What is the answer to (2x-3y-5)+i(x+2y+1)=0 ?

The answer to (2x-3y-5)+i(x+2y+1)=0 is x=1,y=-1

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(2x-3y-5)+i(x+2y+1)=0

(2 x - 3 y - 5) + i (x + 2 y + 1) = 0

x = 1, y = - 1

Solution steps

(2 x - 3 y - 5) + i (x + 2 y + 1) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(2 x - 3 y - 5) + i (x + 2 y + 1) = 0 + 0 i

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

[2 x - 3 y - 5 = 0 x + 2 y + 1 = 0]

[2 x - 3 y - 5 = 0 x + 2 y + 1 = 0] : x = 1, y = - 1

x = 1, y = - 1

5 x + 12 y i = - 10 + 4 i

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

x + y i = 0

\frac{m}{1 + i} + n = - m - \frac{1}{i}

(a + 3 i) (1 + bi) = 25 - 39 i

What is the answer to (2x-3y-5)+i(x+2y+1)=0 ?
The answer to (2x-3y-5)+i(x+2y+1)=0 is x=1,y=-1