Question

Picture below, help fast please!

Answer 1

Answer:

$x=21,y=25$

Step-by-step explanation:

The given equation is:

$\frac{(5-3i)(x+iy)}{(4-5i)}=(2+i)^2-(3-4i)^2$

Apply difference of two squares on the RHS.

$\frac{(5-3i)(x+iy)}{(4-5i)}=[(2+i)+(3-4i)][(2+i)-(3-4i)$

Simplify the RHS.

$\frac{(5-3i)(x+iy)}{(4-5i)}=(5-3i)(5i-1)$

Expand the RHS

$\frac{(5-3i)(x+iy)}{(4-5i)}=(10+28i)$

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:

$x=21,y=25$