What is the answer to (x+2y+3)+i(3x-y-1)=0 ?

The answer to (x+2y+3)+i(3x-y-1)=0 is x=-1/7 ,y=-10/7

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(x+2y+3)+i(3x-y-1)=0

(x + 2 y + 3) + i (3 x - y - 1) = 0

x = - \frac{1}{7}, y = - \frac{10}{7}

Solution steps

(x + 2 y + 3) + i (3 x - y - 1) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(x + 2 y + 3) + i (3 x - y - 1) = 0 + 0 i

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

[x + 2 y + 3 = 0 3 x - y - 1 = 0]

[x + 2 y + 3 = 0 3 x - y - 1 = 0] : x = - \frac{1}{7}, y = - \frac{10}{7}

x = - \frac{1}{7}, y = - \frac{10}{7}

4 z^{2} + (- 13 + 18 i) z - 5 - 10 i = 0

x^{2} + 4 - 3 i = 0

- 3 ai - (- 1 - i) b = 3 a - 2

i (x) = \frac{1}{x}

x^{2} = - i

What is the answer to (x+2y+3)+i(3x-y-1)=0 ?
The answer to (x+2y+3)+i(3x-y-1)=0 is x=-1/7 ,y=-10/7