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

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### 1 Answer

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

ButÂ `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 = 1We'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` )/10y1 = `(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` .**