Question

Найти Z из уравнения. (6i+2z)(3-4i)-(20+5 iz)(2-i)=42-31i

Answer 1

$18i-24 i^{2}+6z-8iz-40+20i-10iz+5 i^{2}z-42+31i=0$
$69i-24 i^{2} +6z-18iz-82+5 i^{2} z=0$
$6z-18iz+5 i^{2} z=-69i+24 i^{2} +82$
$z(6-18i+5 i^{2} )=-69i+24 i^{2} +82$
z=$\frac{24 i^{2}-69i+82 }{5 i^{2}-18i+6 }$