What are x and y if (5-xi)(y+i)=10+4i?

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

(5-xi)(y+i) = 5y + 5i - xyi -
x`i^2`

But `i^2` = -1

(5-xi)(y+i) = 5y + x + i(5
- xy)

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

5y + x = 10 => x = 10 - 5y

We'll
equate the imaginary parts from both sides:

5 - xy = 4 => -xy
= 4-5 => xy = 1

(10-5y)y = 1

We'll remove
the brackets:

10y - `5y^2` - 1 = 0

`5y^2 - 10y +
1` = 0

We'll apply quadratic formula:

y1 = (10 +
` sqrt(100 - 20) `)/10

y1 = (10+ 4`sqrt5` )/10

y1
= `(5+2sqrt5)/5`

y2 = `(5 - 2sqrt5)/5`

x1 =10 -
5y1 => x1 = `5 - 2sqrt5`

x2 =
`5+2sqrt5`

Therefore, the values of x and y
are: x1 = `5-2sqrt5` ; y1 = `(5+2sqrt5)/5` ; x2 = `5+2sqrt5` ; y2 = `(5-2sqrt5)/5`
.