Answer:
[tex]x=21,y=25[/tex]
Step-by-step explanation:
The given equation is:
[tex]\frac{(5-3i)(x+iy)}{(4-5i)}=(2+i)^2-(3-4i)^2[/tex]
Apply difference of two squares on the RHS.
[tex]\frac{(5-3i)(x+iy)}{(4-5i)}=[(2+i)+(3-4i)][(2+i)-(3-4i)[/tex]
Simplify the RHS.
[tex]\frac{(5-3i)(x+iy)}{(4-5i)}=(5-3i)(5i-1)[/tex]
Expand the RHS
[tex]\frac{(5-3i)(x+iy)}{(4-5i)}=(10+28i)[/tex]
Cross multiply to get:
[tex](5-3i)(x+iy)=(10+28i)(4-5i)[/tex]
Expand both sides
[tex](5x+3y)+(5y-3x)i=180+62i[/tex]
Comparing the complex parts, we obtain:
[tex]5y-3x=62...(1)[/tex]
Comparing the real number parts we get;
[tex]5x+3y=180...(2)[/tex]
Solving equations (1) and (2) simultaneously; we get:
[tex]x=21,y=25[/tex]
