Given:

⇒ (x+iy)(2-3i)=4+i

⇒ x(2-3i)+iy(2-3i)=4+i

⇒ 2x-3xi+2yi-3yi²=4+i

We know that i²=-1

⇒ 2x+(-3x+2y)i-3y(-1)=4+i

⇒ (2x+3y)+(-3x+2y)=4+i

Equating Real and Imaginary parts on both sides, we get

⇒ 2x+3y=4 and -3x+2y=1

On solving we get,

⇒

∴ The real values of x and y are .