What is the answer to (x+4)i-(y-3)j=2y(i+j) ?

The answer to (x+4)i-(y-3)j=2y(i+j) is x=-2,y=1,\quad j\ne 0

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(x+4)i-(y-3)j=2y(i+j)

(x + 4) i - (y - 3) j = 2 y (i + j)

x = - 2, y = 1, j \neq = 0

Solution steps

(x + 4) i - (y - 3) j = 2 y (i + j)

Rewrite

(x + 4) i - (y - 3) j

in standard complex form:

(- j y + 3 j) + (x + 4) i

(- j y + 3 j) + (x + 4) i = 2 y (i + j)

Rewrite

2 y (i + j)

in standard complex form:

2 j y + 2 y i

(- j y + 3 j) + (x + 4) i = 2 j y + 2 y i

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

[- j y + 3 j = 2 j y x + 4 = 2 y]

[- j y + 3 j = 2 j y x + 4 = 2 y] : x = - 2, y = 1, j \neq = 0

x = - 2, y = 1, j \neq = 0

x^{2} + 2 y i + 2 x y i + 2 x - y^{2} + 1 + i = 0

2 i x^{2} + 2 x + 9 i = 0

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

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

\frac{2 - 5 i}{4 - 7 i} = \frac{a}{b}

What is the answer to (x+4)i-(y-3)j=2y(i+j) ?
The answer to (x+4)i-(y-3)j=2y(i+j) is x=-2,y=1,\quad j\ne 0