# What are x and y if (6-yi)(x+2i)=12-5i?

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To determine x and y, we'll have to perform the multiplication from the left side:

(6-yi)(x+2i) = 6x + 12i - xyi - 2y`i^(2)`

But `i^(2)` = -1

(6-yi)(x+2i) = 6x+2y + i(12 - xy)

Now, we'll equate the real parts from both sides:

6x+2y = 12 => 3x + y = 6 => y = 6 - 3x

We'll equate the imaginary parts from both sides:

12 - xy = -5 => xy = 12+5 => xy = 17

x(6 - 3x) = 17

We'll remove the brackets:

6x - 3`x^(2)` - 17 = 0

3`x^(2)` - 6x + 17 = 0

We'll apply quadratic formula:

x1 = (6 + `sqrt(36 - 204)` )/6

x1 = (6 + 2i`sqrt(42)` )/6

x1 = (3 + i`sqrt(42)` )/3

x2 = (3 - i`sqrt(42)` )/3

y1 = 6 - 3x1 => y1 = 6 - 3 - i`sqrt(42)`

y1 = 3 - i`sqrt(42)`

y2 = 6 - 3 + i`sqrt(42)`

y2 = 3 + i`sqrt(42)`

**Therefore, the values of x and y are: x1 = (3+i`sqrt(42)` )/3 ; y1 = 3 - i`sqrt(42)` ; x2 = (3 - i`sqrt(42)` )/3 ; y2 = 3 + i`sqrt(42)` .**