Question

(4+5i)x+(3-2i)y-i^2+6i

Answer 1

$(4+5i)x+(3-2i)y-i^2+6i=0\\\\we\:know,\\i^2=-1,use\:it\\\\(4+5i)x+(3-2i)y-(-1)+6i\\=4x+5ix+3y-2iy+1+6i\\=(4x+3y+1)+(5x-2y+6)i\\\\real\:part=(4x+3y+1)\\imarginary\:part=(5x-2y+6)\\\\(4x+3y+1)\times2=0-----(1)\\(5x-2y+6)\times3=0------(2)\\--------------------\\23x=20\\x=\frac{23}{20},y=\frac{-56}{30}$