = 7] : x = 1, y = - 2

x = 1, y = - 2

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

x, (i x - i y)^{2} = 2

(x + i y)^{2} + 2 (x - i y) = - 1 + 6 i

z^{2} = - 2 - 23 i

\frac{i}{z} = - 6

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