Question

Picture below, help fast please!

Answer 1

`x=21,y=25`

Explanation:

The given equation is:

`((5-3i)(x+iy))/((4-5i))=(2+i)^2-(3-4i)^2`

Apply difference of two squares on the RHS.

`((5-3i)(x+iy))/((4-5i))=[(2+i)+(3-4i)][(2+i)-(3-4i)`

Simplify the RHS.

`((5-3i)(x+iy))/((4-5i))=(5-3i)(5i-1)`

Expand the RHS

`((5-3i)(x+iy))/((4-5i))=(10+28i)`

Cross multiply to get:

`(5-3i)(x+iy)=(10+28i)(4-5i)`

Expand both sides

`(5x+3y)+(5y-3x)i=180+62i`

Comparing the complex parts, we obtain:

`5y-3x=62...(1)`

Comparing the real number parts we get;

`5x+3y=180...(2)`

Solving equations (1) and (2) simultaneously; we get:

`x=21,y=25`