What is the answer to 2ix_{2}=-x_{3} ?

The answer to 2ix_{2}=-x_{3} is x_{2}=0,x_{3}=0

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2ix_{2}=-x_{3}

2 i x_{2} = - x_{3}

x_{2} = 0, x_{3} = 0

Solution steps

2 i x_{2} = - x_{3}

Rewrite

2 i x_{2}

in standard complex form:

0 + 2 x_{2} i

0 + 2 x_{2} i = - x_{3}

Rewrite

- x_{3}

in standard complex form:

- x_{3} + 0 i

0 + 2 x_{2} i = - x_{3} + 0 i

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

[0 = - x_{3} 2 x_{2} = 0]

[0 = - x_{3} 2 x_{2} = 0] : x_{2} = 0, x_{3} = 0

x_{2} = 0, x_{3} = 0

z^{2} - (5 - 2 i) z + (8 - 20 i) = 0

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

(4 - i) z + 5 w = 3 i

- \frac{1}{z - i} - (- 2) i - z = 0

z = - 7 + i 4 z + 3 w

What is the answer to 2ix_{2}=-x_{3} ?
The answer to 2ix_{2}=-x_{3} is x_{2}=0,x_{3}=0