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

The answer to (1-i)x+(3+i)y=-3+3i is x=-3,y=0

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

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

x = - 3, y = 0

Solution steps

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

Rewrite

(1 - i) x + (3 + i) y

in standard complex form:

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

(x + 3 y) + (- x + y) i = - 3 + 3 i

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

[x + 3 y = - 3 - x + y = 3]

[x + 3 y = - 3 - x + y = 3] : x = - 3, y = 0

x = - 3, y = 0

(2 + i) x + 7 y = 1 + 2 i, (1 - i) x - i y = 4 - i

x + y i + (x - y i)^{2} = 4

i (x) = \frac{x - 1}{x + 3}

3 a^{2} - 3 b^{2} - i 6 ab = g (a + bi)

a - bi = ai + b i^{2}

What is the answer to (1-i)x+(3+i)y=-3+3i ?
The answer to (1-i)x+(3+i)y=-3+3i is x=-3,y=0